3.7.82 \(\int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [682]

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

[Out]

2/5*I*(e*cos(d*x+c))^(3/2)/d/(a+I*a*tan(d*x+c))^(1/2)+16/15*I*(e*cos(d*x+c))^(3/2)*sec(d*x+c)^2/d/(a+I*a*tan(d
*x+c))^(1/2)-8/15*I*(e*cos(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^(1/2)/a/d

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Rubi [A]
time = 0.21, antiderivative size = 126, normalized size of antiderivative = 1.00, 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 = {3596, 3583, 3578, 3569} \begin {gather*} -\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)}} \end {gather*}

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 3569

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]

Rule 3578

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*S
ec[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 3583

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 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]

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*}

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Mathematica [A]
time = 0.42, size = 63, normalized size = 0.50 \begin {gather*} -\frac {i e^2 (-15+\cos (2 (c+d x))+4 i \sin (2 (c+d x)))}{15 d \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/15*I)*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 = 0.92, size = 100, normalized size = 0.79

method	result	size
default	\(\frac {2 \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}} \left (3 i \left (\cos ^{3}\left (d x +c \right )\right )+3 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 i \cos \left (d x +c \right )+8 \sin \left (d x +c \right )\right )}{15 d \cos \left (d x +c \right ) a}\)	\(100\)

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,method=_RETURNVERBOSE)

[Out]

2/15/d*(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*si
n(d*x+c)+4*I*cos(d*x+c)+8*sin(d*x+c))/cos(d*x+c)/a

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Maxima [A]
time = 0.60, size = 129, normalized size = 1.02 \begin {gather*} \frac {{\left (3 i \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 5 i \, \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 \, \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 \, \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \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 \, \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 )} e^{\frac {3}{2}}}{30 \, \sqrt {a} 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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.32, size = 82, normalized size = 0.65 \begin {gather*} \frac {\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (3 i \, e^{\frac {3}{2}} - 5 i \, e^{\left (4 i \, d x + 4 i \, c + \frac {3}{2}\right )} + 30 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 {5}{2} i \, d x - \frac {5}{2} i \, c\right )}}{30 \, a 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))^(1/2),x, algorithm="fricas")

[Out]

1/30*sqrt(2)*sqrt(1/2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(3*I*e^(3/2) - 5*I*e^(4*I*d*x + 4*I*c + 3/2) + 30*I*e
^(2*I*d*x + 2*I*c + 3/2))*sqrt(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)] 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))**(1/2),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))^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.13, size = 100, normalized size = 0.79 \begin {gather*} \frac {e\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (35\,\sin \left (c+d\,x\right )+3\,\sin \left (3\,c+3\,d\,x\right )+\cos \left (c+d\,x\right )\,25{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}\right )}{30\,a\,d} \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)^(1/2),x)

[Out]

(e*(e*cos(c + d*x))^(1/2)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(cos
(c + d*x)*25i + 35*sin(c + d*x) + cos(3*c + 3*d*x)*3i + 3*sin(3*c + 3*d*x)))/(30*a*d)

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