3.222 \(\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{15/2}} \, dx\)

Optimal. Leaf size=187 \[ \frac {2 a^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{33 d e^8}+\frac {2 a^4 \sin (c+d x)}{33 d e^7 \sqrt {e \sec (c+d x)}}+\frac {2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}} \]

[Out]

2/55*a^4*sin(d*x+c)/d/e^5/(e*sec(d*x+c))^(5/2)+2/33*a^4*sin(d*x+c)/d/e^7/(e*sec(d*x+c))^(1/2)+2/33*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^8-4/15*I*a*(a+I*a*tan(d*x+c))^3/d/(e*sec(d*x+c))^(15/2)-4/55*I*(a^4+I*a^4*tan(d*x+c))/d/e^2/(e*sec(
d*x+c))^(11/2)

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Rubi [A]  time = 0.17, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3496, 3769, 3771, 2641} \[ \frac {2 a^4 \sin (c+d x)}{33 d e^7 \sqrt {e \sec (c+d x)}}+\frac {2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac {2 a^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{33 d e^8}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(15/2),x]

[Out]

(2*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(33*d*e^8) + (2*a^4*Sin[c + d*x])/(5
5*d*e^5*(e*Sec[c + d*x])^(5/2)) + (2*a^4*Sin[c + d*x])/(33*d*e^7*Sqrt[e*Sec[c + d*x]]) - (((4*I)/15)*a*(a + I*
a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(15/2)) - (((4*I)/55)*(a^4 + I*a^4*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x
])^(11/2))

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))^4}{(e \sec (c+d x))^{15/2}} \, dx &=-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}+\frac {a^2 \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{11/2}} \, dx}{5 e^2}\\ &=-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac {\left (7 a^4\right ) \int \frac {1}{(e \sec (c+d x))^{7/2}} \, dx}{55 e^4}\\ &=\frac {2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac {a^4 \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{11 e^6}\\ &=\frac {2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac {2 a^4 \sin (c+d x)}{33 d e^7 \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac {a^4 \int \sqrt {e \sec (c+d x)} \, dx}{33 e^8}\\ &=\frac {2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac {2 a^4 \sin (c+d x)}{33 d e^7 \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac {\left (a^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{33 e^8}\\ &=\frac {2 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)}}{33 d e^8}+\frac {2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac {2 a^4 \sin (c+d x)}{33 d e^7 \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 2.32, size = 155, normalized size = 0.83 \[ -\frac {i a^4 \sqrt {e \sec (c+d x)} (\cos (4 (c+2 d x))+i \sin (4 (c+2 d x))) \left (-54 i \sin (2 (c+d x))-37 i \sin (4 (c+d x))+112 \cos (2 (c+d x))+48 \cos (4 (c+d x))+40 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\sin (4 (c+d x))+i \cos (4 (c+d x)))+64\right )}{660 d e^8 (\cos (d x)+i \sin (d x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(15/2),x]

[Out]

((-1/660*I)*a^4*Sqrt[e*Sec[c + d*x]]*(64 + 112*Cos[2*(c + d*x)] + 48*Cos[4*(c + d*x)] - (54*I)*Sin[2*(c + d*x)
] - (37*I)*Sin[4*(c + d*x)] + 40*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(I*Cos[4*(c + d*x)] + Sin[4*(c +
d*x)]))*(Cos[4*(c + 2*d*x)] + I*Sin[4*(c + 2*d*x)]))/(d*e^8*(Cos[d*x] + I*Sin[d*x])^4)

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ \frac {1320 \, d e^{8} {\rm integral}\left (-\frac {i \, \sqrt {2} a^{4} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{33 \, d e^{8}}, x\right ) + \sqrt {2} {\left (-11 i \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 58 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 128 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 166 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 85 i \, a^{4}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{1320 \, d e^{8}} \]

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))^(15/2),x, algorithm="fricas")

[Out]

1/1320*(1320*d*e^8*integral(-1/33*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^8), x) + sqrt(2)*(-11*I*a^4*e^(8*I*d*x + 8*I*c) - 58*I*a^4*e^(6*I*d*x + 6*I*c) - 128*I*a^4*e^(4*I*d*x + 4*I*
c) - 166*I*a^4*e^(2*I*d*x + 2*I*c) - 85*I*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
^8)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {15}{2}}}\,{d x} \]

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))^(15/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(15/2), x)

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maple [A]  time = 1.15, size = 232, normalized size = 1.24 \[ -\frac {2 a^{4} \left (88 i \left (\cos ^{8}\left (d x +c \right )\right )-88 \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )-60 i \left (\cos ^{6}\left (d x +c \right )\right )+16 \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )-5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right )-3 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-5 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{165 d \cos \left (d x +c \right )^{8} \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {15}{2}}} \]

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))^(15/2),x)

[Out]

-2/165*a^4/d*(88*I*cos(d*x+c)^8-88*sin(d*x+c)*cos(d*x+c)^7-60*I*cos(d*x+c)^6+16*cos(d*x+c)^5*sin(d*x+c)-5*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)-
3*cos(d*x+c)^3*sin(d*x+c)-5*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)-5*cos(d*x+c)*sin(d*x+c))/cos(d*x+c)^8/(e/cos(d*x+c))^(15/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {15}{2}}}\,{d x} \]

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))^(15/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(15/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{15/2}} \,d x \]

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))^(15/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(15/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**(15/2),x)

[Out]

Timed out

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