\(\int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \, dx\) [835]

Optimal result

Integrand size = 45, antiderivative size = 213 \[ \int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {i A-B}{f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {(3 i A-2 B) \sqrt {a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (3 i A-2 B) \sqrt {a+i a \tan (e+f x)}}{15 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 (3 i A-2 B) \sqrt {a+i a \tan (e+f x)}}{15 a c^2 f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3669, 79, 47, 37} \[ \int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {2 (-2 B+3 i A) \sqrt {a+i a \tan (e+f x)}}{15 a c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {2 (-2 B+3 i A) \sqrt {a+i a \tan (e+f x)}}{15 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {(-2 B+3 i A) \sqrt {a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}+\frac {-B+i A}{f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \]

[In]

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

[Out]

(I*A - B)/(f*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(5/2)) - (((3*I)*A - 2*B)*Sqrt[a + I*a*Tan[e +
f*x]])/(5*a*f*(c - I*c*Tan[e + f*x])^(5/2)) - (2*((3*I)*A - 2*B)*Sqrt[a + I*a*Tan[e + f*x]])/(15*a*c*f*(c - I*
c*Tan[e + f*x])^(3/2)) - (2*((3*I)*A - 2*B)*Sqrt[a + I*a*Tan[e + f*x]])/(15*a*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)^{3/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {i A-B}{f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}+\frac {((3 A+2 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 )}{f} \\ & = \frac {i A-B}{f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {(3 i A-2 B) \sqrt {a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}+\frac {(2 (3 A+2 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 f} \\ & = \frac {i A-B}{f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {(3 i A-2 B) \sqrt {a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (3 i A-2 B) \sqrt {a+i a \tan (e+f x)}}{15 a c f (c-i c \tan (e+f x))^{3/2}}+\frac {(2 (3 A+2 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 c f} \\ & = \frac {i A-B}{f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {(3 i A-2 B) \sqrt {a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (3 i A-2 B) \sqrt {a+i a \tan (e+f x)}}{15 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 (3 i A-2 B) \sqrt {a+i a \tan (e+f x)}}{15 a c^2 f \sqrt {c-i c \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.24 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.54 \[ \int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {6 i A+B-(3 A+2 i B) \tan (e+f x)+(12 i A-8 B) \tan ^2(e+f x)+(6 A+4 i B) \tan ^3(e+f x)}{15 c^2 f (i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

((6*I)*A + B - (3*A + (2*I)*B)*Tan[e + f*x] + ((12*I)*A - 8*B)*Tan[e + f*x]^2 + (6*A + (4*I)*B)*Tan[e + f*x]^3
)/(15*c^2*f*(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 = 0.47 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.69

method	result	size
risch	\(-\frac {3 i A \,{\mathrm e}^{6 i \left (f x +e \right )}+3 B \,{\mathrm e}^{6 i \left (f x +e \right )}+15 i A \,{\mathrm e}^{4 i \left (f x +e \right )}+5 B \,{\mathrm e}^{4 i \left (f x +e \right )}+45 i A \,{\mathrm e}^{2 i \left (f x +e \right )}-15 B \,{\mathrm e}^{2 i \left (f x +e \right )}-15 i A +15 B}{120 c^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) f}\)	\(148\)
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (4 i B \tan \left (f x +e \right )^{5}+12 i A \tan \left (f x +e \right )^{4}+6 A \tan \left (f x +e \right )^{5}+2 i B \tan \left (f x +e \right )^{3}-8 B \tan \left (f x +e \right )^{4}+18 i A \tan \left (f x +e \right )^{2}+3 A \tan \left (f x +e \right )^{3}-2 i \tan \left (f x +e \right ) B -7 B \tan \left (f x +e \right )^{2}+6 i A -3 A \tan \left (f x +e \right )+B \right )}{15 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4} \left (i-\tan \left (f x +e \right )\right )^{2}}\)	\(184\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (4 i B \tan \left (f x +e \right )^{5}+12 i A \tan \left (f x +e \right )^{4}+6 A \tan \left (f x +e \right )^{5}+2 i B \tan \left (f x +e \right )^{3}-8 B \tan \left (f x +e \right )^{4}+18 i A \tan \left (f x +e \right )^{2}+3 A \tan \left (f x +e \right )^{3}-2 i \tan \left (f x +e \right ) B -7 B \tan \left (f x +e \right )^{2}+6 i A -3 A \tan \left (f x +e \right )+B \right )}{15 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4} \left (i-\tan \left (f x +e \right )\right )^{2}}\)	\(184\)
parts	\(\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (4 i \tan \left (f x +e \right )^{4}+2 \tan \left (f x +e \right )^{5}+6 i \tan \left (f x +e \right )^{2}+\tan \left (f x +e \right )^{3}+2 i-\tan \left (f x +e \right )\right )}{5 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4} \left (i-\tan \left (f x +e \right )\right )^{2}}+\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (1+\tan \left (f x +e \right )^{2}\right ) \left (8 i \tan \left (f x +e \right )^{2}+4 \tan \left (f x +e \right )^{3}-i-2 \tan \left (f x +e \right )\right )}{15 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4} \left (i-\tan \left (f x +e \right )\right )^{2}}\)	\(230\)

[In]

int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/120/c^2/(a*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))^(1/2)/(c/(exp(2*I*(f*x+e))+1))^(1/2)/(exp(2*I*(f*x+e))+1)
*(3*I*A*exp(6*I*(f*x+e))+3*B*exp(6*I*(f*x+e))+15*I*A*exp(4*I*(f*x+e))+5*B*exp(4*I*(f*x+e))+45*I*A*exp(2*I*(f*x
+e))-15*B*exp(2*I*(f*x+e))-15*I*A+15*B)/f

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.76 \[ \int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {{\left (3 \, {\left (i \, A + B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 2 \, {\left (9 i \, A + 4 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, {\left (6 i \, A - B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 8 \, {\left (-6 i \, A - B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + 30 i \, A e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (-6 i \, A - B\right )} e^{\left (i \, f x + i \, e\right )} - 15 i \, A + 15 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-i \, f x - i \, e\right )}}{120 \, a c^{3} f} \]

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/120*(3*(I*A + B)*e^(8*I*f*x + 8*I*e) + 2*(9*I*A + 4*B)*e^(6*I*f*x + 6*I*e) + 10*(6*I*A - B)*e^(4*I*f*x + 4*
I*e) + 8*(-6*I*A - B)*e^(3*I*f*x + 3*I*e) + 30*I*A*e^(2*I*f*x + 2*I*e) + 8*(-6*I*A - B)*e^(I*f*x + I*e) - 15*I
*A + 15*B)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-I*f*x - I*e)/(a*c^3*f)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(1/2)/(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]

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

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.34 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.87 \[ \int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,45{}\mathrm {i}-15\,B+A\,\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}+20\,B\,\cos \left (2\,e+2\,f\,x\right )+3\,B\,\cos \left (4\,e+4\,f\,x\right )-30\,A\,\sin \left (2\,e+2\,f\,x\right )-3\,A\,\sin \left (4\,e+4\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,10{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}\right )}{120\,a\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

[In]

int((A + B*tan(e + f*x))/((a + a*tan(e + f*x)*1i)^(1/2)*(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*45i - 15*B + A*cos(4*e +
4*f*x)*3i + 20*B*cos(2*e + 2*f*x) + 3*B*cos(4*e + 4*f*x) - 30*A*sin(2*e + 2*f*x) - 3*A*sin(4*e + 4*f*x) - B*si
n(2*e + 2*f*x)*10i + B*sin(4*e + 4*f*x)*3i))/(120*a*c^2*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(c
os(2*e + 2*f*x) + 1))^(1/2))