3.410 \(\int \frac{\tan ^{\frac{7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=600 \[ \frac{a (A b-a B) \tan ^{\frac{5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b-5 a^3 B-13 a b^2 B+9 A b^3\right ) \tan ^{\frac{3}{2}}(c+d x)}{4 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{a^{3/2} \left (6 a^2 A b^3+3 a^4 A b-46 a^3 b^2 B-15 a^5 B-63 a b^4 B+35 A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{7/2} d \left (a^2+b^2\right )^3}-\frac{\left (3 a^3 A b-31 a^2 b^2 B-15 a^4 B+11 a A b^3-8 b^4 B\right ) \sqrt{\tan (c+d x)}}{4 b^3 d \left (a^2+b^2\right )^2}-\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}+\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]

[Out]

((3*a^2*b*(A - B) - b^3*(A - B) - a^3*(A + B) + 3*a*b^2*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt
[2]*(a^2 + b^2)^3*d) - ((3*a^2*b*(A - B) - b^3*(A - B) - a^3*(A + B) + 3*a*b^2*(A + B))*ArcTan[1 + Sqrt[2]*Sqr
t[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) + (a^(3/2)*(3*a^4*A*b + 6*a^2*A*b^3 + 35*A*b^5 - 15*a^5*B - 46*a^3
*b^2*B - 63*a*b^4*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*b^(7/2)*(a^2 + b^2)^3*d) - ((a^3*(A - B)
- 3*a*b^2*(A - B) + 3*a^2*b*(A + B) - b^3*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqr
t[2]*(a^2 + b^2)^3*d) + ((a^3*(A - B) - 3*a*b^2*(A - B) + 3*a^2*b*(A + B) - b^3*(A + B))*Log[1 + Sqrt[2]*Sqrt[
Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) - ((3*a^3*A*b + 11*a*A*b^3 - 15*a^4*B - 31*a^2*b^2*
B - 8*b^4*B)*Sqrt[Tan[c + d*x]])/(4*b^3*(a^2 + b^2)^2*d) + (a*(A*b - a*B)*Tan[c + d*x]^(5/2))/(2*b*(a^2 + b^2)
*d*(a + b*Tan[c + d*x])^2) + (a*(a^2*A*b + 9*A*b^3 - 5*a^3*B - 13*a*b^2*B)*Tan[c + d*x]^(3/2))/(4*b^2*(a^2 + b
^2)^2*d*(a + b*Tan[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.72452, antiderivative size = 600, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 14, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.424, Rules used = {3605, 3645, 3647, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac{a (A b-a B) \tan ^{\frac{5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b-5 a^3 B-13 a b^2 B+9 A b^3\right ) \tan ^{\frac{3}{2}}(c+d x)}{4 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{a^{3/2} \left (6 a^2 A b^3+3 a^4 A b-46 a^3 b^2 B-15 a^5 B-63 a b^4 B+35 A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{7/2} d \left (a^2+b^2\right )^3}-\frac{\left (3 a^3 A b-31 a^2 b^2 B-15 a^4 B+11 a A b^3-8 b^4 B\right ) \sqrt{\tan (c+d x)}}{4 b^3 d \left (a^2+b^2\right )^2}-\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}+\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^(7/2)*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^3,x]

[Out]

((3*a^2*b*(A - B) - b^3*(A - B) - a^3*(A + B) + 3*a*b^2*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt
[2]*(a^2 + b^2)^3*d) - ((3*a^2*b*(A - B) - b^3*(A - B) - a^3*(A + B) + 3*a*b^2*(A + B))*ArcTan[1 + Sqrt[2]*Sqr
t[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) + (a^(3/2)*(3*a^4*A*b + 6*a^2*A*b^3 + 35*A*b^5 - 15*a^5*B - 46*a^3
*b^2*B - 63*a*b^4*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*b^(7/2)*(a^2 + b^2)^3*d) - ((a^3*(A - B)
- 3*a*b^2*(A - B) + 3*a^2*b*(A + B) - b^3*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqr
t[2]*(a^2 + b^2)^3*d) + ((a^3*(A - B) - 3*a*b^2*(A - B) + 3*a^2*b*(A + B) - b^3*(A + B))*Log[1 + Sqrt[2]*Sqrt[
Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) - ((3*a^3*A*b + 11*a*A*b^3 - 15*a^4*B - 31*a^2*b^2*
B - 8*b^4*B)*Sqrt[Tan[c + d*x]])/(4*b^3*(a^2 + b^2)^2*d) + (a*(A*b - a*B)*Tan[c + d*x]^(5/2))/(2*b*(a^2 + b^2)
*d*(a + b*Tan[c + d*x])^2) + (a*(a^2*A*b + 9*A*b^3 - 5*a^3*B - 13*a*b^2*B)*Tan[c + d*x]^(3/2))/(4*b^2*(a^2 + b
^2)^2*d*(a + b*Tan[c + d*x]))

Rule 3605

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e
+ f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

Rule 3645

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*d^2 + c*(c*C - B*d))*(a + b*T
an[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3653

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[((c + d*Tan[e + f*x])^n*(1 + Tan[e + f*x]^2))/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\tan ^{\frac{7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx &=\frac{a (A b-a B) \tan ^{\frac{5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\int \frac{\tan ^{\frac{3}{2}}(c+d x) \left (-\frac{5}{2} a (A b-a B)+2 b (A b-a B) \tan (c+d x)-\frac{1}{2} \left (a A b-5 a^2 B-4 b^2 B\right ) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^2} \, dx}{2 b \left (a^2+b^2\right )}\\ &=\frac{a (A b-a B) \tan ^{\frac{5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b+9 A b^3-5 a^3 B-13 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{\sqrt{\tan (c+d x)} \left (-\frac{3}{4} a \left (a^2 A b+9 A b^3-5 a^3 B-13 a b^2 B\right )-2 b^2 \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)-\frac{1}{4} \left (3 a^3 A b+11 a A b^3-15 a^4 B-31 a^2 b^2 B-8 b^4 B\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 b^2 \left (a^2+b^2\right )^2}\\ &=-\frac{\left (3 a^3 A b+11 a A b^3-15 a^4 B-31 a^2 b^2 B-8 b^4 B\right ) \sqrt{\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}+\frac{a (A b-a B) \tan ^{\frac{5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b+9 A b^3-5 a^3 B-13 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{\frac{1}{8} a \left (3 a^3 A b+11 a A b^3-15 a^4 B-31 a^2 b^2 B-8 b^4 B\right )-b^3 \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)+\frac{1}{8} \left (3 a^4 A b+3 a^2 A b^3+8 A b^5-15 a^5 B-31 a^3 b^2 B-24 a b^4 B\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{b^3 \left (a^2+b^2\right )^2}\\ &=-\frac{\left (3 a^3 A b+11 a A b^3-15 a^4 B-31 a^2 b^2 B-8 b^4 B\right ) \sqrt{\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}+\frac{a (A b-a B) \tan ^{\frac{5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b+9 A b^3-5 a^3 B-13 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{b^3 \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right )-b^3 \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{b^3 \left (a^2+b^2\right )^3}+\frac{\left (a^2 \left (3 a^4 A b+6 a^2 A b^3+35 A b^5-15 a^5 B-46 a^3 b^2 B-63 a b^4 B\right )\right ) \int \frac{1+\tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{8 b^3 \left (a^2+b^2\right )^3}\\ &=-\frac{\left (3 a^3 A b+11 a A b^3-15 a^4 B-31 a^2 b^2 B-8 b^4 B\right ) \sqrt{\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}+\frac{a (A b-a B) \tan ^{\frac{5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b+9 A b^3-5 a^3 B-13 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{2 \operatorname{Subst}\left (\int \frac{b^3 \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right )-b^3 \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{b^3 \left (a^2+b^2\right )^3 d}+\frac{\left (a^2 \left (3 a^4 A b+6 a^2 A b^3+35 A b^5-15 a^5 B-46 a^3 b^2 B-63 a b^4 B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{8 b^3 \left (a^2+b^2\right )^3 d}\\ &=-\frac{\left (3 a^3 A b+11 a A b^3-15 a^4 B-31 a^2 b^2 B-8 b^4 B\right ) \sqrt{\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}+\frac{a (A b-a B) \tan ^{\frac{5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b+9 A b^3-5 a^3 B-13 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\left (a^2 \left (3 a^4 A b+6 a^2 A b^3+35 A b^5-15 a^5 B-46 a^3 b^2 B-63 a b^4 B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 b^3 \left (a^2+b^2\right )^3 d}-\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}+\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}\\ &=\frac{a^{3/2} \left (3 a^4 A b+6 a^2 A b^3+35 A b^5-15 a^5 B-46 a^3 b^2 B-63 a b^4 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{7/2} \left (a^2+b^2\right )^3 d}-\frac{\left (3 a^3 A b+11 a A b^3-15 a^4 B-31 a^2 b^2 B-8 b^4 B\right ) \sqrt{\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}+\frac{a (A b-a B) \tan ^{\frac{5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b+9 A b^3-5 a^3 B-13 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}-\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}-\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}\\ &=\frac{a^{3/2} \left (3 a^4 A b+6 a^2 A b^3+35 A b^5-15 a^5 B-46 a^3 b^2 B-63 a b^4 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{7/2} \left (a^2+b^2\right )^3 d}-\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{\left (3 a^3 A b+11 a A b^3-15 a^4 B-31 a^2 b^2 B-8 b^4 B\right ) \sqrt{\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}+\frac{a (A b-a B) \tan ^{\frac{5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b+9 A b^3-5 a^3 B-13 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}\\ &=\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{a^{3/2} \left (3 a^4 A b+6 a^2 A b^3+35 A b^5-15 a^5 B-46 a^3 b^2 B-63 a b^4 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{7/2} \left (a^2+b^2\right )^3 d}-\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{\left (3 a^3 A b+11 a A b^3-15 a^4 B-31 a^2 b^2 B-8 b^4 B\right ) \sqrt{\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}+\frac{a (A b-a B) \tan ^{\frac{5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b+9 A b^3-5 a^3 B-13 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [B]  time = 6.31232, size = 1563, normalized size = 2.6 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^(7/2)*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^3,x]

[Out]

-(((-1)^(1/4)*a^3*A*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/((a^2 + b^2)^3*d)) - (3*(-1)^(3/4)*a^2*A*b*ArcTan[(
-1)^(3/4)*Sqrt[Tan[c + d*x]]])/((a^2 + b^2)^3*d) + (3*(-1)^(1/4)*a*A*b^2*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]
)/((a^2 + b^2)^3*d) + ((-1)^(3/4)*A*b^3*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/((a^2 + b^2)^3*d) + ((-1)^(3/4)
*a^3*B*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/((a^2 + b^2)^3*d) - (3*(-1)^(1/4)*a^2*b*B*ArcTan[(-1)^(3/4)*Sqrt
[Tan[c + d*x]]])/((a^2 + b^2)^3*d) - (3*(-1)^(3/4)*a*b^2*B*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/((a^2 + b^2)
^3*d) + ((-1)^(1/4)*b^3*B*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/((a^2 + b^2)^3*d) + (3*a^(11/2)*A*ArcTan[(Sqr
t[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*b^(5/2)*(a^2 + b^2)^3*d) + (3*a^(7/2)*A*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x
]])/Sqrt[a]])/(2*Sqrt[b]*(a^2 + b^2)^3*d) + (35*a^(3/2)*A*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]
)/(4*(a^2 + b^2)^3*d) - (15*a^(13/2)*B*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*b^(7/2)*(a^2 + b^2)^3*
d) - (23*a^(9/2)*B*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(2*b^(3/2)*(a^2 + b^2)^3*d) - (63*a^(5/2)*Sqr
t[b]*B*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*(a^2 + b^2)^3*d) - ((-1)^(1/4)*a^3*A*ArcTanh[(-1)^(3/4
)*Sqrt[Tan[c + d*x]]])/((a^2 + b^2)^3*d) + (3*(-1)^(3/4)*a^2*A*b*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/((a^2
+ b^2)^3*d) + (3*(-1)^(1/4)*a*A*b^2*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/((a^2 + b^2)^3*d) - ((-1)^(3/4)*A
*b^3*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/((a^2 + b^2)^3*d) - ((-1)^(3/4)*a^3*B*ArcTanh[(-1)^(3/4)*Sqrt[Tan
[c + d*x]]])/((a^2 + b^2)^3*d) - (3*(-1)^(1/4)*a^2*b*B*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/((a^2 + b^2)^3*
d) + (3*(-1)^(3/4)*a*b^2*B*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/((a^2 + b^2)^3*d) + ((-1)^(1/4)*b^3*B*ArcTa
nh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/((a^2 + b^2)^3*d) - (2*a*A*Sqrt[Tan[c + d*x]])/(b^2*d*(a + b*Tan[c + d*x])^
2) + (a^3*A*Sqrt[Tan[c + d*x]])/(2*b^2*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (10*a^2*B*Sqrt[Tan[c + d*x]])/(
b^3*d*(a + b*Tan[c + d*x])^2) + (2*B*Sqrt[Tan[c + d*x]])/(3*b*d*(a + b*Tan[c + d*x])^2) - (5*a^4*B*Sqrt[Tan[c
+ d*x]])/(2*b^3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) - (8*a^2*B*Sqrt[Tan[c + d*x]])/(3*b*(a^2 + b^2)*d*(a + b
*Tan[c + d*x])^2) - (2*b*B*Sqrt[Tan[c + d*x]])/(3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) - (2*A*Tan[c + d*x]^(3
/2))/(b*d*(a + b*Tan[c + d*x])^2) + (10*a*B*Tan[c + d*x]^(3/2))/(b^2*d*(a + b*Tan[c + d*x])^2) + (2*B*Tan[c +
d*x]^(5/2))/(b*d*(a + b*Tan[c + d*x])^2) + (3*a^2*A*Sqrt[Tan[c + d*x]])/(4*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x]
)) + (3*a^4*A*Sqrt[Tan[c + d*x]])/(4*b^2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x])) + (2*A*b^2*Sqrt[Tan[c + d*x]])/
((a^2 + b^2)^2*d*(a + b*Tan[c + d*x])) - (15*a^5*B*Sqrt[Tan[c + d*x]])/(4*b^3*(a^2 + b^2)^2*d*(a + b*Tan[c + d
*x])) - (31*a^3*B*Sqrt[Tan[c + d*x]])/(4*b*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x])) - (6*a*b*B*Sqrt[Tan[c + d*x]]
)/((a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

________________________________________________________________________________________

Maple [B]  time = 0.064, size = 1864, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(7/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x)

[Out]

-3/2/d/(a^2+b^2)^3*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a*b^2+3/4/d/(a^2+b^2)^3*B*2^(1/2)*ln((1+2^(1/2
)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a^2*b-3/4/d/(a^2+b^2)^3*A*2^(1/2)*ln((
1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a^2*b-3/2/d/(a^2+b^2)^3*A*2^(1
/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^2*b-3/2/d/(a^2+b^2)^3*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a
^2*b-3/4/d/(a^2+b^2)^3*B*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*
x+c)))*a*b^2-5/4/d*a^6/b/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(3/2)*A-9/2/d*a^4*b/(a^2+b^2)^3/(a+b*tan(d*
x+c))^2*tan(d*x+c)^(3/2)*A-13/4/d*a^2*b^3/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(3/2)*A+9/4/d*a^7/b^2/(a^2
+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(3/2)*B+17/4/d*a^3*b^2/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(3/2)*B
-3/4/d*a^7/b^2/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*A*tan(d*x+c)^(1/2)-3/2/d/(a^2+b^2)^3*A*2^(1/2)*arctan(1+2^(1/2)*
tan(d*x+c)^(1/2))*a*b^2-3/2/d/(a^2+b^2)^3*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a*b^2-3/4/d/(a^2+b^2)^
3*A*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a*b^2+1/2/d/(a
^2+b^2)^3*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^3+1/2/d/(a^2+b^2)^3*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d
*x+c)^(1/2))*a^3+1/4/d/(a^2+b^2)^3*A*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^
(1/2)+tan(d*x+c)))*a^3-1/2/d/(a^2+b^2)^3*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^3+1/2/d/(a^2+b^2)^3*B*
2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^3+1/2/d/(a^2+b^2)^3*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))
*a^3+1/2/d/(a^2+b^2)^3*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*b^3+1/4/d/(a^2+b^2)^3*B*2^(1/2)*ln((1-2^(
1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a^3+13/2/d*a^5/(a^2+b^2)^3/(a+b*tan
(d*x+c))^2*tan(d*x+c)^(3/2)*B-7/2/d*a^5/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*A*tan(d*x+c)^(1/2)+3/2/d*a^4/(a^2+b^2)^
3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*A-1/2/d/(a^2+b^2)^3*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c
)^(1/2))*b^3-1/4/d/(a^2+b^2)^3*B*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2
)+tan(d*x+c)))*b^3+1/4/d/(a^2+b^2)^3*A*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c
)^(1/2)+tan(d*x+c)))*b^3+1/2/d/(a^2+b^2)^3*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^3+3/4/d*a^6/b^2/(a^2
+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*A+3/2/d/(a^2+b^2)^3*B*2^(1/2)*arctan(-1+2^(1/2)*tan
(d*x+c)^(1/2))*a^2*b+35/4/d*a^2*b^2/(a^2+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*A-15/4/d*a^
7/b^3/(a^2+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*B-23/2/d*a^5/b/(a^2+b^2)^3/(a*b)^(1/2)*ar
ctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*B-3/2/d/(a^2+b^2)^3*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a*b^2-6
3/4/d*a^3*b/(a^2+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*B+3/2/d/(a^2+b^2)^3*B*2^(1/2)*arcta
n(1+2^(1/2)*tan(d*x+c)^(1/2))*a^2*b-11/4/d*a^3*b^2/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*A*tan(d*x+c)^(1/2)+7/4/d*a^8
/b^3/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*B*tan(d*x+c)^(1/2)+11/2/d*a^6/b/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*B*tan(d*x+c
)^(1/2)+15/4/d*a^4*b/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*B*tan(d*x+c)^(1/2)+2/d*B/b^3*tan(d*x+c)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(7/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(7/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(7/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**3,x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(7/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out