3.3.38 \(\int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx\) [238]

Optimal. Leaf size=90 \[ -\frac {2 e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{3 a^2 d}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

-2/3*e^2*(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+4/3*I*e^2*(e*sec(d*x+c))^(1/2)/d/(a^2+I*a^2*tan(d*x+c))

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Rubi [A]
time = 0.06, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3581, 3856, 2720} \begin {gather*} -\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)}}{3 a^2 d}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(3*a^2*d) + (((4*I)/3)*e^2*Sqrt[e*S
ec[c + d*x]])/(d*(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 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 {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx &=\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {e^2 \int \sqrt {e \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{3 d \left (a^2+i a^2 \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}{3 a^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)}}{3 a^2 d}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.47, size = 101, normalized size = 1.12 \begin {gather*} \frac {2 (e \sec (c+d x))^{5/2} \left (-2 i \cos (c+d x)+\sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (c+d x)+i \sin (c+d x))\right ) (\cos (c+d x)+i \sin (c+d x))}{3 a^2 d (-i+\tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.71, size = 201, normalized size = 2.23

method	result	size
default	\(-\frac {2 \left (i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-2 i \left (\cos ^{2}\left (d x +c \right )\right )-2 \sin \left (d x +c \right ) \cos \left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{3 a^{2} d \sin \left (d x +c \right )^{4}}\)	\(201\)

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))^2,x,method=_RETURNVERBOSE)

[Out]

-2/3/a^2/d*(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)+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)-2*I*cos(d*x+c)^2-2*sin(d*x+c)*cos(d*x+c))*(e/cos(d*x+c))^(5/2)*(-1+cos(d*x+c))^2*(1+cos(d*x+c))^2*cos
(d*x+c)^2/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((e*sec(d*x+c))^(5/2)/(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 = 0.09, size = 88, normalized size = 0.98 \begin {gather*} -\frac {2 \, {\left (-i \, \sqrt {2} e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \frac {\sqrt {2} {\left (-i \, e^{\frac {5}{2}} - i \, e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\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\right )}}{3 \, a^{2} d} \end {gather*}

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))^2,x, algorithm="fricas")

[Out]

-2/3*(-I*sqrt(2)*e^(2*I*d*x + 2*I*c + 5/2)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(-I*e^(5/2) -
I*e^(2*I*d*x + 2*I*c + 5/2))*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)/(a^2
*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \end {gather*}

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))**2,x)

[Out]

-Integral((e*sec(c + d*x))**(5/2)/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), 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((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\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((e/cos(c + d*x))^(5/2)/(a + a*tan(c + d*x)*1i)^2,x)

[Out]

int((e/cos(c + d*x))^(5/2)/(a + a*tan(c + d*x)*1i)^2, x)

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