3.198 \(\int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=116 \[ \frac {2 a^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{7 d e^4}+\frac {2 a^2 \sin (c+d x)}{7 d e^3 \sqrt {e \sec (c+d x)}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}} \]

[Out]

2/7*a^2*sin(d*x+c)/d/e^3/(e*sec(d*x+c))^(1/2)+2/7*a^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic
F(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(e*sec(d*x+c))^(1/2)/d/e^4-4/7*I*(a^2+I*a^2*tan(d*x+c))/d/(e*se
c(d*x+c))^(7/2)

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Rubi [A]  time = 0.09, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3496, 3769, 3771, 2641} \[ \frac {2 a^2 \sin (c+d x)}{7 d e^3 \sqrt {e \sec (c+d x)}}+\frac {2 a^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{7 d e^4}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^2/(e*Sec[c + d*x])^(7/2),x]

[Out]

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

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]

Rule 3496

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && 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]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{7/2}} \, dx &=-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}+\frac {\left (3 a^2\right ) \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{7 e^2}\\ &=\frac {2 a^2 \sin (c+d x)}{7 d e^3 \sqrt {e \sec (c+d x)}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}+\frac {a^2 \int \sqrt {e \sec (c+d x)} \, dx}{7 e^4}\\ &=\frac {2 a^2 \sin (c+d x)}{7 d e^3 \sqrt {e \sec (c+d x)}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}+\frac {\left (a^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{7 e^4}\\ &=\frac {2 a^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{7 d e^4}+\frac {2 a^2 \sin (c+d x)}{7 d e^3 \sqrt {e \sec (c+d x)}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 1.04, size = 133, normalized size = 1.15 \[ \frac {a^2 \sqrt {e \sec (c+d x)} (\cos (2 (c+2 d x))+i \sin (2 (c+2 d x))) \left (-\sin (2 (c+d x))-2 i \cos (2 (c+d x))+2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (2 (c+d x))-i \sin (2 (c+d x)))-2 i\right )}{7 d e^4 (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^2/(e*Sec[c + d*x])^(7/2),x]

[Out]

(a^2*Sqrt[e*Sec[c + d*x]]*(-2*I - (2*I)*Cos[2*(c + d*x)] + 2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos
[2*(c + d*x)] - I*Sin[2*(c + d*x)]) - Sin[2*(c + d*x)])*(Cos[2*(c + 2*d*x)] + I*Sin[2*(c + 2*d*x)]))/(7*d*e^4*
(Cos[d*x] + I*Sin[d*x])^2)

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ \frac {14 \, d e^{4} {\rm integral}\left (-\frac {i \, \sqrt {2} a^{2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{7 \, d e^{4}}, x\right ) + \sqrt {2} {\left (-i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, a^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{14 \, d e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/14*(14*d*e^4*integral(-1/7*I*sqrt(2)*a^2*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I*c)/(d*e^4),
x) + sqrt(2)*(-I*a^2*e^(4*I*d*x + 4*I*c) - 4*I*a^2*e^(2*I*d*x + 2*I*c) - 3*I*a^2)*sqrt(e/(e^(2*I*d*x + 2*I*c)
+ 1))*e^(1/2*I*d*x + 1/2*I*c))/(d*e^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}{\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^2/(e*sec(d*x + c))^(7/2), x)

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maple [A]  time = 0.85, size = 189, normalized size = 1.63 \[ -\frac {2 a^{2} \left (2 i \left (\cos ^{4}\left (d x +c \right )\right )-i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right )-2 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-\cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{7 d \cos \left (d x +c \right )^{4} \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(7/2),x)

[Out]

-2/7*a^2/d*(2*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)*EllipticF(I*(-1+cos(
d*x+c))/sin(d*x+c),I)*cos(d*x+c)-2*cos(d*x+c)^3*sin(d*x+c)-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)-cos(d*x+c)*sin(d*x+c))/cos(d*x+c)^4/(e/cos(d*x+c))^(7/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}{\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^2/(e*sec(d*x + c))^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^2/(e/cos(c + d*x))^(7/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^2/(e/cos(c + d*x))^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \left (- \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\right )\, dx + \int \frac {\tan ^{2}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx + \int \left (- \frac {2 i \tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**2/(e*sec(d*x+c))**(7/2),x)

[Out]

-a**2*(Integral(-1/(e*sec(c + d*x))**(7/2), x) + Integral(tan(c + d*x)**2/(e*sec(c + d*x))**(7/2), x) + Integr
al(-2*I*tan(c + d*x)/(e*sec(c + d*x))**(7/2), x))

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