3.260 \(\int \frac{(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=163 \[ -\frac{2 e^3 \sin (c+d x)}{77 a^4 d \sqrt{e \sec (c+d x)}}-\frac{4 i e^4}{77 d \left (a^4+i a^4 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}-\frac{2 e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{77 a^4 d}+\frac{4 i e^2 \sqrt{e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3} \]

[Out]

(-2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(77*a^4*d) - (2*e^3*Sin[c + d*x])/(
77*a^4*d*Sqrt[e*Sec[c + d*x]]) + (((4*I)/11)*e^2*Sqrt[e*Sec[c + d*x]])/(a*d*(a + I*a*Tan[c + d*x])^3) - (((4*I
)/77)*e^4)/(d*(e*Sec[c + d*x])^(3/2)*(a^4 + I*a^4*Tan[c + d*x]))

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Rubi [A]  time = 0.143827, antiderivative size = 163, normalized size of antiderivative = 1., 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, 3769, 3771, 2641} \[ -\frac{2 e^3 \sin (c+d x)}{77 a^4 d \sqrt{e \sec (c+d x)}}-\frac{4 i e^4}{77 d \left (a^4+i a^4 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}-\frac{2 e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{77 a^4 d}+\frac{4 i e^2 \sqrt{e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(5/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(77*a^4*d) - (2*e^3*Sin[c + d*x])/(
77*a^4*d*Sqrt[e*Sec[c + d*x]]) + (((4*I)/11)*e^2*Sqrt[e*Sec[c + d*x]])/(a*d*(a + I*a*Tan[c + d*x])^3) - (((4*I
)/77)*e^4)/(d*(e*Sec[c + d*x])^(3/2)*(a^4 + I*a^4*Tan[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 3769

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 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]

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]

Rubi steps

\begin{align*} \int \frac{(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx &=\frac{4 i e^2 \sqrt{e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac{e^2 \int \frac{\sqrt{e \sec (c+d x)}}{(a+i a \tan (c+d x))^2} \, dx}{11 a^2}\\ &=\frac{4 i e^2 \sqrt{e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac{4 i e^4}{77 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}-\frac{\left (3 e^4\right ) \int \frac{1}{(e \sec (c+d x))^{3/2}} \, dx}{77 a^4}\\ &=-\frac{2 e^3 \sin (c+d x)}{77 a^4 d \sqrt{e \sec (c+d x)}}+\frac{4 i e^2 \sqrt{e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac{4 i e^4}{77 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}-\frac{e^2 \int \sqrt{e \sec (c+d x)} \, dx}{77 a^4}\\ &=-\frac{2 e^3 \sin (c+d x)}{77 a^4 d \sqrt{e \sec (c+d x)}}+\frac{4 i e^2 \sqrt{e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac{4 i e^4}{77 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}-\frac{\left (e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{77 a^4}\\ &=-\frac{2 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)}}{77 a^4 d}-\frac{2 e^3 \sin (c+d x)}{77 a^4 d \sqrt{e \sec (c+d x)}}+\frac{4 i e^2 \sqrt{e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac{4 i e^4}{77 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.556159, size = 144, normalized size = 0.88 \[ \frac{\sec ^2(c+d x) (e \sec (c+d x))^{5/2} (\cos (c+d x)+i \sin (c+d x)) \left (3 \sin (c+d x)+3 \sin (3 (c+d x))+37 i \cos (c+d x)+11 i \cos (3 (c+d x))-4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))\right )}{154 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(5/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(Sec[c + d*x]^2*(e*Sec[c + d*x])^(5/2)*(Cos[c + d*x] + I*Sin[c + d*x])*((37*I)*Cos[c + d*x] + (11*I)*Cos[3*(c
+ d*x)] + 3*Sin[c + d*x] - 4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x
)]) + 3*Sin[3*(c + d*x)]))/(154*a^4*d*(-I + Tan[c + d*x])^4)

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Maple [A]  time = 0.329, size = 216, normalized size = 1.3 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{77\,{a}^{4}d} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{5}{2}}} \left ( 56\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+56\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) -44\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}-i\cos \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) -i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) -16\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) } \end{align*}

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))^4,x)

[Out]

2/77/a^4/d*(e/cos(d*x+c))^(5/2)*cos(d*x+c)^2*(56*I*cos(d*x+c)^6+56*cos(d*x+c)^5*sin(d*x+c)-44*I*cos(d*x+c)^4-I
*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I
)-I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)-16*cos
(d*x+c)^3*sin(d*x+c)-cos(d*x+c)*sin(d*x+c))

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

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))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (154 \, a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )}{\rm integral}\left (\frac{i \, \sqrt{2} 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 )}}{77 \, a^{4} d}, x\right ) + \sqrt{2}{\left (4 i \, e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 17 i \, e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 20 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i \, 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 )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{154 \, a^{4} d} \end{align*}

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))^4,x, algorithm="fricas")

[Out]

1/154*(154*a^4*d*e^(6*I*d*x + 6*I*c)*integral(1/77*I*sqrt(2)*e^2*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)*(4*I*e^2*e^(6*I*d*x + 6*I*c) + 17*I*e^2*e^(4*I*d*x + 4*I*c) + 20*I*e^2*e^(
2*I*d*x + 2*I*c) + 7*I*e^2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-6*I*d*x - 6*I*c)/(a
^4*d)

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

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))**4,x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sec \left (d x + c\right )\right )^{\frac{5}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}\,{d x} \end{align*}

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))^4,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(5/2)/(I*a*tan(d*x + c) + a)^4, x)