3.682 \(\int \frac{(e \cos (c+d x))^{3/2}}{\sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=126 \[ -\frac{8 i \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{3/2}}{15 a d}+\frac{2 i (e \cos (c+d x))^{3/2}}{5 d \sqrt{a+i a \tan (c+d x)}}+\frac{16 i \sec ^2(c+d x) (e \cos (c+d x))^{3/2}}{15 d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

(((2*I)/5)*(e*Cos[c + d*x])^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((16*I)/15)*(e*Cos[c + d*x])^(3/2)*Sec[c
+ d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((8*I)/15)*(e*Cos[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d
)

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Rubi [A]  time = 0.312992, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3515, 3502, 3497, 3488} \[ -\frac{8 i \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{3/2}}{15 a d}+\frac{2 i (e \cos (c+d x))^{3/2}}{5 d \sqrt{a+i a \tan (c+d x)}}+\frac{16 i \sec ^2(c+d x) (e \cos (c+d x))^{3/2}}{15 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)/5)*(e*Cos[c + d*x])^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((16*I)/15)*(e*Cos[c + d*x])^(3/2)*Sec[c
+ d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((8*I)/15)*(e*Cos[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d
)

Rule 3515

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 3502

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(b*f*(m + 2*n)), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), 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] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rule 3497

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)/(a*f*m), x] + Dist[(a*(m + n))/(m*d^2), Int[(d*Sec[e + f*x])^(m + 2)*(
a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && LtQ[m, -
1] && IntegersQ[2*m, 2*n]

Rule 3488

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)/(a*f*m), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

Rubi steps

\begin{align*} \int \frac{(e \cos (c+d x))^{3/2}}{\sqrt{a+i a \tan (c+d x)}} \, 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} \sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{2 i (e \cos (c+d x))^{3/2}}{5 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (4 (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{5 a}\\ &=\frac{2 i (e \cos (c+d x))^{3/2}}{5 d \sqrt{a+i a \tan (c+d x)}}-\frac{8 i (e \cos (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}{15 a d}+\frac{\left (8 (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}\right ) \int \frac{\sqrt{e \sec (c+d x)}}{\sqrt{a+i a \tan (c+d x)}} \, dx}{15 e^2}\\ &=\frac{2 i (e \cos (c+d x))^{3/2}}{5 d \sqrt{a+i a \tan (c+d x)}}+\frac{16 i (e \cos (c+d x))^{3/2} \sec ^2(c+d x)}{15 d \sqrt{a+i a \tan (c+d x)}}-\frac{8 i (e \cos (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}{15 a d}\\ \end{align*}

Mathematica [A]  time = 0.335607, size = 63, normalized size = 0.5 \[ -\frac{i e^2 (4 i \sin (2 (c+d x))+\cos (2 (c+d x))-15)}{15 d \sqrt{a+i a \tan (c+d x)} \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/15)*e^2*(-15 + Cos[2*(c + d*x)] + (4*I)*Sin[2*(c + d*x)]))/(d*Sqrt[e*Cos[c + d*x]]*Sqrt[a + I*a*Tan[c + d
*x]])

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Maple [A]  time = 0.35, size = 100, normalized size = 0.8 \begin{align*}{\frac{6\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+6\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +8\,i\cos \left ( dx+c \right ) +16\,\sin \left ( dx+c \right ) }{15\,ad\cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( e\cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}} \end{align*}

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))^(1/2),x)

[Out]

2/15/d/a*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(e*cos(d*x+c))^(3/2)*(3*I*cos(d*x+c)^3+3*cos(d*x+c)^2*
sin(d*x+c)+4*I*cos(d*x+c)+8*sin(d*x+c))/cos(d*x+c)

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Maxima [A]  time = 3.56851, size = 184, normalized size = 1.46 \begin{align*} \frac{{\left (3 i \, e \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) - 5 i \, e \cos \left (\frac{3}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 30 i \, e \cos \left (\frac{1}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 3 \, e \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 5 \, e \sin \left (\frac{3}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 30 \, e \sin \left (\frac{1}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right )\right )} \sqrt{e}}{30 \, \sqrt{a} d} \end{align*}

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))^(1/2),x, algorithm="maxima")

[Out]

1/30*(3*I*e*cos(5/2*d*x + 5/2*c) - 5*I*e*cos(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 30*I*e
*cos(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 3*e*sin(5/2*d*x + 5/2*c) + 5*e*sin(3/5*arctan2
(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 30*e*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c
))))*sqrt(e)/(sqrt(a)*d)

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Fricas [A]  time = 2.0634, size = 252, normalized size = 2. \begin{align*} \frac{\sqrt{2} \sqrt{\frac{1}{2}}{\left (-5 i \, e e^{\left (4 i \, d x + 4 i \, c\right )} + 30 i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, e\right )} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac{5}{2} i \, d x - \frac{5}{2} i \, c\right )}}{30 \, a d} \end{align*}

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))^(1/2),x, algorithm="fricas")

[Out]

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

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

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))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

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))^(1/2),x, algorithm="giac")

[Out]

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