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

Optimal. Leaf size=175 \[ \frac {26 a^3 e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{21 d}+\frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{21 d}+\frac {26 i \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}{63 d}+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d} \]

[Out]

26/35*I*a^3*(e*sec(d*x+c))^(5/2)/d+26/21*a^3*e*(e*sec(d*x+c))^(3/2)*sin(d*x+c)/d+26/21*a^3*e^2*(cos(1/2*d*x+1/
2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(e*sec(d*x+c))^(1/2)/d
+2/9*I*a*(e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^2/d+26/63*I*(e*sec(d*x+c))^(5/2)*(a^3+I*a^3*tan(d*x+c))/d

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Rubi [A]  time = 0.19, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3498, 3486, 3768, 3771, 2641} \[ \frac {26 a^3 e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{21 d}+\frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{21 d}+\frac {26 i \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}{63 d}+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(26*a^3*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(21*d) + (((26*I)/35)*a^3*(e*Se
c[c + d*x])^(5/2))/d + (26*a^3*e*(e*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(21*d) + (((2*I)/9)*a*(e*Sec[c + d*x])^(
5/2)*(a + I*a*Tan[c + d*x])^2)/d + (((26*I)/63)*(e*Sec[c + d*x])^(5/2)*(a^3 + I*a^3*Tan[c + d*x]))/d

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3486

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

Rule 3498

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 - 1))/(f*(m + n - 1)), x] + Dist[(a*(m + 2*n - 2))/(m + n - 1), 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] &&
GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx &=\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {1}{9} (13 a) \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2 \, dx\\ &=\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {1}{7} \left (13 a^2\right ) \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx\\ &=\frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {1}{7} \left (13 a^3\right ) \int (e \sec (c+d x))^{5/2} \, dx\\ &=\frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e (e \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {1}{21} \left (13 a^3 e^2\right ) \int \sqrt {e \sec (c+d x)} \, dx\\ &=\frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e (e \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {1}{21} \left (13 a^3 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {26 a^3 e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{21 d}+\frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e (e \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}\\ \end {align*}

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Mathematica [A]  time = 1.93, size = 89, normalized size = 0.51 \[ \frac {a^3 \sec ^2(c+d x) (e \sec (c+d x))^{5/2} \left (-150 \sin (2 (c+d x))+195 \sin (4 (c+d x))+1008 i \cos (2 (c+d x))+1560 \cos ^{\frac {9}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+728 i\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[c + d*x]^2*(e*Sec[c + d*x])^(5/2)*(728*I + (1008*I)*Cos[2*(c + d*x)] + 1560*Cos[c + d*x]^(9/2)*Ellipt
icF[(c + d*x)/2, 2] - 150*Sin[2*(c + d*x)] + 195*Sin[4*(c + d*x)]))/(1260*d)

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (-390 i \, a^{3} e^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 2316 i \, a^{3} e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 2912 i \, a^{3} e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 1716 i \, a^{3} e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 390 i \, a^{3} e^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 315 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} {\rm integral}\left (-\frac {13 i \, \sqrt {2} a^{3} e^{2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{21 \, d}, x\right )}{315 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(sqrt(2)*(-390*I*a^3*e^2*e^(8*I*d*x + 8*I*c) + 2316*I*a^3*e^2*e^(6*I*d*x + 6*I*c) + 2912*I*a^3*e^2*e^(4*
I*d*x + 4*I*c) + 1716*I*a^3*e^2*e^(2*I*d*x + 2*I*c) + 390*I*a^3*e^2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*
I*d*x + 1/2*I*c) + 315*(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I
*d*x + 2*I*c) + d)*integral(-13/21*I*sqrt(2)*a^3*e^2*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I*c
)/d, x))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c)
+ d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.03, size = 229, normalized size = 1.31 \[ \frac {2 a^{3} \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (195 i \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+195 i \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+195 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+252 i \left (\cos ^{2}\left (d x +c \right )\right )-135 \cos \left (d x +c \right ) \sin \left (d x +c \right )-35 i\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{315 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*a^3/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(195*I*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)^5*(
1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+195*I*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(
d*x+c)^4*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+195*cos(d*x+c)^3*sin(d*x+c)+252*I*cos(d*x+
c)^2-135*cos(d*x+c)*sin(d*x+c)-35*I)*(e/cos(d*x+c))^(5/2)/cos(d*x+c)^2/sin(d*x+c)^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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