\(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx\) [841]

Optimal result

Integrand size = 45, antiderivative size = 152 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {-i A-B}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {i A}{3 c f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {2 A \tan (e+f x)}{3 a c f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3669, 79, 47, 39} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {B+i A}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {2 A \tan (e+f x)}{3 a c f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}+\frac {i A}{3 c f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

-1/3*(I*A + B)/(f*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2)) + ((I/3)*A)/(c*f*(a + I*a*Tan[e +
f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]]) + (2*A*Tan[e + f*x])/(3*a*c*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c
*Tan[e + f*x]])

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 3669

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((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)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, 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 {A+B x}{(a+i a x)^{5/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {i A+B}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {(a A) \text {Subst}\left (\int \frac {1}{(a+i a x)^{5/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {i A+B}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {i A}{3 c f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(2 A) \text {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 f} \\ & = -\frac {i A+B}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {i A}{3 c f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {2 A \tan (e+f x)}{3 a c f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.92 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.52 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\cos ^2(e+f x) \left (-B+3 A \tan (e+f x)+2 A \tan ^3(e+f x)\right )}{3 a c f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(Cos[e + f*x]^2*(-B + 3*A*Tan[e + f*x] + 2*A*Tan[e + f*x]^3))/(3*a*c*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c
*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 A \tan \left (f x +e \right )^{5}+5 A \tan \left (f x +e \right )^{3}-B \tan \left (f x +e \right )^{2}+3 A \tan \left (f x +e \right )-B \right )}{3 f \,a^{2} c^{2} \left (i-\tan \left (f x +e \right )\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}\)	\(113\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 A \tan \left (f x +e \right )^{5}+5 A \tan \left (f x +e \right )^{3}-B \tan \left (f x +e \right )^{2}+3 A \tan \left (f x +e \right )-B \right )}{3 f \,a^{2} c^{2} \left (i-\tan \left (f x +e \right )\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}\)	\(113\)
parts	\(-\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (1+\tan \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right ) \left (2 \tan \left (f x +e \right )^{2}+3\right )}{3 f \,a^{2} c^{2} \left (i+\tan \left (f x +e \right )\right )^{3} \left (i-\tan \left (f x +e \right )\right )^{3}}+\frac {B \left (1+\tan \left (f x +e \right )^{2}\right ) \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}}{3 f \,c^{2} a^{2} \left (i+\tan \left (f x +e \right )\right )^{3} \left (i-\tan \left (f x +e \right )\right )^{3}}\)	\(174\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.99 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {{\left ({\left (-i \, A - B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} - 2 \, {\left (5 i \, A + 2 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 8 \, B e^{\left (5 i \, f x + 5 i \, e\right )} - 6 \, B e^{\left (4 i \, f x + 4 i \, e\right )} + 8 \, B e^{\left (3 i \, f x + 3 i \, e\right )} - 2 \, {\left (-5 i \, A + 2 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-3 i \, f x - 3 i \, e\right )}}{24 \, a^{2} c^{2} f} \]

[In]

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

[Out]

1/24*((-I*A - B)*e^(8*I*f*x + 8*I*e) - 2*(5*I*A + 2*B)*e^(6*I*f*x + 6*I*e) + 8*B*e^(5*I*f*x + 5*I*e) - 6*B*e^(
4*I*f*x + 4*I*e) + 8*B*e^(3*I*f*x + 3*I*e) - 2*(-5*I*A + 2*B)*e^(2*I*f*x + 2*I*e) + I*A - B)*sqrt(a/(e^(2*I*f*
x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-3*I*f*x - 3*I*e)/(a^2*c^2*f)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.32 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {{\left (3 \, {\left (3 i \, A - B\right )} \cos \left (2 \, f x + 2 \, e\right ) - 3 \, {\left (3 \, A + i \, B\right )} \sin \left (2 \, f x + 2 \, e\right ) - 2 \, B\right )} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3 \, {\left (-3 i \, A - B\right )} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (3 \, {\left (3 \, A + i \, B\right )} \cos \left (2 \, f x + 2 \, e\right ) + 3 \, {\left (3 i \, A - B\right )} \sin \left (2 \, f x + 2 \, e\right ) + 2 \, A\right )} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3 \, {\left (3 \, A - i \, B\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )}{24 \, a^{\frac {3}{2}} c^{\frac {3}{2}} f} \]

[In]

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

[Out]

1/24*((3*(3*I*A - B)*cos(2*f*x + 2*e) - 3*(3*A + I*B)*sin(2*f*x + 2*e) - 2*B)*cos(3/2*arctan2(sin(2*f*x + 2*e)
, cos(2*f*x + 2*e))) + 3*(-3*I*A - B)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (3*(3*A + I*B)*co
s(2*f*x + 2*e) + 3*(3*I*A - B)*sin(2*f*x + 2*e) + 2*A)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
3*(3*A - I*B)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))/(a^(3/2)*c^(3/2)*f)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.38 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.30 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (10\,A\,\sin \left (2\,e+2\,f\,x\right )-3\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,8{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-4\,B\,\cos \left (2\,e+2\,f\,x\right )-B\,\cos \left (4\,e+4\,f\,x\right )-A\,9{}\mathrm {i}+A\,\sin \left (4\,e+4\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,2{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}\right )}{24\,a^2\,c\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

[In]

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

[Out]

(((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(A*cos(2*e + 2*f*x)*8i - 3*B
- A*9i + A*cos(4*e + 4*f*x)*1i - 4*B*cos(2*e + 2*f*x) - B*cos(4*e + 4*f*x) + 10*A*sin(2*e + 2*f*x) + A*sin(4*e
+ 4*f*x) + B*sin(2*e + 2*f*x)*2i + B*sin(4*e + 4*f*x)*1i))/(24*a^2*c*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))