3.257 \(\int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=163 \[ -\frac {42 e^6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {42 e^5 \sin (c+d x) \sqrt {e \sec (c+d x)}}{5 a^4 d}-\frac {28 i e^4 (e \sec (c+d x))^{3/2}}{5 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3} \]

[Out]

-42/5*e^6*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^4/d/cos(d*x+
c)^(1/2)/(e*sec(d*x+c))^(1/2)+42/5*e^5*sin(d*x+c)*(e*sec(d*x+c))^(1/2)/a^4/d+4/5*I*e^2*(e*sec(d*x+c))^(7/2)/a/
d/(a+I*a*tan(d*x+c))^3-28/5*I*e^4*(e*sec(d*x+c))^(3/2)/d/(a^4+I*a^4*tan(d*x+c))

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Rubi [A]  time = 0.15, antiderivative size = 163, 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 = {3500, 3768, 3771, 2639} \[ \frac {42 e^5 \sin (c+d x) \sqrt {e \sec (c+d x)}}{5 a^4 d}-\frac {28 i e^4 (e \sec (c+d x))^{3/2}}{5 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {42 e^6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(11/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-42*e^6*EllipticE[(c + d*x)/2, 2])/(5*a^4*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (42*e^5*Sqrt[e*Sec[c +
d*x]]*Sin[c + d*x])/(5*a^4*d) + (((4*I)/5)*e^2*(e*Sec[c + d*x])^(7/2))/(a*d*(a + I*a*Tan[c + d*x])^3) - (((28
*I)/5)*e^4*(e*Sec[c + d*x])^(3/2))/(d*(a^4 + I*a^4*Tan[c + d*x]))

Rule 2639

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

Rule 3500

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 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 \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^4} \, dx &=\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {\left (7 e^2\right ) \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx}{5 a^2}\\ &=\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {28 i e^4 (e \sec (c+d x))^{3/2}}{5 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\left (21 e^4\right ) \int (e \sec (c+d x))^{3/2} \, dx}{5 a^4}\\ &=\frac {42 e^5 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {28 i e^4 (e \sec (c+d x))^{3/2}}{5 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {\left (21 e^6\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{5 a^4}\\ &=\frac {42 e^5 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {28 i e^4 (e \sec (c+d x))^{3/2}}{5 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {\left (21 e^6\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 a^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=-\frac {42 e^6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {42 e^5 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {28 i e^4 (e \sec (c+d x))^{3/2}}{5 d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [C]  time = 0.63, size = 106, normalized size = 0.65 \[ -\frac {2 i e^5 e^{-3 i (c+d x)} \left (21 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )-7 e^{2 i (c+d x)}-2\right ) \sqrt {e \sec (c+d x)}}{5 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(11/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(((-2*I)/5)*e^5*(-2 - 7*E^((2*I)*(c + d*x)) + 21*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeomet
ric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])*Sqrt[e*Sec[c + d*x]])/(a^4*d*E^((3*I)*(c + d*x)))

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fricas [F]  time = 1.11, size = 0, normalized size = 0.00 \[ \frac {{\left (5 \, a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} {\rm integral}\left (\frac {21 i \, \sqrt {2} e^{5} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{5 \, a^{4} d}, x\right ) + \sqrt {2} {\left (-42 i \, e^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 28 i \, e^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i \, e^{5}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{5 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/5*(5*a^4*d*e^(3*I*d*x + 3*I*c)*integral(21/5*I*sqrt(2)*e^5*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x +
1/2*I*c)/(a^4*d), x) + sqrt(2)*(-42*I*e^5*e^(4*I*d*x + 4*I*c) - 28*I*e^5*e^(2*I*d*x + 2*I*c) + 4*I*e^5)*sqrt(e
/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-3*I*d*x - 3*I*c)/(a^4*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(11/2)/(I*a*tan(d*x + c) + a)^4, x)

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maple [B]  time = 1.44, size = 379, normalized size = 2.33 \[ -\frac {2 \left (21 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-21 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-8 i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+21 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-21 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+8 \left (\cos ^{4}\left (d x +c \right )\right )+20 i \cos \left (d x +c \right ) \sin \left (d x +c \right )-24 \left (\cos ^{2}\left (d x +c \right )\right )+21 \cos \left (d x +c \right )-5\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {11}{2}} \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2}}{5 a^{4} d \sin \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^4,x)

[Out]

-2/5/a^4/d*(21*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)*EllipticF(I*
(-1+cos(d*x+c))/sin(d*x+c),I)-21*I*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)-8*I*cos(d*x+c)^3*sin(d*x+c)+21*I*(1/(1+cos(d*x+c)))^(1/2)*(cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-21*I*EllipticE(I*(-1+cos(d*
x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+8*cos(d*x+c)^4+20*I*
cos(d*x+c)*sin(d*x+c)-24*cos(d*x+c)^2+21*cos(d*x+c)-5)*cos(d*x+c)^5*(e/cos(d*x+c))^(11/2)*(1+cos(d*x+c))^2*(-1
+cos(d*x+c))^2/sin(d*x+c)^5

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{11/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(11/2)/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

int((e/cos(c + d*x))^(11/2)/(a + a*tan(c + d*x)*1i)^4, 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))**(11/2)/(a+I*a*tan(d*x+c))**4,x)

[Out]

Timed out

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