3.165 \(\int \frac{(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=137 \[ \frac{(2+2 i) a^{3/2} (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a (3 B+4 i A) \sqrt{a+i a \tan (c+d x)}}{3 d \sqrt{\tan (c+d x)}}-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]
[Out]
((2 + 2*I)*a^(3/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*
a*A*Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) - (2*a*((4*I)*A + 3*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3
*d*Sqrt[Tan[c + d*x]])
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Rubi [A] time = 0.365525, antiderivative size = 137, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used =
{3593, 3598, 12, 3544, 205} \[ \frac{(2+2 i) a^{3/2} (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a (3 B+4 i A) \sqrt{a+i a \tan (c+d x)}}{3 d \sqrt{\tan (c+d x)}}-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2),x]
[Out]
((2 + 2*I)*a^(3/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*
a*A*Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) - (2*a*((4*I)*A + 3*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3
*d*Sqrt[Tan[c + d*x]])
Rule 3593
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^2*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^
(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m -
1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ
[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]
Rule 3598
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*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 3544
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
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{(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac{5}{2}}(c+d x)} \, dx &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2}{3} \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{1}{2} a (4 i A+3 B)-\frac{1}{2} a (2 A-3 i B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a (4 i A+3 B) \sqrt{a+i a \tan (c+d x)}}{3 d \sqrt{\tan (c+d x)}}+\frac{4 \int -\frac{3 a^2 (A-i B) \sqrt{a+i a \tan (c+d x)}}{2 \sqrt{\tan (c+d x)}} \, dx}{3 a}\\ &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a (4 i A+3 B) \sqrt{a+i a \tan (c+d x)}}{3 d \sqrt{\tan (c+d x)}}-(2 a (A-i B)) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a (4 i A+3 B) \sqrt{a+i a \tan (c+d x)}}{3 d \sqrt{\tan (c+d x)}}+\frac{\left (4 a^3 (i A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac{(2+2 i) a^{3/2} (i A+B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a (4 i A+3 B) \sqrt{a+i a \tan (c+d x)}}{3 d \sqrt{\tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 5.60928, size = 221, normalized size = 1.61 \[ \frac{a e^{-3 i (c+d x)} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right )^2 (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)} \left (e^{i (c+d x)} \sqrt{1-e^{2 i (c+d x)}} \left (i A \left (-3+5 e^{2 i (c+d x)}\right )+3 B \left (-1+e^{2 i (c+d x)}\right )\right )+3 (B+i A) \left (-1+e^{2 i (c+d x)}\right )^2 \sin ^{-1}\left (e^{i (c+d x)}\right )\right )}{3 d \left (1-e^{2 i (c+d x)}\right )^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2),x]
[Out]
(a*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))^2*(E^(I*(c + d*
x))*Sqrt[1 - E^((2*I)*(c + d*x))]*(3*B*(-1 + E^((2*I)*(c + d*x))) + I*A*(-3 + 5*E^((2*I)*(c + d*x)))) + 3*(I*A
+ B)*(-1 + E^((2*I)*(c + d*x)))^2*ArcSin[E^(I*(c + d*x))])*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(3
*d*E^((3*I)*(c + d*x))*(1 - E^((2*I)*(c + d*x)))^(5/2))
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Maple [B] time = 0.042, size = 618, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x)
[Out]
-1/6/d*(a*(1+I*tan(d*x+c)))^(1/2)*a/tan(d*x+c)^(3/2)*(-12*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*ta
n(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^2*a+3*I*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/
2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a+16*I*
A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+12*A*ln(1/2*(2*I*a*tan(d*x+c)+2*(a
*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^2*a+6*I*ln(1/2*(2*I*a*
tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^2*a-3*(
I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan
(d*x+c)+I))*tan(d*x+c)^2*a+12*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+6*ln
(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*
x+c)^2*a+4*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2))/(I*a)^(1/2)/(-I*a)^(1/2)/(a*tan(d
*x+c)*(1+I*tan(d*x+c)))^(1/2)
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Maxima [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+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [B] time = 1.7478, size = 1418, normalized size = 10.35 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="fricas")
[Out]
1/6*(4*sqrt(2)*((5*A - 3*I*B)*a*e^(4*I*d*x + 4*I*c) + 2*A*a*e^(2*I*d*x + 2*I*c) - 3*(A - I*B)*a)*sqrt(a/(e^(2*
I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 3*sqrt((-8
*I*A^2 - 16*A*B + 8*I*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((2*I*A
+ 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c
) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + I*sqrt((-8*I*A^2 - 16*A*B + 8*I*B^2)*a^3/d^2)*d*e^(2*I*d*x
+ 2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*a)) + 3*sqrt((-8*I*A^2 - 16*A*B + 8*I*B^2)*a^3/d^2)*(d*e^(4*I*d
*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*a
)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*
c) - I*sqrt((-8*I*A^2 - 16*A*B + 8*I*B^2)*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*
a)))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)
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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((a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(5/2),x)
[Out]
Timed out
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Giac [A] time = 1.52234, size = 261, normalized size = 1.91 \begin{align*} \frac{-\left (i - 1\right ) \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{4} +{\left (-\left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} + \left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4}\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} \sqrt{i \, a \tan \left (d x + c\right ) + a} B}{{\left (-2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a + 10 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} - 18 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} + 14 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} - 4 i \, a^{5}\right )} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="giac")
[Out]
(-(I - 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)^2*a^4 + (-(2*I + 2)*(I*a*tan(d*x +
c) + a)^2*a^3 + (2*I + 2)*(I*a*tan(d*x + c) + a)*a^4)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*tan(d
*x + c) + a)*B)/((-2*I*(I*a*tan(d*x + c) + a)^4*a + 10*I*(I*a*tan(d*x + c) + a)^3*a^2 - 18*I*(I*a*tan(d*x + c)
+ a)^2*a^3 + 14*I*(I*a*tan(d*x + c) + a)*a^4 - 4*I*a^5)*d)