3.2.99 \(\int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{9/2}} \, dx\) [199]

Optimal. Leaf size=116 \[ \frac {2 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 a^2 \sin (c+d x)}{9 d e^3 (e \sec (c+d x))^{3/2}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}} \]

[Out]

2/9*a^2*sin(d*x+c)/d/e^3/(e*sec(d*x+c))^(3/2)+2/3*a^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic
E(sin(1/2*d*x+1/2*c),2^(1/2))/d/e^4/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)-4/9*I*(a^2+I*a^2*tan(d*x+c))/d/(e*se
c(d*x+c))^(9/2)

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Rubi [A]
time = 0.07, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3577, 3854, 3856, 2719} \begin {gather*} \frac {2 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 a^2 \sin (c+d x)}{9 d e^3 (e \sec (c+d x))^{3/2}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^2/(e*Sec[c + d*x])^(9/2),x]

[Out]

(2*a^2*EllipticE[(c + d*x)/2, 2])/(3*d*e^4*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*a^2*Sin[c + d*x])/(9*
d*e^3*(e*Sec[c + d*x])^(3/2)) - (((4*I)/9)*(a^2 + I*a^2*Tan[c + d*x]))/(d*(e*Sec[c + d*x])^(9/2))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 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))^2}{(e \sec (c+d x))^{9/2}} \, dx &=-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}}+\frac {\left (5 a^2\right ) \int \frac {1}{(e \sec (c+d x))^{5/2}} \, dx}{9 e^2}\\ &=\frac {2 a^2 \sin (c+d x)}{9 d e^3 (e \sec (c+d x))^{3/2}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}}+\frac {a^2 \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{3 e^4}\\ &=\frac {2 a^2 \sin (c+d x)}{9 d e^3 (e \sec (c+d x))^{3/2}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}}+\frac {a^2 \int \sqrt {\cos (c+d x)} \, dx}{3 e^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=\frac {2 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 a^2 \sin (c+d x)}{9 d e^3 (e \sec (c+d x))^{3/2}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 1.76, size = 133, normalized size = 1.15 \begin {gather*} \frac {i a^2 \left (9-4 e^{2 i (c+d x)}-e^{4 i (c+d x)}-\frac {8 e^{2 i (c+d x)} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}\right )}{18 \sqrt {2} d e^4 \sqrt {\frac {e e^{i (c+d x)}}{1+e^{2 i (c+d x)}}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^2/(e*Sec[c + d*x])^(9/2),x]

[Out]

((I/18)*a^2*(9 - 4*E^((2*I)*(c + d*x)) - E^((4*I)*(c + d*x)) - (8*E^((2*I)*(c + d*x))*Hypergeometric2F1[1/2, 3
/4, 7/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))]))/(Sqrt[2]*d*e^4*Sqrt[(e*E^(I*(c + d*x)))/(1 + E
^((2*I)*(c + d*x)))])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (126 ) = 252\).
time = 0.53, size = 351, normalized size = 3.03

method	result	size
risch	\(-\frac {i \left ({\mathrm e}^{4 i \left (d x +c \right )}+4 \,{\mathrm e}^{2 i \left (d x +c \right )}+15\right ) a^{2} \sqrt {2}}{36 d \,e^{4} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i \left (-\frac {2 \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}{e \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i \EllipticE \left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i \EllipticF \left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}}\right ) a^{2} \sqrt {2}\, \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{3 d \,e^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\)	\(337\)
default	\(\frac {2 a^{2} \left (-2 i \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-2 \left (\cos ^{6}\left (d x +c \right )\right )+3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+\cos ^{4}\left (d x +c \right )-2 \left (\cos ^{2}\left (d x +c \right )\right )+3 \cos \left (d x +c \right )\right )}{9 d \cos \left (d x +c \right )^{5} \sin \left (d x +c \right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}}}\)	\(351\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(9/2),x,method=_RETURNVERBOSE)

[Out]

2/9*a^2/d*(-2*I*cos(d*x+c)^5*sin(d*x+c)+3*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*Ellipti
cF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)-3*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)-2*cos(d*x+c)^6+3*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)*sin(d*x+c)-3*I*(1/(1+cos(d
*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+cos(d*x+c
)^4-2*cos(d*x+c)^2+3*cos(d*x+c))/cos(d*x+c)^5/sin(d*x+c)/(e/cos(d*x+c))^(9/2)

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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))^2/(e*sec(d*x+c))^(9/2),x, algorithm="maxima")

[Out]

e^(-9/2)*integrate((I*a*tan(d*x + c) + a)^2/sec(d*x + c)^(9/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 122, normalized size = 1.05 \begin {gather*} \frac {{\left (24 i \, \sqrt {2} a^{2} e^{\left (i \, d x + i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \frac {\sqrt {2} {\left (-i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 5 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 9 i \, a^{2}\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 (-i \, d x - i \, c - \frac {9}{2}\right )}}{36 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(9/2),x, algorithm="fricas")

[Out]

1/36*(24*I*sqrt(2)*a^2*e^(I*d*x + I*c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))) + s
qrt(2)*(-I*a^2*e^(6*I*d*x + 6*I*c) - 5*I*a^2*e^(4*I*d*x + 4*I*c) + 5*I*a^2*e^(2*I*d*x + 2*I*c) + 9*I*a^2)*e^(1
/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1))*e^(-I*d*x - I*c - 9/2)/d

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**2/(e*sec(d*x+c))**(9/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3879 deep

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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))^2/(e*sec(d*x+c))^(9/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^2*e^(-9/2)/sec(d*x + c)^(9/2), 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 )}^2}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{9/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^2/(e/cos(c + d*x))^(9/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^2/(e/cos(c + d*x))^(9/2), x)

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