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

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

[Out]

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

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Rubi [A]  time = 0.103749, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3515, 3486, 3768, 3771, 2641} \[ \frac{2 i a}{5 d (e \cos (c+d x))^{5/2}}+\frac{2 a \cos ^{\frac{5}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d (e \cos (c+d x))^{5/2}}+\frac{2 a \sin (c+d x) \cos (c+d x)}{3 d (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3515

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

Rule 3486

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 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]

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

Mathematica [A]  time = 0.495286, size = 57, normalized size = 0.59 \[ \frac{a \left (5 \sin (2 (c+d x))+10 \cos ^{\frac{5}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+6 i\right )}{15 d (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(6*I + 10*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 5*Sin[2*(c + d*x)]))/(15*d*(e*Cos[c + d*x])^(5/2))

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Maple [B]  time = 4.727, size = 283, normalized size = 3. \begin{align*}{\frac{2\,a}{15\,{e}^{2}d} \left ( -20\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+20\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-20\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) -5\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +10\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) +3\,i\sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) \left ( 4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{-1} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/
e^2*(-20*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*s
in(1/2*d*x+1/2*c)^4+20*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2
*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-20*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2
*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+10*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c
)+3*I*sin(1/2*d*x+1/2*c))*a/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{\frac{1}{2}}{\left (-20 i \, a e^{\left (5 i \, d x + 5 i \, c\right )} + 48 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} + 20 i \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} e^{\left (-\frac{1}{2} i \, d x - \frac{1}{2} i \, c\right )} + 15 \,{\left (d e^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3}\right )}{\rm integral}\left (-\frac{2 i \, \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} a e^{\left (-\frac{1}{2} i \, d x - \frac{1}{2} i \, c\right )}}{3 \,{\left (d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3}\right )}}, x\right )}{15 \,{\left (d e^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(sqrt(1/2)*(-20*I*a*e^(5*I*d*x + 5*I*c) + 48*I*a*e^(3*I*d*x + 3*I*c) + 20*I*a*e^(I*d*x + I*c))*sqrt(e*e^(
2*I*d*x + 2*I*c) + e)*e^(-1/2*I*d*x - 1/2*I*c) + 15*(d*e^3*e^(6*I*d*x + 6*I*c) + 3*d*e^3*e^(4*I*d*x + 4*I*c) +
3*d*e^3*e^(2*I*d*x + 2*I*c) + d*e^3)*integral(-2/3*I*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*a*e^(-1/2*I*d*
x - 1/2*I*c)/(d*e^3*e^(2*I*d*x + 2*I*c) + d*e^3), x))/(d*e^3*e^(6*I*d*x + 6*I*c) + 3*d*e^3*e^(4*I*d*x + 4*I*c)
+ 3*d*e^3*e^(2*I*d*x + 2*I*c) + d*e^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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