3.3.42 \(\int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx\) [242]

Optimal. Leaf size=150 \[ \frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{33 a^2 d e^2}+\frac {2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac {10 \sin (c+d x)}{33 a^2 d e \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{11 d (e \sec (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

2/11*e*sin(d*x+c)/a^2/d/(e*sec(d*x+c))^(5/2)+10/33*sin(d*x+c)/a^2/d/e/(e*sec(d*x+c))^(1/2)+10/33*(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)
/a^2/d/e^2+4/11*I*e^2/d/(e*sec(d*x+c))^(7/2)/(a^2+I*a^2*tan(d*x+c))

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Rubi [A]
time = 0.10, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3581, 3854, 3856, 2720} \begin {gather*} \frac {4 i e^2}{11 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}+\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{33 a^2 d e^2}+\frac {2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac {10 \sin (c+d x)}{33 a^2 d e \sqrt {e \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(33*a^2*d*e^2) + (2*e*Sin[c + d*x])/(11
*a^2*d*(e*Sec[c + d*x])^(5/2)) + (10*Sin[c + d*x])/(33*a^2*d*e*Sqrt[e*Sec[c + d*x]]) + (((4*I)/11)*e^2)/(d*(e*
Sec[c + d*x])^(7/2)*(a^2 + I*a^2*Tan[c + d*x]))

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 3581

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*d^2*
(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + 2*n))), x] - Dist[d^2*((m - 2)/(b^2*(m + 2*n)
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && 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 {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx &=\frac {4 i e^2}{11 d (e \sec (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (7 e^2\right ) \int \frac {1}{(e \sec (c+d x))^{7/2}} \, dx}{11 a^2}\\ &=\frac {2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac {4 i e^2}{11 d (e \sec (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {5 \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{11 a^2}\\ &=\frac {2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac {10 \sin (c+d x)}{33 a^2 d e \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{11 d (e \sec (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {5 \int \sqrt {e \sec (c+d x)} \, dx}{33 a^2 e^2}\\ &=\frac {2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac {10 \sin (c+d x)}{33 a^2 d e \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{11 d (e \sec (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (5 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{33 a^2 e^2}\\ &=\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{33 a^2 d e^2}+\frac {2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac {10 \sin (c+d x)}{33 a^2 d e \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{11 d (e \sec (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.67, size = 134, normalized size = 0.89 \begin {gather*} -\frac {\sec ^4(c+d x) \left (28 i+24 i \cos (2 (c+d x))-4 i \cos (4 (c+d x))+40 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-6 \sin (2 (c+d x))+7 \sin (4 (c+d x))\right )}{132 a^2 d (e \sec (c+d x))^{3/2} (-i+\tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/132*(Sec[c + d*x]^4*(28*I + (24*I)*Cos[2*(c + d*x)] - (4*I)*Cos[4*(c + d*x)] + 40*Sqrt[Cos[c + d*x]]*Ellipt
icF[(c + d*x)/2, 2]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) - 6*Sin[2*(c + d*x)] + 7*Sin[4*(c + d*x)]))/(a^2*d
*(e*Sec[c + d*x])^(3/2)*(-I + Tan[c + d*x])^2)

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Maple [A]
time = 0.90, size = 234, normalized size = 1.56

method	result	size
default	\(\frac {2 \cos \left (d x +c \right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{2} \left (6 i \left (\cos ^{6}\left (d x +c \right )\right )+6 \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+3 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+5 \sin \left (d x +c \right ) \cos \left (d x +c \right )\right )}{33 a^{2} d \,e^{3} \sin \left (d x +c \right )^{4}}\)	\(234\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/33/a^2/d*cos(d*x+c)*(e/cos(d*x+c))^(3/2)*(-1+cos(d*x+c))^2*(1+cos(d*x+c))^2*(6*I*cos(d*x+c)^6+6*sin(d*x+c)*c
os(d*x+c)^5+5*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)+3*sin(d*x+c)*cos(d*x+c)^3+5*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*Ell
ipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*sin(d*x+c)*cos(d*x+c))/e^3/sin(d*x+c)^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 117, normalized size = 0.78 \begin {gather*} \frac {{\left (-80 i \, \sqrt {2} e^{\left (6 i \, d x + 6 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \frac {\sqrt {2} {\left (-11 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 30 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 56 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 18 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\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 (-6 i \, d x - 6 i \, c - \frac {3}{2}\right )}}{264 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/264*(-80*I*sqrt(2)*e^(6*I*d*x + 6*I*c)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(-11*I*e^(8*I*d
*x + 8*I*c) + 30*I*e^(6*I*d*x + 6*I*c) + 56*I*e^(4*I*d*x + 4*I*c) + 18*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(1/2*I*d
*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1))*e^(-6*I*d*x - 6*I*c - 3/2)/(a^2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (c + d x \right )} - 2 i \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan {\left (c + d x \right )} - \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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