3.429 \(\int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=209 \[ -\frac {256 i \sqrt {a+i a \tan (c+d x)}}{385 a^2 d e^2 \sqrt {e \sec (c+d x)}}-\frac {96 i \sqrt {a+i a \tan (c+d x)}}{385 a^2 d (e \sec (c+d x))^{5/2}}+\frac {128 i}{385 a d e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {16 i}{77 a d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{5/2}} \]

[Out]

16/77*I/a/d/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2)+128/385*I/a/d/e^2/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*
x+c))^(1/2)-96/385*I*(a+I*a*tan(d*x+c))^(1/2)/a^2/d/(e*sec(d*x+c))^(5/2)-256/385*I*(a+I*a*tan(d*x+c))^(1/2)/a^
2/d/e^2/(e*sec(d*x+c))^(1/2)+2/11*I/d/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.39, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3502, 3497, 3488} \[ -\frac {256 i \sqrt {a+i a \tan (c+d x)}}{385 a^2 d e^2 \sqrt {e \sec (c+d x)}}-\frac {96 i \sqrt {a+i a \tan (c+d x)}}{385 a^2 d (e \sec (c+d x))^{5/2}}+\frac {128 i}{385 a d e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {16 i}{77 a d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

((2*I)/11)/(d*(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^(3/2)) + ((16*I)/77)/(a*d*(e*Sec[c + d*x])^(5/2)*S
qrt[a + I*a*Tan[c + d*x]]) + ((128*I)/385)/(a*d*e^2*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((96*I
)/385)*Sqrt[a + I*a*Tan[c + d*x]])/(a^2*d*(e*Sec[c + d*x])^(5/2)) - (((256*I)/385)*Sqrt[a + I*a*Tan[c + d*x]])
/(a^2*d*e^2*Sqrt[e*Sec[c + d*x]])

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]

Rule 3497

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] + Dist[(a*(m + n))/(m*d^2), Int[(d*Sec[e + f*x])^(m + 2)*(
a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && LtQ[m, -
1] && IntegersQ[2*m, 2*n]

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]

Rubi steps

\begin {align*} \int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {2 i}{11 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^{3/2}}+\frac {8 \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx}{11 a}\\ &=\frac {2 i}{11 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^{3/2}}+\frac {16 i}{77 a d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}+\frac {48 \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx}{77 a^2}\\ &=\frac {2 i}{11 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^{3/2}}+\frac {16 i}{77 a d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}-\frac {96 i \sqrt {a+i a \tan (c+d x)}}{385 a^2 d (e \sec (c+d x))^{5/2}}+\frac {192 \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{385 a e^2}\\ &=\frac {2 i}{11 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^{3/2}}+\frac {16 i}{77 a d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}+\frac {128 i}{385 a d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {96 i \sqrt {a+i a \tan (c+d x)}}{385 a^2 d (e \sec (c+d x))^{5/2}}+\frac {128 \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}} \, dx}{385 a^2 e^2}\\ &=\frac {2 i}{11 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^{3/2}}+\frac {16 i}{77 a d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}+\frac {128 i}{385 a d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {96 i \sqrt {a+i a \tan (c+d x)}}{385 a^2 d (e \sec (c+d x))^{5/2}}-\frac {256 i \sqrt {a+i a \tan (c+d x)}}{385 a^2 d e^2 \sqrt {e \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 100, normalized size = 0.48 \[ -\frac {(e \sec (c+d x))^{3/2} (880 i \sin (2 (c+d x))+56 i \sin (4 (c+d x))+660 \cos (2 (c+d x))+21 \cos (4 (c+d x))-385)}{1540 a d e^4 (\tan (c+d x)-i) \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

-1/1540*((e*Sec[c + d*x])^(3/2)*(-385 + 660*Cos[2*(c + d*x)] + 21*Cos[4*(c + d*x)] + (880*I)*Sin[2*(c + d*x)]
+ (56*I)*Sin[4*(c + d*x)]))/(a*d*e^4*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [A]  time = 0.66, size = 111, normalized size = 0.53 \[ \frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-77 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 1617 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 770 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 990 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 255 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 35 i\right )} e^{\left (-\frac {11}{2} i \, d x - \frac {11}{2} i \, c\right )}}{3080 \, a^{2} d e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/3080*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^(10*I*d*x + 10*I*c) - 1617
*I*e^(8*I*d*x + 8*I*c) - 770*I*e^(6*I*d*x + 6*I*c) + 990*I*e^(4*I*d*x + 4*I*c) + 255*I*e^(2*I*d*x + 2*I*c) + 3
5*I)*e^(-11/2*I*d*x - 11/2*I*c)/(a^2*d*e^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((e*sec(d*x + c))^(5/2)*(I*a*tan(d*x + c) + a)^(3/2)), x)

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maple [A]  time = 1.17, size = 142, normalized size = 0.68 \[ \frac {2 \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (70 i \left (\cos ^{6}\left (d x +c \right )\right )+70 \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )+5 i \left (\cos ^{4}\left (d x +c \right )\right )+40 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+16 i \left (\cos ^{2}\left (d x +c \right )\right )+64 \cos \left (d x +c \right ) \sin \left (d x +c \right )-128 i\right )}{385 d \,e^{5} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

2/385/d*cos(d*x+c)^3*(e/cos(d*x+c))^(5/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(70*I*cos(d*x+c)^6+70
*cos(d*x+c)^5*sin(d*x+c)+5*I*cos(d*x+c)^4+40*cos(d*x+c)^3*sin(d*x+c)+16*I*cos(d*x+c)^2+64*cos(d*x+c)*sin(d*x+c
)-128*I)/e^5/a^2

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maxima [A]  time = 1.16, size = 226, normalized size = 1.08 \[ \frac {35 i \, \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 220 i \, \cos \left (\frac {7}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) - 77 i \, \cos \left (\frac {5}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 770 i \, \cos \left (\frac {3}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) - 1540 i \, \cos \left (\frac {1}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 35 \, \sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 220 \, \sin \left (\frac {7}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 77 \, \sin \left (\frac {5}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 770 \, \sin \left (\frac {3}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 1540 \, \sin \left (\frac {1}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right )}{3080 \, a^{\frac {3}{2}} d e^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/3080*(35*I*cos(11/2*d*x + 11/2*c) + 220*I*cos(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))
- 77*I*cos(5/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 770*I*cos(3/11*arctan2(sin(11/2*d*x
+ 11/2*c), cos(11/2*d*x + 11/2*c))) - 1540*I*cos(1/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))
) + 35*sin(11/2*d*x + 11/2*c) + 220*sin(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 77*sin
(5/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 770*sin(3/11*arctan2(sin(11/2*d*x + 11/2*c),
cos(11/2*d*x + 11/2*c))) + 1540*sin(1/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))))/(a^(3/2)*d*
e^(5/2))

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mupad [B]  time = 4.57, size = 127, normalized size = 0.61 \[ \frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (2310\,\sin \left (c+d\,x\right )+297\,\sin \left (3\,c+3\,d\,x\right )+35\,\sin \left (5\,c+5\,d\,x\right )-\cos \left (c+d\,x\right )\,770{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,143{}\mathrm {i}+\cos \left (5\,c+5\,d\,x\right )\,35{}\mathrm {i}\right )}{3080\,a\,d\,e^3\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e/cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i)^(3/2)),x)

[Out]

((e/cos(c + d*x))^(1/2)*(2310*sin(c + d*x) - cos(c + d*x)*770i + cos(3*c + 3*d*x)*143i + cos(5*c + 5*d*x)*35i
+ 297*sin(3*c + 3*d*x) + 35*sin(5*c + 5*d*x)))/(3080*a*d*e^3*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))
/(cos(2*c + 2*d*x) + 1))^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))**(5/2)/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

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