\(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx\) [842]

Optimal result

Integrand size = 45, antiderivative size = 269 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {i A-B}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i A-B}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {(4 i A-B) \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (4 i A-B) \sqrt {a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 (4 i A-B) \sqrt {a+i a \tan (e+f x)}}{15 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

-2/15*(4*I*A-B)*(a+I*a*tan(f*x+e))^(1/2)/a^2/c^2/f/(c-I*c*tan(f*x+e))^(1/2)+1/3*(4*I*A-B)/a/f/(a+I*a*tan(f*x+e
))^(1/2)/(c-I*c*tan(f*x+e))^(5/2)-1/5*(4*I*A-B)*(a+I*a*tan(f*x+e))^(1/2)/a^2/f/(c-I*c*tan(f*x+e))^(5/2)+1/3*(I
*A-B)/f/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(5/2)-2/15*(4*I*A-B)*(a+I*a*tan(f*x+e))^(1/2)/a^2/c/f/(c-I
*c*tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3669, 79, 47, 37} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {2 (-B+4 i A) \sqrt {a+i a \tan (e+f x)}}{15 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {2 (-B+4 i A) \sqrt {a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {(-B+4 i A) \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {-B+i A}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {-B+4 i A}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \]

[In]

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

[Out]

(I*A - B)/(3*f*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2)) + ((4*I)*A - B)/(3*a*f*Sqrt[a + I*a*
Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(5/2)) - (((4*I)*A - B)*Sqrt[a + I*a*Tan[e + f*x]])/(5*a^2*f*(c - I*c*Tan
[e + f*x])^(5/2)) - (2*((4*I)*A - B)*Sqrt[a + I*a*Tan[e + f*x]])/(15*a^2*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - (
2*((4*I)*A - B)*Sqrt[a + I*a*Tan[e + f*x]])/(15*a^2*c^2*f*Sqrt[c - I*c*Tan[e + f*x]])

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 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)^{7/2}} \, 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))^{5/2}}+\frac {((4 A+i B) c) \text {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} (c-i c x)^{7/2}} \, 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))^{5/2}}+\frac {4 i A-B}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}+\frac {((4 A+i B) c) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{a f} \\ & = \frac {i A-B}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i A-B}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {(4 i A-B) \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {(2 (4 A+i B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 a f} \\ & = \frac {i A-B}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i A-B}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {(4 i A-B) \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (4 i A-B) \sqrt {a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}+\frac {(2 (4 A+i B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a c f} \\ & = \frac {i A-B}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i A-B}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {(4 i A-B) \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (4 i A-B) \sqrt {a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 (4 i A-B) \sqrt {a+i a \tan (e+f x)}}{15 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.75 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.57 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {3 (A-i B)+(12 i A-3 B) \tan (e+f x)+3 (4 A+i B) \tan ^2(e+f x)+(8 i A-2 B) \tan ^3(e+f x)+(8 A+2 i B) \tan ^4(e+f x)}{15 a c^2 f (-i+\tan (e+f x)) (i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(3*(A - I*B) + ((12*I)*A - 3*B)*Tan[e + f*x] + 3*(4*A + I*B)*Tan[e + f*x]^2 + ((8*I)*A - 2*B)*Tan[e + f*x]^3 +
(8*A + (2*I)*B)*Tan[e + f*x]^4)/(15*a*c^2*f*(-I + Tan[e + f*x])*(I + Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x
]]*Sqrt[c - I*c*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i A \tan \left (f x +e \right )^{6}-2 i B \tan \left (f x +e \right )^{5}-2 B \tan \left (f x +e \right )^{6}+20 i A \tan \left (f x +e \right )^{4}-8 A \tan \left (f x +e \right )^{5}-5 i B \tan \left (f x +e \right )^{3}-5 B \tan \left (f x +e \right )^{4}+15 i A \tan \left (f x +e \right )^{2}-20 A \tan \left (f x +e \right )^{3}-3 i \tan \left (f x +e \right ) B +3 i A -12 A \tan \left (f x +e \right )+3 B \right )}{15 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{4} \left (i-\tan \left (f x +e \right )\right )^{3}}\)	\(199\)
default	\(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i A \tan \left (f x +e \right )^{6}-2 i B \tan \left (f x +e \right )^{5}-2 B \tan \left (f x +e \right )^{6}+20 i A \tan \left (f x +e \right )^{4}-8 A \tan \left (f x +e \right )^{5}-5 i B \tan \left (f x +e \right )^{3}-5 B \tan \left (f x +e \right )^{4}+15 i A \tan \left (f x +e \right )^{2}-20 A \tan \left (f x +e \right )^{3}-3 i \tan \left (f x +e \right ) B +3 i A -12 A \tan \left (f x +e \right )+3 B \right )}{15 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{4} \left (i-\tan \left (f x +e \right )\right )^{3}}\)	\(199\)
parts	\(-\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i \tan \left (f x +e \right )^{5}+8 \tan \left (f x +e \right )^{6}+20 i \tan \left (f x +e \right )^{3}+20 \tan \left (f x +e \right )^{4}+12 i \tan \left (f x +e \right )+15 \tan \left (f x +e \right )^{2}+3\right )}{15 f \,a^{2} c^{3} \left (i-\tan \left (f x +e \right )\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 i \tan \left (f x +e \right )^{5}+2 \tan \left (f x +e \right )^{6}+5 i \tan \left (f x +e \right )^{3}+5 \tan \left (f x +e \right )^{4}+3 i \tan \left (f x +e \right )-3\right )}{15 f \,a^{2} c^{3} \left (i-\tan \left (f x +e \right )\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\)	\(253\)

[In]

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

[Out]

1/15*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)/a^2/c^3*(8*I*A*tan(f*x+e)^6-2*I*B*tan(f*x+e)^5
-2*B*tan(f*x+e)^6+20*I*A*tan(f*x+e)^4-8*A*tan(f*x+e)^5-5*I*B*tan(f*x+e)^3-5*B*tan(f*x+e)^4+15*I*A*tan(f*x+e)^2
-20*A*tan(f*x+e)^3-3*I*B*tan(f*x+e)+3*I*A-12*A*tan(f*x+e)+3*B)/(I+tan(f*x+e))^4/(I-tan(f*x+e))^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.67 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {{\left (3 \, {\left (i \, A + B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} - {\left (-23 i \, A - 13 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, {\left (11 i \, A + B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 48 \, {\left (-i \, A - B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + 30 \, {\left (i \, A + B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 48 \, {\left (-i \, A - B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + 5 \, {\left (-13 i \, A + 7 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 5 i \, A + 5 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-3 i \, f x - 3 i \, e\right )}}{240 \, a^{2} c^{3} f} \]

[In]

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

[Out]

-1/240*(3*(I*A + B)*e^(10*I*f*x + 10*I*e) - (-23*I*A - 13*B)*e^(8*I*f*x + 8*I*e) + 10*(11*I*A + B)*e^(6*I*f*x
+ 6*I*e) + 48*(-I*A - B)*e^(5*I*f*x + 5*I*e) + 30*(I*A + B)*e^(4*I*f*x + 4*I*e) + 48*(-I*A - B)*e^(3*I*f*x + 3
*I*e) + 5*(-13*I*A + 7*B)*e^(2*I*f*x + 2*I*e) - 5*I*A + 5*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^3*f)

Sympy [F]

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

[In]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

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

[In]

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

Mupad [B] (verification not implemented)

Time = 9.54 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.73 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (40\,A\,\sin \left (2\,e+2\,f\,x\right )+A\,\cos \left (2\,e+2\,f\,x\right )\,20{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-20\,B\,\cos \left (2\,e+2\,f\,x\right )-4\,B\,\cos \left (4\,e+4\,f\,x\right )-A\,45{}\mathrm {i}+4\,A\,\sin \left (4\,e+4\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,10{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}\right )}{120\,a^2\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

[In]

int((A + B*tan(e + f*x))/((a + a*tan(e + f*x)*1i)^(3/2)*(c - c*tan(e + f*x)*1i)^(5/2)),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)*20i - A*4
5i + A*cos(4*e + 4*f*x)*1i - 20*B*cos(2*e + 2*f*x) - 4*B*cos(4*e + 4*f*x) + 40*A*sin(2*e + 2*f*x) + 4*A*sin(4*
e + 4*f*x) + B*sin(2*e + 2*f*x)*10i + B*sin(4*e + 4*f*x)*1i))/(120*a^2*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))