3.216 \(\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=146 \[ -\frac {10 i a^4 \sqrt {e \sec (c+d x)}}{d e^2}-\frac {2 i \left (a^4+i a^4 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{d e^2}-\frac {10 a^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{d e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}} \]

[Out]

-10*I*a^4*(e*sec(d*x+c))^(1/2)/d/e^2-10*a^4*(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)/d/e^2-4/3*I*a*(a+I*a*tan(d*x+c))^3/d/(e*sec(d*x+c))^
(3/2)-2*I*(e*sec(d*x+c))^(1/2)*(a^4+I*a^4*tan(d*x+c))/d/e^2

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Rubi [A]  time = 0.15, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3496, 3498, 3486, 3771, 2641} \[ -\frac {10 i a^4 \sqrt {e \sec (c+d x)}}{d e^2}-\frac {2 i \left (a^4+i a^4 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{d e^2}-\frac {10 a^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{d e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(3/2),x]

[Out]

((-10*I)*a^4*Sqrt[e*Sec[c + d*x]])/(d*e^2) - (10*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c
+ d*x]])/(d*e^2) - (((4*I)/3)*a*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(3/2)) - ((2*I)*Sqrt[e*Sec[c +
d*x]]*(a^4 + I*a^4*Tan[c + d*x]))/(d*e^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 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 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 3498

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] + Dist[(a*(m + 2*n - 2))/(m + n - 1), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] &&
GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 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))^4}{(e \sec (c+d x))^{3/2}} \, dx &=-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}-\frac {\left (3 a^2\right ) \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2 \, dx}{e^2}\\ &=-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}-\frac {2 i \sqrt {e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )}{d e^2}-\frac {\left (5 a^3\right ) \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x)) \, dx}{e^2}\\ &=-\frac {10 i a^4 \sqrt {e \sec (c+d x)}}{d e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}-\frac {2 i \sqrt {e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )}{d e^2}-\frac {\left (5 a^4\right ) \int \sqrt {e \sec (c+d x)} \, dx}{e^2}\\ &=-\frac {10 i a^4 \sqrt {e \sec (c+d x)}}{d e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}-\frac {2 i \sqrt {e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )}{d e^2}-\frac {\left (5 a^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{e^2}\\ &=-\frac {10 i a^4 \sqrt {e \sec (c+d x)}}{d e^2}-\frac {10 a^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{d e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}-\frac {2 i \sqrt {e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )}{d e^2}\\ \end {align*}

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Mathematica [A]  time = 1.39, size = 130, normalized size = 0.89 \[ \frac {a^4 \sec ^3(c+d x) (\sin (c+5 d x)-i \cos (c+5 d x)) \left (-11 i \sin (2 (c+d x))+19 \cos (2 (c+d x))-30 i \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (c+d x)-i \sin (c+d x))+21\right )}{3 d (\cos (d x)+i \sin (d x))^4 (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(3/2),x]

[Out]

(a^4*Sec[c + d*x]^3*(21 + 19*Cos[2*(c + d*x)] - (30*I)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2]*(Cos[c + d
*x] - I*Sin[c + d*x]) - (11*I)*Sin[2*(c + d*x)])*((-I)*Cos[c + 5*d*x] + Sin[c + 5*d*x]))/(3*d*(e*Sec[c + d*x])
^(3/2)*(Cos[d*x] + I*Sin[d*x])^4)

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (-8 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 42 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 30 i \, a^{4}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 3 \, {\left (d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{2}\right )} {\rm integral}\left (\frac {5 i \, \sqrt {2} a^{4} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{d e^{2}}, x\right )}{3 \, {\left (d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/3*(sqrt(2)*(-8*I*a^4*e^(4*I*d*x + 4*I*c) - 42*I*a^4*e^(2*I*d*x + 2*I*c) - 30*I*a^4)*sqrt(e/(e^(2*I*d*x + 2*I
*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 3*(d*e^2*e^(2*I*d*x + 2*I*c) + d*e^2)*integral(5*I*sqrt(2)*a^4*sqrt(e/(e^(
2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I*c)/(d*e^2), x))/(d*e^2*e^(2*I*d*x + 2*I*c) + d*e^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.97, size = 200, normalized size = 1.37 \[ -\frac {2 a^{4} \left (15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right )+8 i \left (\cos ^{3}\left (d x +c \right )\right )-8 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+12 i \cos \left (d x +c \right )-\sin \left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3} \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*a^4/d*(15*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+15*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)+8*I*cos(d*x+c)^3-8*cos(d*x+c)^2*sin(d*x+c)+12*I*cos(d*x+c)-sin(d*x+c))/cos(d*x+c)^
3/(e/cos(d*x+c))^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(3/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{4} \left (\int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \left (- \frac {6 \tan ^{2}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\right )\, dx + \int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {4 i \tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \left (- \frac {4 i \tan ^{3}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**4/(e*sec(d*x+c))**(3/2),x)

[Out]

a**4*(Integral((e*sec(c + d*x))**(-3/2), x) + Integral(-6*tan(c + d*x)**2/(e*sec(c + d*x))**(3/2), x) + Integr
al(tan(c + d*x)**4/(e*sec(c + d*x))**(3/2), x) + Integral(4*I*tan(c + d*x)/(e*sec(c + d*x))**(3/2), x) + Integ
ral(-4*I*tan(c + d*x)**3/(e*sec(c + d*x))**(3/2), x))

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