3.390 \(\int \frac{(a+b \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=283 \[ \frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{2 a^2 A}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a (a B+2 A b)}{d \sqrt{\tan (c+d x)}} \]

[Out]

((2*a*b*(A - B) + a^2*(A + B) - b^2*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - ((2*a*b*(A
- B) + a^2*(A + B) - b^2*(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) + ((a^2*(A - B) - b^2*(A
- B) - 2*a*b*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - ((a^2*(A - B) - b^2
*(A - B) - 2*a*b*(A + B))*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - (2*a^2*A)/(3*d*T
an[c + d*x]^(3/2)) - (2*a*(2*A*b + a*B))/(d*Sqrt[Tan[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.354313, antiderivative size = 283, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {3604, 3628, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{2 a^2 A}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a (a B+2 A b)}{d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2),x]

[Out]

((2*a*b*(A - B) + a^2*(A + B) - b^2*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - ((2*a*b*(A
- B) + a^2*(A + B) - b^2*(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) + ((a^2*(A - B) - b^2*(A
- B) - 2*a*b*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - ((a^2*(A - B) - b^2
*(A - B) - 2*a*b*(A + B))*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - (2*a^2*A)/(3*d*T
an[c + d*x]^(3/2)) - (2*a*(2*A*b + a*B))/(d*Sqrt[Tan[c + d*x]])

Rule 3604

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

Rule 3628

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 + b^2
)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
+ b^2, 0]

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]

Rubi steps

\begin{align*} \int \frac{(a+b \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac{5}{2}}(c+d x)} \, dx &=-\frac{2 a^2 A}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\int \frac{a (2 A b+a B)-\left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+b^2 B \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 A}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a (2 A b+a B)}{d \sqrt{\tan (c+d x)}}+\int \frac{-a^2 A+A b^2+2 a b B+\left (b^2 B-a (2 A b+a B)\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a^2 A}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a (2 A b+a B)}{d \sqrt{\tan (c+d x)}}+\frac{2 \operatorname{Subst}\left (\int \frac{-a^2 A+A b^2+2 a b B+\left (b^2 B-a (2 A b+a B)\right ) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{2 a^2 A}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a (2 A b+a B)}{d \sqrt{\tan (c+d x)}}-\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}-\frac{\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{2 a^2 A}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a (2 A b+a B)}{d \sqrt{\tan (c+d x)}}+\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (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} d}+\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (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} d}-\frac{\left (2 a b (A-B)+a^2 (A+B)-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 d}-\frac{\left (2 a b (A-B)+a^2 (A+B)-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 d}\\ &=\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{2 a^2 A}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a (2 A b+a B)}{d \sqrt{\tan (c+d x)}}-\frac{\left (2 a b (A-B)+a^2 (A+B)-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} d}+\frac{\left (2 a b (A-B)+a^2 (A+B)-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} d}\\ &=\frac{\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{2 a^2 A}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a (2 A b+a B)}{d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.687083, size = 119, normalized size = 0.42 \[ \frac{2 \left (a^2 (-A)+2 a b B+A b^2\right ) \text{Hypergeometric2F1}\left (-\frac{3}{4},1,\frac{1}{4},-\tan ^2(c+d x)\right )-6 \left (a^2 B+2 a A b-b^2 B\right ) \tan (c+d x) \text{Hypergeometric2F1}\left (-\frac{1}{4},1,\frac{3}{4},-\tan ^2(c+d x)\right )-2 b (2 a B+A b+3 b B \tan (c+d x))}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2),x]

[Out]

(2*(-(a^2*A) + A*b^2 + 2*a*b*B)*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[c + d*x]^2] - 6*(2*a*A*b + a^2*B - b^2*B)
*Hypergeometric2F1[-1/4, 1, 3/4, -Tan[c + d*x]^2]*Tan[c + d*x] - 2*b*(A*b + 2*a*B + 3*b*B*Tan[c + d*x]))/(3*d*
Tan[c + d*x]^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.025, size = 710, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x)

[Out]

-4/d*a/tan(d*x+c)^(1/2)*A*b-2/d*a^2/tan(d*x+c)^(1/2)*B-2/3*a^2*A/d/tan(d*x+c)^(3/2)-1/2/d*a^2*A*arctan(-1+2^(1
/2)*tan(d*x+c)^(1/2))*2^(1/2)+1/2/d*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*b^2-1/4/d*a^2*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)))+1/4/d*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^2-1/2/d*a^2*A*2^(1/2)*arctan(1+2^(1/2)
*tan(d*x+c)^(1/2))+1/2/d*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^2+1/d*B*2^(1/2)*arctan(-1+2^(1/2)*tan(
d*x+c)^(1/2))*a*b+1/2/d*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+1/d*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a*b-1/2/d*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-1/d*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a
*b-1/d*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a*b-1/4/d*a^2*B*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)))*2^(1/2)+1/4/d*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^2-1/2/d*a^2*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+1/2/d*B*2^
(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*b^2-1/2/d*a^2*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+1/2/d*B*2
^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^2

________________________________________________________________________________________

Maxima [A]  time = 1.69539, size = 335, normalized size = 1.18 \begin{align*} -\frac{6 \, \sqrt{2}{\left ({\left (A + B\right )} a^{2} + 2 \,{\left (A - B\right )} a b -{\left (A + B\right )} b^{2}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 6 \, \sqrt{2}{\left ({\left (A + B\right )} a^{2} + 2 \,{\left (A - B\right )} a b -{\left (A + B\right )} b^{2}\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 3 \, \sqrt{2}{\left ({\left (A - B\right )} a^{2} - 2 \,{\left (A + B\right )} a b -{\left (A - B\right )} b^{2}\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 3 \, \sqrt{2}{\left ({\left (A - B\right )} a^{2} - 2 \,{\left (A + B\right )} a b -{\left (A - B\right )} b^{2}\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \frac{8 \,{\left (A a^{2} + 3 \,{\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )\right )}}{\tan \left (d x + c\right )^{\frac{3}{2}}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

-1/12*(6*sqrt(2)*((A + B)*a^2 + 2*(A - B)*a*b - (A + B)*b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)
))) + 6*sqrt(2)*((A + B)*a^2 + 2*(A - B)*a*b - (A + B)*b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)
))) + 3*sqrt(2)*((A - B)*a^2 - 2*(A + B)*a*b - (A - B)*b^2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)
- 3*sqrt(2)*((A - B)*a^2 - 2*(A + B)*a*b - (A - B)*b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) +
8*(A*a^2 + 3*(B*a^2 + 2*A*a*b)*tan(d*x + c))/tan(d*x + c)^(3/2))/d

________________________________________________________________________________________

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((a+b*tan(d*x+c))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \tan{\left (c + d x \right )}\right ) \left (a + b \tan{\left (c + d x \right )}\right )^{2}}{\tan ^{\frac{5}{2}}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**2*(A+B*tan(d*x+c))/tan(d*x+c)**(5/2),x)

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**2/tan(c + d*x)**(5/2), x)

________________________________________________________________________________________

Giac [A]  time = 1.51921, size = 473, normalized size = 1.67 \begin{align*} -\frac{{\left (\sqrt{2} A a^{2} + \sqrt{2} B a^{2} + 2 \, \sqrt{2} A a b - 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} - \sqrt{2} B b^{2}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \, d} - \frac{{\left (\sqrt{2} A a^{2} + \sqrt{2} B a^{2} + 2 \, \sqrt{2} A a b - 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} - \sqrt{2} B b^{2}\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \, d} - \frac{{\left (\sqrt{2} A a^{2} - \sqrt{2} B a^{2} - 2 \, \sqrt{2} A a b - 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} + \sqrt{2} B b^{2}\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \, d} + \frac{{\left (\sqrt{2} A a^{2} - \sqrt{2} B a^{2} - 2 \, \sqrt{2} A a b - 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} + \sqrt{2} B b^{2}\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \, d} - \frac{2 \,{\left (3 \, B a^{2} \tan \left (d x + c\right ) + 6 \, A a b \tan \left (d x + c\right ) + A a^{2}\right )}}{3 \, d \tan \left (d x + c\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="giac")

[Out]

-1/2*(sqrt(2)*A*a^2 + sqrt(2)*B*a^2 + 2*sqrt(2)*A*a*b - 2*sqrt(2)*B*a*b - sqrt(2)*A*b^2 - sqrt(2)*B*b^2)*arcta
n(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c))))/d - 1/2*(sqrt(2)*A*a^2 + sqrt(2)*B*a^2 + 2*sqrt(2)*A*a*b - 2*s
qrt(2)*B*a*b - sqrt(2)*A*b^2 - sqrt(2)*B*b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c))))/d - 1/4*(s
qrt(2)*A*a^2 - sqrt(2)*B*a^2 - 2*sqrt(2)*A*a*b - 2*sqrt(2)*B*a*b - sqrt(2)*A*b^2 + sqrt(2)*B*b^2)*log(sqrt(2)*
sqrt(tan(d*x + c)) + tan(d*x + c) + 1)/d + 1/4*(sqrt(2)*A*a^2 - sqrt(2)*B*a^2 - 2*sqrt(2)*A*a*b - 2*sqrt(2)*B*
a*b - sqrt(2)*A*b^2 + sqrt(2)*B*b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)/d - 2/3*(3*B*a^2*tan(
d*x + c) + 6*A*a*b*tan(d*x + c) + A*a^2)/(d*tan(d*x + c)^(3/2))