3.3.16 \(\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{3/2}} \, dx\) [216]

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

[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 = 0.12, 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 = {3577, 3579, 3567, 3856, 2720} \begin {gather*} -\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}} \end {gather*}

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 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 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 3577

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 3579

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 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))^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 = 0.96, size = 130, normalized size = 0.89 \begin {gather*} \frac {a^4 \sec ^3(c+d x) \left (21+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))-11 i \sin (2 (c+d x))\right ) (-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} \end {gather*}

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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Maple [A]
time = 0.55, size = 198, normalized size = 1.36

method	result	size
default	\(\frac {2 a^{4} \left (-15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{2}\left (d x +c \right )\right )-15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-8 i \left (\cos ^{3}\left (d x +c \right )\right )+8 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-12 i \cos \left (d x +c \right )+\sin \left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3} \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}\)	\(198\)

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,method=_RETURNVERBOSE)

[Out]

2/3*a^4/d*(-15*I*cos(d*x+c)^2*(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)-15*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)-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 \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))^4/(e*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 122, normalized size = 0.84 \begin {gather*} -\frac {2 \, {\left (\frac {\sqrt {2} {\left (4 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 21 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 15 i \, a^{4}\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}} + 15 \, {\left (-i \, \sqrt {2} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, \sqrt {2} a^{4}\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{3 \, {\left (d e^{\frac {3}{2}} + d e^{\left (2 i \, d x + 2 i \, c + \frac {3}{2}\right )}\right )}} \end {gather*}

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]

-2/3*(sqrt(2)*(4*I*a^4*e^(4*I*d*x + 4*I*c) + 21*I*a^4*e^(2*I*d*x + 2*I*c) + 15*I*a^4)*e^(1/2*I*d*x + 1/2*I*c)/
sqrt(e^(2*I*d*x + 2*I*c) + 1) + 15*(-I*sqrt(2)*a^4*e^(2*I*d*x + 2*I*c) - I*sqrt(2)*a^4)*weierstrassPInverse(-4
, 0, e^(I*d*x + I*c)))/(d*e^(3/2) + d*e^(2*I*d*x + 2*I*c + 3/2))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 ) \end {gather*}

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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