3.7.56 \(\int (e \cos (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx\) [656]

Optimal. Leaf size=90 \[ -\frac {2 i a (e \cos (c+d x))^{5/2}}{5 d}+\frac {6 a (e \cos (c+d x))^{5/2} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a (e \cos (c+d x))^{5/2} \tan (c+d x)}{5 d} \]

[Out]

-2/5*I*a*(e*cos(d*x+c))^(5/2)/d+6/5*a*(e*cos(d*x+c))^(5/2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ell
ipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(5/2)+2/5*a*(e*cos(d*x+c))^(5/2)*tan(d*x+c)/d

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Rubi [A]
time = 0.08, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3596, 3567, 3854, 3856, 2719} \begin {gather*} -\frac {2 i a (e \cos (c+d x))^{5/2}}{5 d}+\frac {6 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (e \cos (c+d x))^{5/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a \tan (c+d x) (e \cos (c+d x))^{5/2}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x]),x]

[Out]

(((-2*I)/5)*a*(e*Cos[c + d*x])^(5/2))/d + (6*a*(e*Cos[c + d*x])^(5/2)*EllipticE[(c + d*x)/2, 2])/(5*d*Cos[c +
d*x]^(5/2)) + (2*a*(e*Cos[c + d*x])^(5/2)*Tan[c + d*x])/(5*d)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 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 (e \cos (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx &=\left ((e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{5/2}} \, dx\\ &=-\frac {2 i a (e \cos (c+d x))^{5/2}}{5 d}+\left (a (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac {1}{(e \sec (c+d x))^{5/2}} \, dx\\ &=-\frac {2 i a (e \cos (c+d x))^{5/2}}{5 d}+\frac {2 a (e \cos (c+d x))^{5/2} \tan (c+d x)}{5 d}+\frac {\left (3 a (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{5 e^2}\\ &=-\frac {2 i a (e \cos (c+d x))^{5/2}}{5 d}+\frac {2 a (e \cos (c+d x))^{5/2} \tan (c+d x)}{5 d}+\frac {\left (3 a (e \cos (c+d x))^{5/2}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 \cos ^{\frac {5}{2}}(c+d x)}\\ &=-\frac {2 i a (e \cos (c+d x))^{5/2}}{5 d}+\frac {6 a (e \cos (c+d x))^{5/2} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a (e \cos (c+d x))^{5/2} \tan (c+d x)}{5 d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 12.66, size = 387, normalized size = 4.30 \begin {gather*} \frac {(e \cos (c+d x))^{5/2} (\cos (d x)-i \sin (d x)) \left (\frac {2 \sqrt {2} e^{-i d x} (-i+\cot (c)) \left (3+3 e^{2 i (c+d x)}+3 \sqrt {1-i e^{i (c+d x)}} \sqrt {e^{i (c+d x)} \left (-i+e^{i (c+d x)}\right )} E\left (\left .\text {ArcSin}\left (\sqrt {-i \cos (c+d x)+\sin (c+d x)}\right )\right |-1\right )-3 \sqrt {1-i e^{i (c+d x)}} \sqrt {e^{i (c+d x)} \left (-i+e^{i (c+d x)}\right )} F\left (\left .\text {ArcSin}\left (\sqrt {-i \cos (c+d x)+\sin (c+d x)}\right )\right |-1\right )+e^{2 i d x} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right )}{5 \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}+\frac {2}{5} \sqrt {\cos (c+d x)} \sin (c) \left (-1+\cos (2 d x) (1-i \cot (c))-6 \cot ^2(c)+i \sin (2 d x)+\cot (c) (5 i+\sin (2 d x))\right )\right ) (a+i a \tan (c+d x))}{2 d \cos ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x]),x]

[Out]

((e*Cos[c + d*x])^(5/2)*(Cos[d*x] - I*Sin[d*x])*((2*Sqrt[2]*(-I + Cot[c])*(3 + 3*E^((2*I)*(c + d*x)) + 3*Sqrt[
1 - I*E^(I*(c + d*x))]*Sqrt[E^(I*(c + d*x))*(-I + E^(I*(c + d*x)))]*EllipticE[ArcSin[Sqrt[(-I)*Cos[c + d*x] +
Sin[c + d*x]]], -1] - 3*Sqrt[1 - I*E^(I*(c + d*x))]*Sqrt[E^(I*(c + d*x))*(-I + E^(I*(c + d*x)))]*EllipticF[Arc
Sin[Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]], -1] + E^((2*I)*d*x)*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2
F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(5*E^(I*d*x)*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]) + (2*S
qrt[Cos[c + d*x]]*Sin[c]*(-1 + Cos[2*d*x]*(1 - I*Cot[c]) - 6*Cot[c]^2 + I*Sin[2*d*x] + Cot[c]*(5*I + Sin[2*d*x
])))/5)*(a + I*a*Tan[c + d*x]))/(2*d*Cos[c + d*x]^(3/2))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (101 ) = 202\).
time = 1.28, size = 205, normalized size = 2.28

method	result	size
default	\(\frac {2 a \,e^{3} \left (8 i \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 i \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+6 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(205\)
risch	\(-\frac {i \left ({\mathrm e}^{2 i \left (d x +c \right )}+7\right ) \sqrt {2}\, e^{2} a \sqrt {e \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) {\mathrm e}^{-i \left (d x +c \right )}}}{10 d}-\frac {3 i \left (-\frac {2 \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}{e \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i \EllipticE \left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i \EllipticF \left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}}\right ) \sqrt {2}\, e^{2} a \sqrt {e \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) {\mathrm e}^{-i \left (d x +c \right )}}\, \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{5 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(320\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2/5/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a*e^3*(8*I*sin(1/2*d*x+1/2*c)^7+8*cos(1/2*d*x+1/2*c
)*sin(1/2*d*x+1/2*c)^6-12*I*sin(1/2*d*x+1/2*c)^5-8*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+6*I*sin(1/2*d*x+1/2
*c)^3+2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2
)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-I*sin(1/2*d*x+1/2*c))/d

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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((e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

e^(5/2)*integrate((I*a*tan(d*x + c) + a)*cos(d*x + c)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 74, normalized size = 0.82 \begin {gather*} \frac {6 i \, \sqrt {2} a e^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {\frac {1}{2}} {\left (5 i \, a e^{\frac {5}{2}} - i \, a e^{\left (2 i \, d x + 2 i \, c + \frac {5}{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 )}}{5 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/5*(6*I*sqrt(2)*a*e^(5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))) + sqrt(1/2)*(5*
I*a*e^(5/2) - I*a*e^(2*I*d*x + 2*I*c + 5/2))*sqrt(e^(2*I*d*x + 2*I*c) + 1)*e^(-1/2*I*d*x - 1/2*I*c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(5/2)*(a+I*a*tan(d*x+c)),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6191 deep

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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))^(5/2)*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)*cos(d*x + c)^(5/2)*e^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i),x)

[Out]

int((e*cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i), x)

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