3.174 \(\int \frac{(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=185 \[ -\frac{2 a^2 (5 B+8 i A) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 (38 A-35 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]

[Out]

((-4 - 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2
*a^2*((8*I)*A + 5*B)*Sqrt[a + I*a*Tan[c + d*x]])/(15*d*Tan[c + d*x]^(3/2)) + (2*a^2*(38*A - (35*I)*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(15*d*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + I*a*Tan[c + d*x])^(3/2))/(5*d*Tan[c + d*x]^(5/2))

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Rubi [A]  time = 0.574996, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{2 a^2 (5 B+8 i A) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 (38 A-35 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

((-4 - 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2
*a^2*((8*I)*A + 5*B)*Sqrt[a + I*a*Tan[c + d*x]])/(15*d*Tan[c + d*x]^(3/2)) + (2*a^2*(38*A - (35*I)*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(15*d*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + I*a*Tan[c + d*x])^(3/2))/(5*d*Tan[c + d*x]^(5/2))

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))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac{7}{2}}(c+d x)} \, dx &=-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{(a+i a \tan (c+d x))^{3/2} \left (\frac{1}{2} a (8 i A+5 B)-\frac{1}{2} a (2 A-5 i B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (8 i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4}{15} \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^2 (38 A-35 i B)-\frac{1}{4} a^2 (22 i A+25 B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (8 i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 (38 A-35 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 \int -\frac{15 a^3 (i A+B) \sqrt{a+i a \tan (c+d x)}}{2 \sqrt{\tan (c+d x)}} \, dx}{15 a}\\ &=-\frac{2 a^2 (8 i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 (38 A-35 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\left (4 a^2 (i A+B)\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a^2 (8 i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 (38 A-35 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{\left (8 a^4 (A-i 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{(4+4 i) a^{5/2} (A-i 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^2 (8 i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 (38 A-35 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 10.7001, size = 323, normalized size = 1.75 \[ -\frac{4 \sqrt{2} e^{-2 i c} \sqrt{e^{i d x}} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} (a+i a \tan (c+d x))^{5/2} \left (e^{i (c+d x)} \sqrt{1-e^{2 i (c+d x)}} \left (i A \left (-35 e^{2 i (c+d x)}+26 e^{4 i (c+d x)}+15\right )+5 B \left (-7 e^{2 i (c+d x)}+4 e^{4 i (c+d x)}+3\right )\right )+15 (B+i A) \left (-1+e^{2 i (c+d x)}\right )^3 \sin ^{-1}\left (e^{i (c+d x)}\right )\right ) (A+B \tan (c+d x))}{15 d \left (1-e^{2 i (c+d x)}\right )^{7/2} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sec ^{\frac{7}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

(-4*Sqrt[2]*Sqrt[E^(I*d*x)]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(E^(I*(c + d*x))
*Sqrt[1 - E^((2*I)*(c + d*x))]*(5*B*(3 - 7*E^((2*I)*(c + d*x)) + 4*E^((4*I)*(c + d*x))) + I*A*(15 - 35*E^((2*I
)*(c + d*x)) + 26*E^((4*I)*(c + d*x)))) + 15*(I*A + B)*(-1 + E^((2*I)*(c + d*x)))^3*ArcSin[E^(I*(c + d*x))])*(
a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(15*d*E^((2*I)*c)*(1 - E^((2*I)*(c + d*x)))^(7/2)*Sqrt[E^(I*
(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sec[c + d*x]^(7/2)*(Cos[d*x] + I*Sin[d*x])^(5/2)*(A*Cos[c + d*x] + B*Sin
[c + d*x]))

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Maple [B]  time = 0.044, size = 709, normalized size = 3.8 \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))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x)

[Out]

-1/15/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(5/2)*(-76*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*
x+c)*(1+I*tan(d*x+c)))^(1/2)+60*I*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)^3*a-15*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)^3*a+70*I*B*(I*a)^(1/2)*(-I*a)^(1/2
)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+60*B*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)^3*a+30*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)^3*a-15*(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
)^3*a+22*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)-30*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)^3*a+10*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*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))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.76325, size = 1642, normalized size = 8.88 \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))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

1/30*(sqrt(2)*((208*I*A + 160*B)*a^2*e^(6*I*d*x + 6*I*c) + (-72*I*A - 120*B)*a^2*e^(4*I*d*x + 4*I*c) + (-160*I
*A - 160*B)*a^2*e^(2*I*d*x + 2*I*c) + (120*I*A + 120*B)*a^2)*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) - 15*sqrt((32*I*A^2 + 64*A*B - 32*I*B^2)*a^5/d^2
)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)*log((sqrt(2)*((4*I*A + 4*B)*
a^2*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*B)*a^2)*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) + sqrt((32*I*A^2 + 64*A*B - 32*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*
I*c))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^2)) + 15*sqrt((32*I*A^2 + 64*A*B - 32*I*B^2)*a^5/d^2)*(d*e^(6*I*d*
x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)*log((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x
+ 2*I*c) + (4*I*A + 4*B)*a^2)*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) - sqrt((32*I*A^2 + 64*A*B - 32*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*
d*x - 2*I*c)/((4*I*A + 4*B)*a^2)))/(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*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))**(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.60115, size = 288, normalized size = 1.56 \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 )}^{3} a^{5} +{\left (-\left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} + \left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5}\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 )}^{5} a - 6 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{2} + 14 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{3} - 16 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{4} + 9 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{5} - 2 \, a^{6}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),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)^3*a^5 + (-(2*I + 2)*(I*a*tan(d
*x + c) + a)^3*a^4 + (2*I + 2)*(I*a*tan(d*x + c) + a)^2*a^5)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*
a*tan(d*x + c) + a)*B)/(((I*a*tan(d*x + c) + a)^5*a - 6*(I*a*tan(d*x + c) + a)^4*a^2 + 14*(I*a*tan(d*x + c) +
a)^3*a^3 - 16*(I*a*tan(d*x + c) + a)^2*a^4 + 9*(I*a*tan(d*x + c) + a)*a^5 - 2*a^6)*d)