3.7.65 \(\int \frac {(e \cos (c+d x))^{3/2}}{(a+i a \tan (c+d x))^2} \, dx\) [665]

Optimal. Leaf size=154 \[ \frac {10 (e \cos (c+d x))^{3/2} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{33 a^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \cos (c+d x) (e \cos (c+d x))^{3/2} \sin (c+d x)}{11 a^2 d}+\frac {10 (e \cos (c+d x))^{3/2} \tan (c+d x)}{33 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{3/2}}{11 d \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

10/33*(e*cos(d*x+c))^(3/2)*(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
))/a^2/d/cos(d*x+c)^(3/2)+2/11*cos(d*x+c)*(e*cos(d*x+c))^(3/2)*sin(d*x+c)/a^2/d+10/33*(e*cos(d*x+c))^(3/2)*tan
(d*x+c)/a^2/d+4/11*I*cos(d*x+c)^2*(e*cos(d*x+c))^(3/2)/d/(a^2+I*a^2*tan(d*x+c))

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Rubi [A]
time = 0.15, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3596, 3581, 3854, 3856, 2720} \begin {gather*} \frac {10 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (e \cos (c+d x))^{3/2}}{33 a^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{3/2}}{11 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {2 \sin (c+d x) \cos (c+d x) (e \cos (c+d x))^{3/2}}{11 a^2 d}+\frac {10 \tan (c+d x) (e \cos (c+d x))^{3/2}}{33 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*(e*Cos[c + d*x])^(3/2)*EllipticF[(c + d*x)/2, 2])/(33*a^2*d*Cos[c + d*x]^(3/2)) + (2*Cos[c + d*x]*(e*Cos[c
+ d*x])^(3/2)*Sin[c + d*x])/(11*a^2*d) + (10*(e*Cos[c + d*x])^(3/2)*Tan[c + d*x])/(33*a^2*d) + (((4*I)/11)*Co
s[c + d*x]^2*(e*Cos[c + d*x])^(3/2))/(d*(a^2 + I*a^2*Tan[c + d*x]))

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 3581

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

Rule 3596

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 3854

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 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 {(e \cos (c+d x))^{3/2}}{(a+i a \tan (c+d x))^2} \, dx &=\left ((e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}\right ) \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx\\ &=\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{3/2}}{11 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (7 e^2 (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}\right ) \int \frac {1}{(e \sec (c+d x))^{7/2}} \, dx}{11 a^2}\\ &=\frac {2 \cos (c+d x) (e \cos (c+d x))^{3/2} \sin (c+d x)}{11 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{3/2}}{11 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (5 (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}\right ) \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{11 a^2}\\ &=\frac {2 \cos (c+d x) (e \cos (c+d x))^{3/2} \sin (c+d x)}{11 a^2 d}+\frac {10 (e \cos (c+d x))^{3/2} \tan (c+d x)}{33 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{3/2}}{11 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (5 (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}\right ) \int \sqrt {e \sec (c+d x)} \, dx}{33 a^2 e^2}\\ &=\frac {2 \cos (c+d x) (e \cos (c+d x))^{3/2} \sin (c+d x)}{11 a^2 d}+\frac {10 (e \cos (c+d x))^{3/2} \tan (c+d x)}{33 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{3/2}}{11 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (5 (e \cos (c+d x))^{3/2}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{33 a^2 \cos ^{\frac {3}{2}}(c+d x)}\\ &=\frac {10 (e \cos (c+d x))^{3/2} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{33 a^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \cos (c+d x) (e \cos (c+d x))^{3/2} \sin (c+d x)}{11 a^2 d}+\frac {10 (e \cos (c+d x))^{3/2} \tan (c+d x)}{33 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{3/2}}{11 d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.76, size = 131, normalized size = 0.85 \begin {gather*} \frac {(e \cos (c+d x))^{3/2} \left (-20 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+\sqrt {\cos (c+d x)} (-28 i \cos (c+d x)+4 i \cos (3 (c+d x))+13 \sin (c+d x)-7 \sin (3 (c+d x)))\right )}{66 a^2 d \cos ^{\frac {7}{2}}(c+d x) (-i+\tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Cos[c + d*x])^(3/2)*(-20*EllipticF[(c + d*x)/2, 2]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + Sqrt[Cos[c +
d*x]]*((-28*I)*Cos[c + d*x] + (4*I)*Cos[3*(c + d*x)] + 13*Sin[c + d*x] - 7*Sin[3*(c + d*x)])))/(66*a^2*d*Cos[c
+ d*x]^(7/2)*(-I + Tan[c + d*x])^2)

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Maple [A]
time = 1.70, size = 315, normalized size = 2.05

method	result	size
default	\(-\frac {2 e^{2} \left (-384 i \left (\sin ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+384 \left (\sin ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1152 i \left (\sin ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-960 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-1440 i \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1008 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+960 i \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-552 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-360 i \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+176 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+72 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-28 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-6 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{33 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(315\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-2/33/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^2*(-384*I*sin(1/2*d*x+1/2*c)^13+384*sin(1/2
*d*x+1/2*c)^12*cos(1/2*d*x+1/2*c)+1152*I*sin(1/2*d*x+1/2*c)^11-960*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)-14
40*I*sin(1/2*d*x+1/2*c)^9+1008*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+960*I*sin(1/2*d*x+1/2*c)^7-552*cos(1/2*
d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-360*I*sin(1/2*d*x+1/2*c)^5+176*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+72*I*si
n(1/2*d*x+1/2*c)^3-28*sin(1/2*d*x+1/2*c)^2*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))-6*I*sin(1/2*d*x+1/2*c))/d

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 112, normalized size = 0.73 \begin {gather*} \frac {{\left (\sqrt {\frac {1}{2}} {\left (3 i \, e^{\frac {3}{2}} - 11 i \, e^{\left (6 i \, d x + 6 i \, c + \frac {3}{2}\right )} + 41 i \, e^{\left (4 i \, d x + 4 i \, c + \frac {3}{2}\right )} + 15 i \, e^{\left (2 i \, d x + 2 i \, c + \frac {3}{2}\right )}\right )} \sqrt {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 )} - 40 i \, \sqrt {2} e^{\left (5 i \, d x + 5 i \, c + \frac {3}{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{132 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/132*(sqrt(1/2)*(3*I*e^(3/2) - 11*I*e^(6*I*d*x + 6*I*c + 3/2) + 41*I*e^(4*I*d*x + 4*I*c + 3/2) + 15*I*e^(2*I*
d*x + 2*I*c + 3/2))*sqrt(e^(2*I*d*x + 2*I*c) + 1)*e^(-1/2*I*d*x - 1/2*I*c) - 40*I*sqrt(2)*e^(5*I*d*x + 5*I*c +
3/2)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))*e^(-5*I*d*x - 5*I*c)/(a^2*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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