3.599 \(\int \frac{\tan ^{\frac{11}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=493 \[ \frac{a^{7/2} \left (102 a^2 b^2+35 a^4+99 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{9/2} d \left (a^2+b^2\right )^3}-\frac{a^2 \left (7 a^2+15 b^2\right ) \tan ^{\frac{5}{2}}(c+d x)}{4 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{\left (67 a^2 b^2+35 a^4+8 b^4\right ) \tan ^{\frac{3}{2}}(c+d x)}{12 b^3 d \left (a^2+b^2\right )^2}+\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{a \left (67 a^2 b^2+35 a^4+24 b^4\right ) \sqrt{\tan (c+d x)}}{4 b^4 d \left (a^2+b^2\right )^2}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]

[Out]

((a + b)*(a^2 - 4*a*b + b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) - ((a + b)*(a^2
- 4*a*b + b^2)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) + (a^(7/2)*(35*a^4 + 102*a^2
*b^2 + 99*b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*b^(9/2)*(a^2 + b^2)^3*d) + ((a - b)*(a^2 + 4*a
*b + b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) - ((a - b)*(a^2 + 4*
a*b + b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) - (a*(35*a^4 + 67*a
^2*b^2 + 24*b^4)*Sqrt[Tan[c + d*x]])/(4*b^4*(a^2 + b^2)^2*d) + ((35*a^4 + 67*a^2*b^2 + 8*b^4)*Tan[c + d*x]^(3/
2))/(12*b^3*(a^2 + b^2)^2*d) - (a^2*Tan[c + d*x]^(7/2))/(2*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) - (a^2*(7*a
^2 + 15*b^2)*Tan[c + d*x]^(5/2))/(4*b^2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.40242, antiderivative size = 493, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 14, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.609, Rules used = {3565, 3645, 3647, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac{a^{7/2} \left (102 a^2 b^2+35 a^4+99 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{9/2} d \left (a^2+b^2\right )^3}-\frac{a^2 \left (7 a^2+15 b^2\right ) \tan ^{\frac{5}{2}}(c+d x)}{4 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{\left (67 a^2 b^2+35 a^4+8 b^4\right ) \tan ^{\frac{3}{2}}(c+d x)}{12 b^3 d \left (a^2+b^2\right )^2}+\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{a \left (67 a^2 b^2+35 a^4+24 b^4\right ) \sqrt{\tan (c+d x)}}{4 b^4 d \left (a^2+b^2\right )^2}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(11/2)/(a + b*Tan[c + d*x])^3,x]

[Out]

((a + b)*(a^2 - 4*a*b + b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) - ((a + b)*(a^2
- 4*a*b + b^2)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) + (a^(7/2)*(35*a^4 + 102*a^2
*b^2 + 99*b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*b^(9/2)*(a^2 + b^2)^3*d) + ((a - b)*(a^2 + 4*a
*b + b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) - ((a - b)*(a^2 + 4*
a*b + b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) - (a*(35*a^4 + 67*a
^2*b^2 + 24*b^4)*Sqrt[Tan[c + d*x]])/(4*b^4*(a^2 + b^2)^2*d) + ((35*a^4 + 67*a^2*b^2 + 8*b^4)*Tan[c + d*x]^(3/
2))/(12*b^3*(a^2 + b^2)^2*d) - (a^2*Tan[c + d*x]^(7/2))/(2*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) - (a^2*(7*a
^2 + 15*b^2)*Tan[c + d*x]^(5/2))/(4*b^2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

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{11}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=-\frac{a^2 \tan ^{\frac{7}{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{5}{2}}(c+d x) \left (\frac{7 a^2}{2}-2 a b \tan (c+d x)+\frac{1}{2} \left (7 a^2+4 b^2\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^2 \tan ^{\frac{7}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a^2 \left (7 a^2+15 b^2\right ) \tan ^{\frac{5}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{\tan ^{\frac{3}{2}}(c+d x) \left (\frac{5}{4} a^2 \left (7 a^2+15 b^2\right )-4 a b^3 \tan (c+d x)+\frac{1}{4} \left (35 a^4+67 a^2 b^2+8 b^4\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 (35 a^4+67 a^2 b^2+8 b^4\right ) \tan ^{\frac{3}{2}}(c+d x)}{12 b^3 \left (a^2+b^2\right )^2 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a^2 \left (7 a^2+15 b^2\right ) \tan ^{\frac{5}{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}{8} a \left (35 a^4+67 a^2 b^2+8 b^4\right )+3 b^3 \left (a^2-b^2\right ) \tan (c+d x)-\frac{3}{8} a \left (35 a^4+67 a^2 b^2+24 b^4\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{3 b^3 \left (a^2+b^2\right )^2}\\ &=-\frac{a \left (35 a^4+67 a^2 b^2+24 b^4\right ) \sqrt{\tan (c+d x)}}{4 b^4 \left (a^2+b^2\right )^2 d}+\frac{\left (35 a^4+67 a^2 b^2+8 b^4\right ) \tan ^{\frac{3}{2}}(c+d x)}{12 b^3 \left (a^2+b^2\right )^2 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a^2 \left (7 a^2+15 b^2\right ) \tan ^{\frac{5}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{2 \int \frac{\frac{3}{16} a^2 \left (35 a^4+67 a^2 b^2+24 b^4\right )+3 a b^5 \tan (c+d x)+\frac{3}{16} \left (35 a^6+67 a^4 b^2+32 a^2 b^4-8 b^6\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{3 b^4 \left (a^2+b^2\right )^2}\\ &=-\frac{a \left (35 a^4+67 a^2 b^2+24 b^4\right ) \sqrt{\tan (c+d x)}}{4 b^4 \left (a^2+b^2\right )^2 d}+\frac{\left (35 a^4+67 a^2 b^2+8 b^4\right ) \tan ^{\frac{3}{2}}(c+d x)}{12 b^3 \left (a^2+b^2\right )^2 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a^2 \left (7 a^2+15 b^2\right ) \tan ^{\frac{5}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{2 \int \frac{-\frac{3}{2} a b^4 \left (a^2-3 b^2\right )+\frac{3}{2} b^5 \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{3 b^4 \left (a^2+b^2\right )^3}+\frac{\left (a^4 \left (35 a^4+102 a^2 b^2+99 b^4\right )\right ) \int \frac{1+\tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{8 b^4 \left (a^2+b^2\right )^3}\\ &=-\frac{a \left (35 a^4+67 a^2 b^2+24 b^4\right ) \sqrt{\tan (c+d x)}}{4 b^4 \left (a^2+b^2\right )^2 d}+\frac{\left (35 a^4+67 a^2 b^2+8 b^4\right ) \tan ^{\frac{3}{2}}(c+d x)}{12 b^3 \left (a^2+b^2\right )^2 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a^2 \left (7 a^2+15 b^2\right ) \tan ^{\frac{5}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{4 \operatorname{Subst}\left (\int \frac{-\frac{3}{2} a b^4 \left (a^2-3 b^2\right )+\frac{3}{2} b^5 \left (3 a^2-b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{3 b^4 \left (a^2+b^2\right )^3 d}+\frac{\left (a^4 \left (35 a^4+102 a^2 b^2+99 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{8 b^4 \left (a^2+b^2\right )^3 d}\\ &=-\frac{a \left (35 a^4+67 a^2 b^2+24 b^4\right ) \sqrt{\tan (c+d x)}}{4 b^4 \left (a^2+b^2\right )^2 d}+\frac{\left (35 a^4+67 a^2 b^2+8 b^4\right ) \tan ^{\frac{3}{2}}(c+d x)}{12 b^3 \left (a^2+b^2\right )^2 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a^2 \left (7 a^2+15 b^2\right ) \tan ^{\frac{5}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\left ((a+b) \left (a^2-4 a b+b^2\right )\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-b) \left (a^2+4 a b+b^2\right )\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^4 \left (35 a^4+102 a^2 b^2+99 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 b^4 \left (a^2+b^2\right )^3 d}\\ &=\frac{a^{7/2} \left (35 a^4+102 a^2 b^2+99 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{9/2} \left (a^2+b^2\right )^3 d}-\frac{a \left (35 a^4+67 a^2 b^2+24 b^4\right ) \sqrt{\tan (c+d x)}}{4 b^4 \left (a^2+b^2\right )^2 d}+\frac{\left (35 a^4+67 a^2 b^2+8 b^4\right ) \tan ^{\frac{3}{2}}(c+d x)}{12 b^3 \left (a^2+b^2\right )^2 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a^2 \left (7 a^2+15 b^2\right ) \tan ^{\frac{5}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\left ((a+b) \left (a^2-4 a b+b^2\right )\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+b) \left (a^2-4 a b+b^2\right )\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-b) \left (a^2+4 a b+b^2\right )\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-b) \left (a^2+4 a b+b^2\right )\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^{7/2} \left (35 a^4+102 a^2 b^2+99 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{9/2} \left (a^2+b^2\right )^3 d}+\frac{(a-b) \left (a^2+4 a b+b^2\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{(a-b) \left (a^2+4 a b+b^2\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{a \left (35 a^4+67 a^2 b^2+24 b^4\right ) \sqrt{\tan (c+d x)}}{4 b^4 \left (a^2+b^2\right )^2 d}+\frac{\left (35 a^4+67 a^2 b^2+8 b^4\right ) \tan ^{\frac{3}{2}}(c+d x)}{12 b^3 \left (a^2+b^2\right )^2 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a^2 \left (7 a^2+15 b^2\right ) \tan ^{\frac{5}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\left ((a+b) \left (a^2-4 a b+b^2\right )\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 ((a+b) \left (a^2-4 a b+b^2\right )\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{(a+b) \left (a^2-4 a b+b^2\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+b) \left (a^2-4 a b+b^2\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^{7/2} \left (35 a^4+102 a^2 b^2+99 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{9/2} \left (a^2+b^2\right )^3 d}+\frac{(a-b) \left (a^2+4 a b+b^2\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{(a-b) \left (a^2+4 a b+b^2\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{a \left (35 a^4+67 a^2 b^2+24 b^4\right ) \sqrt{\tan (c+d x)}}{4 b^4 \left (a^2+b^2\right )^2 d}+\frac{\left (35 a^4+67 a^2 b^2+8 b^4\right ) \tan ^{\frac{3}{2}}(c+d x)}{12 b^3 \left (a^2+b^2\right )^2 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a^2 \left (7 a^2+15 b^2\right ) \tan ^{\frac{5}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 6.39517, size = 723, normalized size = 1.47 \[ \frac{b^2 \tan ^{\frac{13}{2}}(c+d x)}{2 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{-\frac{b \tan ^{\frac{11}{2}}(c+d x)}{d (a+b \tan (c+d x))}+\frac{2 \left (\frac{9 a b \tan ^{\frac{9}{2}}(c+d x)}{2 d (a+b \tan (c+d x))}+\frac{2 \left (-\frac{63 a^2 b \tan ^{\frac{7}{2}}(c+d x)}{4 d (a+b \tan (c+d x))}+\frac{2 \left (\frac{105 a b \left (7 a^2+4 b^2\right ) \tan ^{\frac{5}{2}}(c+d x)}{8 d (a+b \tan (c+d x))}+\frac{2 \left (-\frac{315 a^2 b \left (35 a^2+32 b^2\right ) \tan ^{\frac{3}{2}}(c+d x)}{16 d (a+b \tan (c+d x))}+\frac{2 \left (-\frac{945 a b \left (32 a^2 b^2+35 a^4-4 b^4\right ) \sqrt{\tan (c+d x)}}{32 d (a+b \tan (c+d x))}-\frac{2 \left (\frac{\left (-\frac{945}{128} a^2 b^4 \left (32 a^2 b^2+35 a^4-4 b^4\right )-a \left (\frac{945}{128} a^5 b^2 \left (35 a^2+32 b^2\right )-\frac{945 a b^8}{32}\right )\right ) \sqrt{\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{\frac{2 \left (\frac{945 a^4 b^8}{16}-\frac{945}{256} a^4 b^4 \left (67 a^2 b^2+35 a^4+24 b^4\right )-\frac{945}{256} a^4 b^2 \left (67 a^4 b^2+32 a^2 b^4+35 a^6-8 b^6\right )\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} d \left (a^2+b^2\right )}+\frac{-\frac{\sqrt [4]{-1} \left (\frac{945}{32} a^3 b^6 \left (a^2-3 b^2\right )+\frac{945}{32} i a^2 b^7 \left (3 a^2-b^2\right )\right ) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{\sqrt [4]{-1} \left (\frac{945}{32} a^3 b^6 \left (a^2-3 b^2\right )-\frac{945}{32} i a^2 b^7 \left (3 a^2-b^2\right )\right ) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}}{a^2+b^2}}{a \left (a^2+b^2\right )}\right )}{b}\right )}{b}\right )}{3 b}\right )}{5 b}\right )}{7 b}\right )}{9 b}}{2 a \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(11/2)/(a + b*Tan[c + d*x])^3,x]

[Out]

(b^2*Tan[c + d*x]^(13/2))/(2*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (-((b*Tan[c + d*x]^(11/2))/(d*(a + b*Ta
n[c + d*x]))) + (2*((9*a*b*Tan[c + d*x]^(9/2))/(2*d*(a + b*Tan[c + d*x])) + (2*((-63*a^2*b*Tan[c + d*x]^(7/2))
/(4*d*(a + b*Tan[c + d*x])) + (2*((105*a*b*(7*a^2 + 4*b^2)*Tan[c + d*x]^(5/2))/(8*d*(a + b*Tan[c + d*x])) + (2
*((-315*a^2*b*(35*a^2 + 32*b^2)*Tan[c + d*x]^(3/2))/(16*d*(a + b*Tan[c + d*x])) + (2*((-945*a*b*(35*a^4 + 32*a
^2*b^2 - 4*b^4)*Sqrt[Tan[c + d*x]])/(32*d*(a + b*Tan[c + d*x])) - (2*(((2*((945*a^4*b^8)/16 - (945*a^4*b^4*(35
*a^4 + 67*a^2*b^2 + 24*b^4))/256 - (945*a^4*b^2*(35*a^6 + 67*a^4*b^2 + 32*a^2*b^4 - 8*b^6))/256)*ArcTan[(Sqrt[
b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*(a^2 + b^2)*d) + (-(((-1)^(1/4)*((945*a^3*b^6*(a^2 - 3*b^2))
/32 + ((945*I)/32)*a^2*b^7*(3*a^2 - b^2))*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d) - ((-1)^(1/4)*((945*a^3*b^
6*(a^2 - 3*b^2))/32 - ((945*I)/32)*a^2*b^7*(3*a^2 - b^2))*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d)/(a^2 + b^
2))/(a*(a^2 + b^2)) + (((-945*a^2*b^4*(35*a^4 + 32*a^2*b^2 - 4*b^4))/128 - a*((-945*a*b^8)/32 + (945*a^5*b^2*(
35*a^2 + 32*b^2))/128))*Sqrt[Tan[c + d*x]])/(a*(a^2 + b^2)*d*(a + b*Tan[c + d*x]))))/b))/b))/(3*b)))/(5*b)))/(
7*b)))/(9*b))/(2*a*(a^2 + b^2))

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Maple [B]  time = 0.046, size = 936, normalized size = 1.9 \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)^(11/2)/(a+b*tan(d*x+c))^3,x)

[Out]

2/3/d/b^3*tan(d*x+c)^(3/2)-6/d/b^4*a*tan(d*x+c)^(1/2)-13/4/d*a^8/b^3/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)
^(3/2)-17/2/d*a^6/b/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(3/2)-21/4/d*a^4*b/(a^2+b^2)^3/(a+b*tan(d*x+c))^
2*tan(d*x+c)^(3/2)-11/4/d*a^9/b^4/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(1/2)-15/2/d*a^7/b^2/(a^2+b^2)^3/(
a+b*tan(d*x+c))^2*tan(d*x+c)^(1/2)-19/4/d*a^5/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(1/2)+35/4/d*a^8/b^4/(
a^2+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))+51/2/d*a^6/b^2/(a^2+b^2)^3/(a*b)^(1/2)*arctan(ta
n(d*x+c)^(1/2)*b/(a*b)^(1/2))+99/4/d*a^4/(a^2+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))-1/2/d/
(a^2+b^2)^3*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a^3+3/2/d/(a^2+b^2)^3*arctan(-1+2^(1/2)*tan(d*x+c)^(1/
2))*2^(1/2)*a*b^2-1/4/d/(a^2+b^2)^3*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+3/4/d/(a^2+b^2)^3*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*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^3+3/2/d/(a^2+b^2)^3
*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*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2
)*a^2*b-1/2/d/(a^2+b^2)^3*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*b^3+3/4/d/(a^2+b^2)^3*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-1/4/d/(a^2+b^2)^3*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+3/2/d/(a^2+b^2)^3*2^(1/2)*
arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^2*b-1/2/d/(a^2+b^2)^3*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^3

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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)^(11/2)/(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)^(11/2)/(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)**(11/2)/(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)^(11/2)/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out