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

Optimal. Leaf size=531 \[ -\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{\left (-18 a^2 A b^3+15 a^4 A b+26 a^3 b^2 B-3 a^5 B-3 a b^4 B-A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{b} d \left (a^2+b^2\right )^3}-\frac{(A b-a B) \sqrt{\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{\left (7 a^2 A b-3 a^3 B+5 a b^2 B-A b^3\right ) \sqrt{\tan (c+d x)}}{4 a d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}+\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]

[Out]

-(((a^3*(A - B) - 3*a*b^2*(A - B) + 3*a^2*b*(A + B) - b^3*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sq
rt[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))*ArcTan[1 + Sqrt[2]*
Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) - ((15*a^4*A*b - 18*a^2*A*b^3 - A*b^5 - 3*a^5*B + 26*a^3*b^2*B
- 3*a*b^4*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*a^(3/2)*Sqrt[b]*(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))*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqr
t[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))*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*B)*Sqrt[Tan[c + d*x]])/(2*(a^2 + b^2)*d
*(a + b*Tan[c + d*x])^2) - ((7*a^2*A*b - A*b^3 - 3*a^3*B + 5*a*b^2*B)*Sqrt[Tan[c + d*x]])/(4*a*(a^2 + b^2)^2*d
*(a + b*Tan[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.31205, antiderivative size = 531, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.394, Rules used = {3608, 3649, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ -\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{\left (-18 a^2 A b^3+15 a^4 A b+26 a^3 b^2 B-3 a^5 B-3 a b^4 B-A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{b} d \left (a^2+b^2\right )^3}-\frac{(A b-a B) \sqrt{\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{\left (7 a^2 A b-3 a^3 B+5 a b^2 B-A b^3\right ) \sqrt{\tan (c+d x)}}{4 a d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}+\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^3*(A - B) - 3*a*b^2*(A - B) + 3*a^2*b*(A + B) - b^3*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sq
rt[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))*ArcTan[1 + Sqrt[2]*
Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) - ((15*a^4*A*b - 18*a^2*A*b^3 - A*b^5 - 3*a^5*B + 26*a^3*b^2*B
- 3*a*b^4*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*a^(3/2)*Sqrt[b]*(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))*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqr
t[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))*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*B)*Sqrt[Tan[c + d*x]])/(2*(a^2 + b^2)*d
*(a + b*Tan[c + d*x])^2) - ((7*a^2*A*b - A*b^3 - 3*a^3*B + 5*a*b^2*B)*Sqrt[Tan[c + d*x]])/(4*a*(a^2 + b^2)^2*d
*(a + b*Tan[c + d*x]))

Rule 3608

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

Rule 3649

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*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[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{\sqrt{\tan (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx &=-\frac{(A b-a B) \sqrt{\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{\int \frac{-\frac{1}{2} b (A b-a B)-2 b (a A+b B) \tan (c+d x)+\frac{3}{2} b (A b-a B) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx}{2 b \left (a^2+b^2\right )}\\ &=-\frac{(A b-a B) \sqrt{\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\int \frac{-\frac{1}{4} b \left (9 a^2 A b+A b^3-5 a^3 B+3 a b^2 B\right )-2 a b \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)+\frac{1}{4} b \left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 a b \left (a^2+b^2\right )^2}\\ &=-\frac{(A b-a B) \sqrt{\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\int \frac{-2 a b \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right )-2 a b \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{2 a b \left (a^2+b^2\right )^3}-\frac{\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \int \frac{1+\tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{8 a \left (a^2+b^2\right )^3}\\ &=-\frac{(A b-a B) \sqrt{\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\operatorname{Subst}\left (\int \frac{-2 a b \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right )-2 a b \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a b \left (a^2+b^2\right )^3 d}-\frac{\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{8 a \left (a^2+b^2\right )^3 d}\\ &=-\frac{(A b-a B) \sqrt{\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 a \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{\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{b} \left (a^2+b^2\right )^3 d}-\frac{(A b-a B) \sqrt{\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 a \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{\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 (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{\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{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{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 (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{b} \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 ) \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^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (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{(A b-a B) \sqrt{\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\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}{-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^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}{-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^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (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 (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (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 (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{b} \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 ) \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^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (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{(A b-a B) \sqrt{\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 6.11245, size = 344, normalized size = 0.65 \[ \frac{\frac{2 (a+b \tan (c+d x)) \left (\frac{1}{4} a^{3/2} b^{3/2} \left (a^2+b^2\right ) \left (-7 a^2 A b+3 a^3 B-5 a b^2 B+A b^3\right ) \sqrt{\tan (c+d x)}-(a+b \tan (c+d x)) \left (-\frac{1}{4} a b \left (18 a^2 A b^3-15 a^4 A b-26 a^3 b^2 B+3 a^5 B+3 a b^4 B+A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )+\sqrt [4]{-1} a^{5/2} b^{3/2} \left ((-b+i a)^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )-(b+i a)^3 (A+i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )\right )\right )\right )}{a^{3/2} b^{3/2} \left (a^2+b^2\right )^2}+b (A b-a B) \tan ^{\frac{3}{2}}(c+d x)-(A b-a B) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))}{2 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(A*b - a*B)*Tan[c + d*x]^(3/2) - (A*b - a*B)*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x]) + (2*(a + b*Tan[c + d*
x])*((a^(3/2)*b^(3/2)*(a^2 + b^2)*(-7*a^2*A*b + A*b^3 + 3*a^3*B - 5*a*b^2*B)*Sqrt[Tan[c + d*x]])/4 - (-(a*b*(-
15*a^4*A*b + 18*a^2*A*b^3 + A*b^5 + 3*a^5*B - 26*a^3*b^2*B + 3*a*b^4*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sq
rt[a]])/4 + (-1)^(1/4)*a^(5/2)*b^(3/2)*((I*a - b)^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] - (I*a + b
)^3*(A + I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]))*(a + b*Tan[c + d*x])))/(a^(3/2)*b^(3/2)*(a^2 + b^2)^2))
/(2*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2)

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Maple [B]  time = 0.075, size = 1835, normalized size = 3.5 \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)^(1/2)*(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+1/4/d/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*b^6/
a*tan(d*x+c)^(3/2)*A-3/4/d/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(1/2)*B*a*b^4+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-3/2/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^2-5/4/d/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*b^5*tan(d*x+c)^(3/2)*B-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^3-1/4/d/(a^2+b^
2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(1/2)*A*b^5+3/4/d/(a^2+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(
1/2))*B*b^4-1/4/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+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^3-1/4/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+1/2/d*a^3/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*b^2*tan(d*x+c)^(1/2)*B+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^3-1/2/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^3-1/2/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*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^3-15/4/d*a^3/(a^2+b^2)^3*b/(
a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*A+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+1/4/d/(a^2+b^2)^3/a/(a*b)^(1/2)*arctan(tan(d*x+c)^
(1/2)*b/(a*b)^(1/2))*A*b^5+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)*t
an(d*x+c)^(1/2)+tan(d*x+c)))*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+3/4/d*a^4/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*b*tan(d*x+c)^(3/2)*B-1/2/d
*a^2/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*b^3*tan(d*x+c)^(3/2)*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^2*b-7/4/d*a^3/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*b^
2*tan(d*x+c)^(3/2)*A-3/2/d*a/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*b^4*tan(d*x+c)^(3/2)*A+9/2/d*a/(a^2+b^2)^3*b^3/(a*
b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*A-13/2/d*a^2/(a^2+b^2)^3*b^2/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/
2)*b/(a*b)^(1/2))*B-9/4/d*a^4/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*b*tan(d*x+c)^(1/2)*A-5/2/d*a^2/(a^2+b^2)^3/(a+b*t
an(d*x+c))^2*b^3*tan(d*x+c)^(1/2)*A+5/4/d*a^5/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(1/2)*B+3/4/d*a^4/(a^2
+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*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^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+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

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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)^(1/2)*(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)^(1/2)*(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \tan{\left (c + d x \right )}\right ) \sqrt{\tan{\left (c + d x \right )}}}{\left (a + b \tan{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**3,x)

[Out]

Integral((A + B*tan(c + d*x))*sqrt(tan(c + d*x))/(a + b*tan(c + d*x))**3, x)

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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)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out