3.3.60 \(\int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx\) [260]

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

[Out]

-2/77*e^3*sin(d*x+c)/a^4/d/(e*sec(d*x+c))^(1/2)-2/77*e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip
ticF(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^4/d+4/11*I*e^2*(e*sec(d*x+c))^(1/2)/a
/d/(a+I*a*tan(d*x+c))^3-4/77*I*e^4/d/(e*sec(d*x+c))^(3/2)/(a^4+I*a^4*tan(d*x+c))

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Rubi [A]
time = 0.11, 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 = {3581, 3854, 3856, 2720} \begin {gather*} -\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^3 \sin (c+d x)}{77 a^4 d \sqrt {e \sec (c+d x)}}-\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} \end {gather*}

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 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 {(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*}

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Mathematica [A]
time = 0.70, size = 144, normalized size = 0.88 \begin {gather*} \frac {\sec ^2(c+d x) (e \sec (c+d x))^{5/2} (\cos (c+d x)+i \sin (c+d x)) \left (37 i \cos (c+d x)+11 i \cos (3 (c+d x))+3 \sin (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)))+3 \sin (3 (c+d x))\right )}{154 a^4 d (-i+\tan (c+d x))^4} \end {gather*}

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 = 0.78, size = 244, normalized size = 1.50

method	result	size
default	\(\frac {2 \left (56 i \left (\cos ^{6}\left (d x +c \right )\right )+56 \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )-44 i \left (\cos ^{4}\left (d x +c \right )\right )-i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-16 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-\sin \left (d x +c \right ) \cos \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}{77 a^{4} d \sin \left (d x +c \right )^{4}}\)	\(244\)

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,method=_RETURNVERBOSE)

[Out]

2/77/a^4/d*(56*I*cos(d*x+c)^6+56*sin(d*x+c)*cos(d*x+c)^5-44*I*cos(d*x+c)^4-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)-16*sin(d*x+c)*cos(d*x+c)^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)*
cos(d*x+c))*cos(d*x+c)^2*(e/cos(d*x+c))^(5/2)*(-1+cos(d*x+c))^2*(1+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))^4,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 = 112, normalized size = 0.69 \begin {gather*} \frac {{\left (4 i \, \sqrt {2} e^{\left (6 i \, d x + 6 i \, c + \frac {5}{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \frac {\sqrt {2} {\left (7 i \, e^{\frac {5}{2}} + 4 i \, e^{\left (6 i \, d x + 6 i \, c + \frac {5}{2}\right )} + 17 i \, e^{\left (4 i \, d x + 4 i \, c + \frac {5}{2}\right )} + 20 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 (-6 i \, d x - 6 i \, c\right )}}{154 \, a^{4} 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))^4,x, algorithm="fricas")

[Out]

1/154*(4*I*sqrt(2)*e^(6*I*d*x + 6*I*c + 5/2)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(7*I*e^(5/2
) + 4*I*e^(6*I*d*x + 6*I*c + 5/2) + 17*I*e^(4*I*d*x + 4*I*c + 5/2) + 20*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^(-6*I*d*x - 6*I*c)/(a^4*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 ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \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))**4,x)

[Out]

Integral((e*sec(c + d*x))**(5/2)/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan(c + d*x)**2 + 4*I*tan(c + d*x)
+ 1), x)/a**4

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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))^4,x, algorithm="giac")

[Out]

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

[Out]

int((e/cos(c + d*x))^(5/2)/(a + a*tan(c + d*x)*1i)^4, x)

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