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

Optimal. Leaf size=135 \[ -\frac{2 (3 B+i A) \sqrt{a+i a \tan (c+d x)}}{3 d \sqrt{\tan (c+d x)}}+\frac{(1+i) \sqrt{a} (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

[Out]

((1 + I)*Sqrt[a]*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*A*
Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) - (2*(I*A + 3*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Sqrt[Ta
n[c + d*x]])

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Rubi [A]  time = 0.342461, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3598, 12, 3544, 205} \[ -\frac{2 (3 B+i A) \sqrt{a+i a \tan (c+d x)}}{3 d \sqrt{\tan (c+d x)}}+\frac{(1+i) \sqrt{a} (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + I)*Sqrt[a]*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*A*
Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) - (2*(I*A + 3*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Sqrt[Ta
n[c + d*x]])

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

Mathematica [A]  time = 6.45787, size = 174, normalized size = 1.29 \[ \frac{(B+i A) e^{-i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right ) \sqrt{a+i a \tan (c+d x)}}{d \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}-\frac{2 \sqrt{a+i a \tan (c+d x)} (A \cot (c+d x)+i A+3 B)}{3 d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*A + B)*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]]*Sqrt[a + I*a
*Tan[c + d*x]])/(d*E^(I*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]) - (2*(I*
A + 3*B + A*Cot[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Sqrt[Tan[c + d*x]])

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Maple [B]  time = 0.043, size = 553, normalized size = 4.1 \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))^(1/2)*(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)/tan(d*x+c)^(3/2)*(3*I*A*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)^3*a-12*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)*tan(d*x+c)^2-3*I*B*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+3*B*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)^3*a-8*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)*tan(d*x+c)-4*I*A*tan(d*x+c)^2*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+3*A*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*I*B*(-I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+4*I*A*(-I*a)^(1/2)*(a*tan(d*x+
c)*(1+I*tan(d*x+c)))^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I)/(-I*a)^(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))^(1/2)*(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 = 1.81017, size = 1327, normalized size = 9.83 \begin{align*} \frac{\sqrt{2}{\left (4 \,{\left (2 \, A - 3 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, A e^{\left (2 i \, d x + 2 i \, c\right )} + 12 i \, B\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 3 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{{\left (-2 i \, A^{2} - 4 \, A B + 2 i \, B^{2}\right )} a}{d^{2}}} \log \left (\frac{{\left (\sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + i \, d \sqrt{\frac{{\left (-2 i \, A^{2} - 4 \, A B + 2 i \, B^{2}\right )} a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) + 3 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{{\left (-2 i \, A^{2} - 4 \, A B + 2 i \, B^{2}\right )} a}{d^{2}}} \log \left (\frac{{\left (\sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - i \, d \sqrt{\frac{{\left (-2 i \, A^{2} - 4 \, A B + 2 i \, B^{2}\right )} a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right )}{6 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

1/6*(sqrt(2)*(4*(2*A - 3*I*B)*e^(4*I*d*x + 4*I*c) + 8*A*e^(2*I*d*x + 2*I*c) + 12*I*B)*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*(d*e^(4*I*d*x + 4*I
*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)*a/d^2)*log((sqrt(2)*((I*A + B)*e^(2*I*d*x
+ 2*I*c) + I*A + B)*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*d*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)
/(I*A + B)) + 3*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)*a/d^2)
*log((sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*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*d*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)*a/d^2)*e^(2*I
*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)))/(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))**(1/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.42187, size = 240, normalized size = 1.78 \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 )} a^{4} +{\left (-\left (2 i - 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + \left (2 i - 2\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}{2 \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a - 5 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} + 9 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} - 7 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + 2 \, 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))^(1/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="giac")

[Out]

1/2*((I + 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)*a^4 + (-(2*I - 2)*(I*a*tan(d*x +
c) + a)*a^3 + (2*I - 2)*a^4)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*tan(d*x + c) + a)*B)/(((I*a*t
an(d*x + c) + a)^4*a - 5*(I*a*tan(d*x + c) + a)^3*a^2 + 9*(I*a*tan(d*x + c) + a)^2*a^3 - 7*(I*a*tan(d*x + c) +
a)*a^4 + 2*a^5)*d)