3.435 \(\int \frac{1}{\sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=162 \[ -\frac{32 i \sqrt{a+i a \tan (c+d x)}}{77 a^3 d \sqrt{e \sec (c+d x)}}+\frac{16 i}{77 a^2 d \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{12 i}{77 a d (a+i a \tan (c+d x))^{3/2} \sqrt{e \sec (c+d x)}}+\frac{2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt{e \sec (c+d x)}} \]

[Out]

((2*I)/11)/(d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)) + ((12*I)/77)/(a*d*Sqrt[e*Sec[c + d*x]]*(a +
I*a*Tan[c + d*x])^(3/2)) + ((16*I)/77)/(a^2*d*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((32*I)/77)*
Sqrt[a + I*a*Tan[c + d*x]])/(a^3*d*Sqrt[e*Sec[c + d*x]])

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Rubi [A]  time = 0.301009, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3502, 3488} \[ -\frac{32 i \sqrt{a+i a \tan (c+d x)}}{77 a^3 d \sqrt{e \sec (c+d x)}}+\frac{16 i}{77 a^2 d \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{12 i}{77 a d (a+i a \tan (c+d x))^{3/2} \sqrt{e \sec (c+d x)}}+\frac{2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((2*I)/11)/(d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)) + ((12*I)/77)/(a*d*Sqrt[e*Sec[c + d*x]]*(a +
I*a*Tan[c + d*x])^(3/2)) + ((16*I)/77)/(a^2*d*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((32*I)/77)*
Sqrt[a + I*a*Tan[c + d*x]])/(a^3*d*Sqrt[e*Sec[c + d*x]])

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 3488

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{2 i}{11 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{6 \int \frac{1}{\sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx}{11 a}\\ &=\frac{2 i}{11 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{12 i}{77 a d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{24 \int \frac{1}{\sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}} \, dx}{77 a^2}\\ &=\frac{2 i}{11 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{12 i}{77 a d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{16 i}{77 a^2 d \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{16 \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}} \, dx}{77 a^3}\\ &=\frac{2 i}{11 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{12 i}{77 a d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{16 i}{77 a^2 d \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{32 i \sqrt{a+i a \tan (c+d x)}}{77 a^3 d \sqrt{e \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.400181, size = 102, normalized size = 0.63 \[ \frac{i \sec ^3(c+d x) (-22 i \sin (c+d x)+42 i \sin (3 (c+d x))-55 \cos (c+d x)+35 \cos (3 (c+d x)))}{154 a^2 d (\tan (c+d x)-i)^2 \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((I/154)*Sec[c + d*x]^3*(-55*Cos[c + d*x] + 35*Cos[3*(c + d*x)] - (22*I)*Sin[c + d*x] + (42*I)*Sin[3*(c + d*x)
]))/(a^2*d*Sqrt[e*Sec[c + d*x]]*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.32, size = 140, normalized size = 0.9 \begin{align*}{\frac{2\,\cos \left ( dx+c \right ) \left ( 28\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+28\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) -9\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +2\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+8\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -16\,i \right ) }{77\,d{a}^{3}e}\sqrt{{\frac{e}{\cos \left ( dx+c \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(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

2/77/d/a^3*(e/cos(d*x+c))^(1/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)*(28*I*cos(d*x+c)^6+2
8*cos(d*x+c)^5*sin(d*x+c)-9*I*cos(d*x+c)^4+5*cos(d*x+c)^3*sin(d*x+c)+2*I*cos(d*x+c)^2+8*cos(d*x+c)*sin(d*x+c)-
16*I)/e

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Maxima [A]  time = 1.9415, size = 240, normalized size = 1.48 \begin{align*} \frac{7 i \, \cos \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right ) + 33 i \, \cos \left (\frac{7}{11} \, \arctan \left (\sin \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right ), \cos \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right )\right )\right ) + 77 i \, \cos \left (\frac{3}{11} \, \arctan \left (\sin \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right ), \cos \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right )\right )\right ) - 77 i \, \cos \left (\frac{1}{11} \, \arctan \left (\sin \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right ), \cos \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right )\right )\right ) + 7 \, \sin \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right ) + 33 \, \sin \left (\frac{7}{11} \, \arctan \left (\sin \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right ), \cos \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right )\right )\right ) + 77 \, \sin \left (\frac{3}{11} \, \arctan \left (\sin \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right ), \cos \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right )\right )\right ) + 77 \, \sin \left (\frac{1}{11} \, \arctan \left (\sin \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right ), \cos \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right )\right )\right )}{308 \, a^{\frac{5}{2}} d \sqrt{e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/308*(7*I*cos(11/2*d*x + 11/2*c) + 33*I*cos(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 7
7*I*cos(3/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 77*I*cos(1/11*arctan2(sin(11/2*d*x + 1
1/2*c), cos(11/2*d*x + 11/2*c))) + 7*sin(11/2*d*x + 11/2*c) + 33*sin(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(
11/2*d*x + 11/2*c))) + 77*sin(3/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 77*sin(1/11*arct
an2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))))/(a^(5/2)*d*sqrt(e))

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Fricas [A]  time = 2.12603, size = 271, normalized size = 1.67 \begin{align*} \frac{\sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-77 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 110 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 40 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i\right )} e^{\left (-\frac{11}{2} i \, d x - \frac{11}{2} i \, c\right )}}{308 \, a^{3} d e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/308*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-77*I*e^(8*I*d*x + 8*I*c) + 110*I*e
^(4*I*d*x + 4*I*c) + 40*I*e^(2*I*d*x + 2*I*c) + 7*I)*e^(-11/2*I*d*x - 11/2*I*c)/(a^3*d*e)

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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(1/(e*sec(d*x+c))**(1/2)/(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{1}{\sqrt{e \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(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(e*sec(d*x + c))*(I*a*tan(d*x + c) + a)^(5/2)), x)