\(\int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx\) [1037]

Optimal result

Integrand size = 35, antiderivative size = 90 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {i (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}} \]

[Out]

-1/5*I*(a+I*a*tan(f*x+e))^(3/2)/f/(c-I*c*tan(f*x+e))^(5/2)-1/15*I*(a+I*a*tan(f*x+e))^(3/2)/c/f/(c-I*c*tan(f*x+
e))^(3/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3604, 47, 37} \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {i (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[In]

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

[Out]

((-1/5*I)*(a + I*a*Tan[e + f*x])^(3/2))/(f*(c - I*c*Tan[e + f*x])^(5/2)) - ((I/15)*(a + I*a*Tan[e + f*x])^(3/2
))/(c*f*(c - I*c*Tan[e + f*x])^(3/2))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 3604

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[a*(c/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {i (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {a \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f} \\ & = -\frac {i (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {i (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.67 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {a (1+i \tan (e+f x)) (4 i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}}{15 c^2 f (i+\tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(a*(1 + I*Tan[e + f*x])*(4*I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]])/(15*c^2*f*(I + Tan[e + f*x])^2*Sqrt[c
- I*c*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (4 i+\tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{4}}\)	\(71\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (4 i+\tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{4}}\)	\(71\)
risch	\(-\frac {i a \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (3 \,{\mathrm e}^{4 i \left (f x +e \right )}+5 \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{30 c^{2} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(77\)

[In]

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

[Out]

-1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a/c^3*(1+tan(f*x+e)^2)*(4*I+tan(f*x+e))/(tan(f*
x+e)+I)^4

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {{\left (-3 i \, a e^{\left (7 i \, f x + 7 i \, e\right )} - 8 i \, a e^{\left (5 i \, f x + 5 i \, e\right )} - 5 i \, a e^{\left (3 i \, f x + 3 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{30 \, c^{3} f} \]

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/30*(-3*I*a*e^(7*I*f*x + 7*I*e) - 8*I*a*e^(5*I*f*x + 5*I*e) - 5*I*a*e^(3*I*f*x + 3*I*e))*sqrt(a/(e^(2*I*f*x +
2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^3*f)

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}}}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.24 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {{\left (-3 i \, a \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 5 i \, a \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3 \, a \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 5 \, a \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{30 \, c^{\frac {5}{2}} f} \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.18 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.51 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {a\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (-5\,\sin \left (2\,e+2\,f\,x\right )-3\,\sin \left (4\,e+4\,f\,x\right )+\cos \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}\right )}{30\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

[In]

int((a + a*tan(e + f*x)*1i)^(3/2)/(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

-(a*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(cos(2*e + 2*f*x)*5i + cos
(4*e + 4*f*x)*3i - 5*sin(2*e + 2*f*x) - 3*sin(4*e + 4*f*x)))/(30*c^2*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x
)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2))