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

Optimal. Leaf size=85 \[ \frac{4 i a e \sec (c+d x) \sqrt{e \cos (c+d x)}}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 i \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{3/2}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.212166, antiderivative size = 86, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3515, 3497, 3488} \[ \frac{4 i a \sec ^2(c+d x) (e \cos (c+d x))^{3/2}}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 i \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((4*I)/3)*a*(e*Cos[c + d*x])^(3/2)*Sec[c + d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/3)*(e*Cos[c + d*x
])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/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 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 (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{\sqrt{a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx\\ &=-\frac{2 i (e \cos (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{\left (2 a (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}{3 e^2}\\ &=\frac{4 i a (e \cos (c+d x))^{3/2} \sec ^2(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 i (e \cos (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.203282, size = 56, normalized size = 0.66 \[ \frac{2 e \sqrt{a+i a \tan (c+d x)} \sqrt{e \cos (c+d x)} (2 \sin (c+d x)+i \cos (c+d x))}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.338, size = 70, normalized size = 0.8 \begin{align*}{\frac{2\,i\cos \left ( dx+c \right ) +4\,\sin \left ( dx+c \right ) }{3\,d\cos \left ( dx+c \right ) } \left ( e\cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \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]

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

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Maxima [A]  time = 3.02176, size = 80, normalized size = 0.94 \begin{align*} \frac{{\left (-i \, e \cos \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 3 i \, e \cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + e \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 3 \, e \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} \sqrt{a} \sqrt{e}}{3 \, 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]

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

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Fricas [A]  time = 2.06136, size = 204, normalized size = 2.4 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e}{\left (-i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, e\right )} \sqrt{\frac{a}{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 )}}{3 \, 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]

1/3*sqrt(2)*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*(-I*e*e^(2*I*d*x + 2*I*c) + 3*I*e)*sqrt(a/(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(-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(-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, algorithm="giac")

[Out]

Timed out