3.2.91 \(\int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{5/2}} \, dx\) [191]

Optimal. Leaf size=96 \[ -\frac {2 i a}{5 d (e \sec (c+d x))^{5/2}}+\frac {6 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 a \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}} \]

[Out]

-2/5*I*a/d/(e*sec(d*x+c))^(5/2)+2/5*a*sin(d*x+c)/d/e/(e*sec(d*x+c))^(3/2)+6/5*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/c
os(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/e^2/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3567, 3854, 3856, 2719} \begin {gather*} \frac {6 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {2 i a}{5 d (e \sec (c+d x))^{5/2}}+\frac {2 a \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2719

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

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.78, size = 99, normalized size = 1.03 \begin {gather*} -\frac {a \left (2+2 \cos (2 (c+d x))-2 \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-3 i \sin (2 (c+d x))\right ) (-i+\tan (c+d x))}{5 d e^2 \sqrt {e \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(a*(2 + 2*Cos[2*(c + d*x)] - 2*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(
c + d*x))] - (3*I)*Sin[2*(c + d*x)])*(-I + Tan[c + d*x]))/(d*e^2*Sqrt[e*Sec[c + d*x]])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 338 vs. \(2 (107 ) = 214\).
time = 0.46, size = 339, normalized size = 3.53

method	result	size
risch	\(-\frac {i \left ({\mathrm e}^{2 i \left (d x +c \right )}+7\right ) a \sqrt {2}}{10 d \,e^{2} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {3 i \left (-\frac {2 \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}{e \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i \EllipticE \left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i \EllipticF \left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}}\right ) a \sqrt {2}\, \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{5 d \,e^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\)	\(320\)
default	\(-\frac {2 a \left (3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+\cos ^{4}\left (d x +c \right )+2 \left (\cos ^{2}\left (d x +c \right )\right )-3 \cos \left (d x +c \right )\right )}{5 d \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}\)	\(339\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))/(e*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/5*a/d*(3*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c
),I)*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)+3*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*Ell
ipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(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)+I*sin(d*x+c)*cos(d*x+c)^3+cos(d*x+c)^4+2*cos(d*x+c)^2-
3*cos(d*x+c))/(e/cos(d*x+c))^(5/2)/sin(d*x+c)/cos(d*x+c)^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 100, normalized size = 1.04 \begin {gather*} \frac {{\left (12 i \, \sqrt {2} a e^{\left (i \, d x + i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \frac {\sqrt {2} {\left (-i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, a\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 (-i \, d x - i \, c - \frac {5}{2}\right )}}{10 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(12*I*sqrt(2)*a*e^(I*d*x + I*c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))) + sqr
t(2)*(-I*a*e^(4*I*d*x + 4*I*c) + 4*I*a*e^(2*I*d*x + 2*I*c) + 5*I*a)*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x +
2*I*c) + 1))*e^(-I*d*x - I*c - 5/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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