3.5.26 \(\int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [426]

Optimal. Leaf size=80 \[ \frac {2 i \sqrt {e \sec (c+d x)}}{5 d (a+i a \tan (c+d x))^{3/2}}+\frac {4 i \sqrt {e \sec (c+d x)}}{5 a d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

4/5*I*(e*sec(d*x+c))^(1/2)/a/d/(a+I*a*tan(d*x+c))^(1/2)+2/5*I*(e*sec(d*x+c))^(1/2)/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.09, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3583, 3569} \begin {gather*} \frac {4 i \sqrt {e \sec (c+d x)}}{5 a d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i \sqrt {e \sec (c+d x)}}{5 d (a+i a \tan (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)/5)*Sqrt[e*Sec[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((4*I)/5)*Sqrt[e*Sec[c + d*x]])/(a*d*Sqrt
[a + I*a*Tan[c + d*x]])

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 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]

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 63, normalized size = 0.79 \begin {gather*} \frac {2 \sqrt {e \sec (c+d x)} (3+2 i \tan (c+d x))}{5 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.86, size = 101, normalized size = 1.26

method	result	size
default	\(-\frac {2 i \sqrt {\frac {e}{\cos \left (d x +c \right )}}\, \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \left (2 i \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )-2 \left (\cos ^{3}\left (d x +c \right )\right )+2 i \sin \left (d x +c \right )-\cos \left (d x +c \right )\right )}{5 d \,a^{2}}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.56, size = 79, normalized size = 0.99 \begin {gather*} \frac {{\left (i \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 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 ) + \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \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 {1}{2}}}{5 \, a^{\frac {3}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 76, normalized size = 0.95 \begin {gather*} \frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (i \, e^{\frac {1}{2}} + 5 i \, e^{\left (4 i \, d x + 4 i \, c + \frac {1}{2}\right )} + 6 i \, e^{\left (2 i \, d x + 2 i \, c + \frac {1}{2}\right )}\right )} e^{\left (-\frac {5}{2} i \, d x - \frac {5}{2} i \, c\right )}}{5 \, a^{2} d \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {e \sec {\left (c + d x \right )}}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(1/2)/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral(sqrt(e*sec(c + d*x))/(I*a*(tan(c + d*x) - I))**(3/2), x)

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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*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(e^(1/2)*sqrt(sec(d*x + c))/(I*a*tan(d*x + c) + a)^(3/2), x)

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Mupad [B]
time = 3.94, size = 84, normalized size = 1.05 \begin {gather*} \frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+\sin \left (2\,c+2\,d\,x\right )+5{}\mathrm {i}\right )}{5\,a\,d\,\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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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