3.375 \(\int \frac{\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=122 \[ \frac{3 i \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{3 i \sec (c+d x)}{16 a d (a+i a \tan (c+d x))^{3/2}}+\frac{i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

(((3*I)/16)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*a^(5/2)*d) + ((I/4)
*Sec[c + d*x])/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((3*I)/16)*Sec[c + d*x])/(a*d*(a + I*a*Tan[c + d*x])^(3/2))

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Rubi [A]  time = 0.115477, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3502, 3489, 206} \[ \frac{3 i \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{3 i \sec (c+d x)}{16 a d (a+i a \tan (c+d x))^{3/2}}+\frac{i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)/16)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*a^(5/2)*d) + ((I/4)
*Sec[c + d*x])/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((3*I)/16)*Sec[c + d*x])/(a*d*(a + I*a*Tan[c + d*x])^(3/2))

Rule 3502

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(b*f*(m + 2*n)), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rule 3489

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-2*a)/(b*f), Subst[
Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^
2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 \int \frac{\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx}{8 a}\\ &=\frac{i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 i \sec (c+d x)}{16 a d (a+i a \tan (c+d x))^{3/2}}+\frac{3 \int \frac{\sec (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx}{32 a^2}\\ &=\frac{i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 i \sec (c+d x)}{16 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{2-a x^2} \, dx,x,\frac{\sec (c+d x)}{\sqrt{a+i a \tan (c+d x)}}\right )}{16 a^2 d}\\ &=\frac{3 i \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 i \sec (c+d x)}{16 a d (a+i a \tan (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.751814, size = 121, normalized size = 0.99 \[ -\frac{i \sec ^3(c+d x) \left (3 i \sin (2 (c+d x))+7 \cos (2 (c+d x))+3 e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )+7\right )}{32 a^2 d (\tan (c+d x)-i)^2 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/32)*Sec[c + d*x]^3*(7 + 3*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c
+ d*x))]] + 7*Cos[2*(c + d*x)] + (3*I)*Sin[2*(c + d*x)]))/(a^2*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*
x]])

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Maple [B]  time = 0.283, size = 346, normalized size = 2.8 \begin{align*}{\frac{1}{64\,d{a}^{3}} \left ( 64\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}+3\,i\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i+\sin \left ( dx+c \right ) \right ) }{2\,\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \sqrt{2}\cos \left ( dx+c \right ) +64\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+3\,i\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i+\sin \left ( dx+c \right ) \right ) }{2\,\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \sqrt{2}+3\,\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ( 1/2\,{\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i+\sin \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \sqrt{2}\sin \left ( dx+c \right ) -24\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -12\,i\cos \left ( dx+c \right ) \right ) \sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

1/64/d/a^3*(64*I*cos(d*x+c)^5+3*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I+sin(
d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)*cos(d*x+c)+64*sin(d*x+c)*cos(d*x+c)^4+3*I*(-2
*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(co
s(d*x+c)+1))^(1/2))*2^(1/2)+3*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I+sin(d*x+
c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)*sin(d*x+c)-24*I*cos(d*x+c)^3+8*cos(d*x+c)^2*sin(d
*x+c)-12*I*cos(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)/(I*a*tan(d*x + c) + a)^(5/2), x)

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Fricas [B]  time = 2.11828, size = 807, normalized size = 6.61 \begin{align*} \frac{{\left (3 i \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left ({\left (2 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{1}{a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3 i \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-{\left (2 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{1}{a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (5 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 7 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{32 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(3*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log((2*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(I*
d*x + I*c) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x -
I*c)) - 3*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-(2*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^
(I*d*x + I*c) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x
- I*c)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(5*I*e^(4*I*d*x + 4*I*c) + 7*I*e^(2*I*d*x + 2*I*c) + 2*I)
*e^(I*d*x + I*c))*e^(-5*I*d*x - 5*I*c)/(a^3*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(sec(d*x+c)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)/(I*a*tan(d*x + c) + a)^(5/2), x)