3.3.12 \(\int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx\) [212]

Optimal. Leaf size=186 \[ \frac {2 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{11 d e^8}+\frac {6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac {2 a^3 \sin (c+d x)}{11 d e^7 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}} \]

[Out]

6/55*a^3*sin(d*x+c)/d/e^5/(e*sec(d*x+c))^(5/2)+2/11*a^3*sin(d*x+c)/d/e^7/(e*sec(d*x+c))^(1/2)+2/11*a^3*(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/e^8-2/15*I*(a+I*a*tan(d*x+c))^3/d/(e*sec(d*x+c))^(15/2)-12/55*I*(a^3+I*a^3*tan(d*x+c))/d/e^2/(e*sec(d
*x+c))^(11/2)

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Rubi [A]
time = 0.15, antiderivative size = 186, 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 = {3578, 3577, 3854, 3856, 2720} \begin {gather*} \frac {2 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{11 d e^8}+\frac {2 a^3 \sin (c+d x)}{11 d e^7 \sqrt {e \sec (c+d x)}}+\frac {6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(15/2),x]

[Out]

(2*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(11*d*e^8) + (6*a^3*Sin[c + d*x])/(5
5*d*e^5*(e*Sec[c + d*x])^(5/2)) + (2*a^3*Sin[c + d*x])/(11*d*e^7*Sqrt[e*Sec[c + d*x]]) - (((2*I)/15)*(a + I*a*
Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(15/2)) - (((12*I)/55)*(a^3 + I*a^3*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x]
)^(11/2))

Rule 2720

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

Rule 3577

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(d
*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] - Dist[b^2*((m + 2*n - 2)/(d^2*m)), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3578

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*S
ec[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 3854

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

Rule 3856

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 \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx &=-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}+\frac {(3 a) \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{11/2}} \, dx}{5 e^2}\\ &=-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac {\left (21 a^3\right ) \int \frac {1}{(e \sec (c+d x))^{7/2}} \, dx}{55 e^4}\\ &=\frac {6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac {\left (3 a^3\right ) \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{11 e^6}\\ &=\frac {6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac {2 a^3 \sin (c+d x)}{11 d e^7 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac {a^3 \int \sqrt {e \sec (c+d x)} \, dx}{11 e^8}\\ &=\frac {6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac {2 a^3 \sin (c+d x)}{11 d e^7 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac {\left (a^3 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{11 e^8}\\ &=\frac {2 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{11 d e^8}+\frac {6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac {2 a^3 \sin (c+d x)}{11 d e^7 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}\\ \end {align*}

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Mathematica [A]
time = 1.71, size = 170, normalized size = 0.91 \begin {gather*} \frac {a^3 \sqrt {e \sec (c+d x)} \left (-332 i \cos (c+d x)-154 i \cos (3 (c+d x))+22 i \cos (5 (c+d x))-114 \sin (c+d x)+240 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (3 (c+d x))-i \sin (3 (c+d x)))-81 \sin (3 (c+d x))+33 \sin (5 (c+d x))\right ) (\cos (3 (c+2 d x))+i \sin (3 (c+2 d x)))}{1320 d e^8 (\cos (d x)+i \sin (d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(15/2),x]

[Out]

(a^3*Sqrt[e*Sec[c + d*x]]*((-332*I)*Cos[c + d*x] - (154*I)*Cos[3*(c + d*x)] + (22*I)*Cos[5*(c + d*x)] - 114*Si
n[c + d*x] + 240*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[3*(c + d*x)] - I*Sin[3*(c + d*x)]) - 81*Sin
[3*(c + d*x)] + 33*Sin[5*(c + d*x)])*(Cos[3*(c + 2*d*x)] + I*Sin[3*(c + 2*d*x)]))/(1320*d*e^8*(Cos[d*x] + I*Si
n[d*x])^3)

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Maple [A]
time = 0.70, size = 232, normalized size = 1.25

method	result	size
default	\(\frac {2 a^{3} \left (-44 i \left (\cos ^{8}\left (d x +c \right )\right )+44 \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )+15 i \left (\cos ^{6}\left (d x +c \right )\right )+7 \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )+15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+9 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+15 \sin \left (d x +c \right ) \cos \left (d x +c \right )\right )}{165 d \cos \left (d x +c \right )^{8} \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {15}{2}}}\)	\(232\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(15/2),x,method=_RETURNVERBOSE)

[Out]

2/165*a^3/d*(-44*I*cos(d*x+c)^8+44*sin(d*x+c)*cos(d*x+c)^7+15*I*cos(d*x+c)^6+7*sin(d*x+c)*cos(d*x+c)^5+15*I*(1
/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+
9*sin(d*x+c)*cos(d*x+c)^3+15*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(
d*x+c))/sin(d*x+c),I)+15*sin(d*x+c)*cos(d*x+c))/cos(d*x+c)^8/(e/cos(d*x+c))^(15/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(15/2),x, algorithm="maxima")

[Out]

e^(-15/2)*integrate((I*a*tan(d*x + c) + a)^3/sec(d*x + c)^(15/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 147, normalized size = 0.79 \begin {gather*} \frac {{\left (-480 i \, \sqrt {2} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \frac {\sqrt {2} {\left (-11 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 73 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 218 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 446 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 235 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 55 i \, a^{3}\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c - \frac {15}{2}\right )}}{2640 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(15/2),x, algorithm="fricas")

[Out]

1/2640*(-480*I*sqrt(2)*a^3*e^(2*I*d*x + 2*I*c)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(-11*I*a^
3*e^(10*I*d*x + 10*I*c) - 73*I*a^3*e^(8*I*d*x + 8*I*c) - 218*I*a^3*e^(6*I*d*x + 6*I*c) - 446*I*a^3*e^(4*I*d*x
+ 4*I*c) - 235*I*a^3*e^(2*I*d*x + 2*I*c) + 55*I*a^3)*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1))*e^
(-2*I*d*x - 2*I*c - 15/2)/d

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**3/(e*sec(d*x+c))**(15/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(15/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^3*e^(-15/2)/sec(d*x + c)^(15/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{15/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(15/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(15/2), x)

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