3.5.33 \(\int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [433]

Optimal. Leaf size=80 \[ \frac {2 i (e \sec (c+d x))^{3/2}}{7 d (a+i a \tan (c+d x))^{5/2}}+\frac {4 i (e \sec (c+d x))^{3/2}}{21 a d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

2/7*I*(e*sec(d*x+c))^(3/2)/d/(a+I*a*tan(d*x+c))^(5/2)+4/21*I*(e*sec(d*x+c))^(3/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)

________________________________________________________________________________________

Rubi [A]
time = 0.11, 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 (e \sec (c+d x))^{3/2}}{21 a d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i (e \sec (c+d x))^{3/2}}{7 d (a+i a \tan (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)/7)*(e*Sec[c + d*x])^(3/2))/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((4*I)/21)*(e*Sec[c + d*x])^(3/2))/(a*d
*(a + I*a*Tan[c + d*x])^(3/2))

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 {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac {2 i (e \sec (c+d x))^{3/2}}{7 d (a+i a \tan (c+d x))^{5/2}}+\frac {2 \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx}{7 a}\\ &=\frac {2 i (e \sec (c+d x))^{3/2}}{7 d (a+i a \tan (c+d x))^{5/2}}+\frac {4 i (e \sec (c+d x))^{3/2}}{21 a d (a+i a \tan (c+d x))^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.26, size = 63, normalized size = 0.79 \begin {gather*} \frac {2 (e \sec (c+d x))^{3/2} (-5 i+2 \tan (c+d x))}{21 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*Sec[c + d*x])^(3/2)*(-5*I + 2*Tan[c + d*x]))/(21*a^2*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

________________________________________________________________________________________

Maple [A]
time = 1.07, size = 112, normalized size = 1.40

method	result	size
default	\(\frac {2 i \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \left (-12 i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+12 \left (\cos ^{4}\left (d x +c \right )\right )-i \cos \left (d x +c \right ) \sin \left (d x +c \right )-5 \left (\cos ^{2}\left (d x +c \right )\right )-2\right )}{21 d \,a^{3}}\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/21*I/d*(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)^2*(-12*I*cos(d*x+c)^3*
sin(d*x+c)+12*cos(d*x+c)^4-I*cos(d*x+c)*sin(d*x+c)-5*cos(d*x+c)^2-2)/a^3

________________________________________________________________________________________

Maxima [A]
time = 0.55, size = 81, normalized size = 1.01 \begin {gather*} \frac {{\left (3 i \, \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 i \, \cos \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 3 \, \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, \sin \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right )\right )} e^{\frac {3}{2}}}{21 \, a^{\frac {5}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/21*(3*I*cos(7/2*d*x + 7/2*c) + 7*I*cos(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 3*sin(7/2*
d*x + 7/2*c) + 7*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*e^(3/2)/(a^(5/2)*d)

________________________________________________________________________________________

Fricas [A]
time = 0.40, size = 76, normalized size = 0.95 \begin {gather*} \frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (3 i \, e^{\frac {3}{2}} + 7 i \, e^{\left (4 i \, d x + 4 i \, c + \frac {3}{2}\right )} + 10 i \, e^{\left (2 i \, d x + 2 i \, c + \frac {3}{2}\right )}\right )} e^{\left (-\frac {7}{2} i \, d x - \frac {7}{2} i \, c\right )}}{21 \, a^{3} 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))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/21*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(3*I*e^(3/2) + 7*I*e^(4*I*d*x + 4*I*c + 3/2) + 10*I*e^(2*I*d*x + 2*I*c
+ 3/2))*e^(-7/2*I*d*x - 7/2*I*c)/(a^3*d*sqrt(e^(2*I*d*x + 2*I*c) + 1))

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(3/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral((e*sec(c + d*x))**(3/2)/(I*a*(tan(c + d*x) - I))**(5/2), x)

________________________________________________________________________________________

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))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 4.23, size = 102, normalized size = 1.28 \begin {gather*} \frac {e\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (7\,\sin \left (c+d\,x\right )+3\,\sin \left (3\,c+3\,d\,x\right )+\cos \left (c+d\,x\right )\,7{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}\right )}{21\,a^2\,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))^(3/2)/(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

(e*(e/cos(c + d*x))^(1/2)*(cos(c + d*x)*7i + 7*sin(c + d*x) + cos(3*c + 3*d*x)*3i + 3*sin(3*c + 3*d*x)))/(21*a
^2*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))

________________________________________________________________________________________