\(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx\) [840]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, 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, 37} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {B+i A}{f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(B+2 i A) \sqrt {c-i c \tan (e+f x)}}{3 a c f \sqrt {a+i a \tan (e+f x)}}+\frac {(B+2 i A) \sqrt {c-i c \tan (e+f x)}}{3 c f (a+i a \tan (e+f x))^{3/2}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 3.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.64 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {A+i B+(-2 i A-B) \tan (e+f x)+(2 A-i B) \tan ^2(e+f x)}{3 a f (-i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i A \tan \left (f x +e \right )^{4}-i B \tan \left (f x +e \right )^{3}+B \tan \left (f x +e \right )^{4}+3 i A \tan \left (f x +e \right )^{2}+2 A \tan \left (f x +e \right )^{3}-i \tan \left (f x +e \right ) B +i A +2 A \tan \left (f x +e \right )-B \right )}{3 f \,a^{2} c \left (i-\tan \left (f x +e \right )\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}\)	\(152\)
default	\(\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i A \tan \left (f x +e \right )^{4}-i B \tan \left (f x +e \right )^{3}+B \tan \left (f x +e \right )^{4}+3 i A \tan \left (f x +e \right )^{2}+2 A \tan \left (f x +e \right )^{3}-i \tan \left (f x +e \right ) B +i A +2 A \tan \left (f x +e \right )-B \right )}{3 f \,a^{2} c \left (i-\tan \left (f x +e \right )\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}\)	\(152\)
parts	\(\frac {A \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i \tan \left (f x +e \right )^{3}-2 \tan \left (f x +e \right )^{4}+2 i \tan \left (f x +e \right )-3 \tan \left (f x +e \right )^{2}-1\right )}{3 f \,a^{2} c \left (i-\tan \left (f x +e \right )\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {i B \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (i \tan \left (f x +e \right )^{3}-\tan \left (f x +e \right )^{4}+i \tan \left (f x +e \right )+1\right )}{3 f \,a^{2} c \left (i-\tan \left (f x +e \right )\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}\)	\(211\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.95 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {{\left (3 \, {\left (i \, A + B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 4 \, {\left (i \, A - B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + 3 \, {\left (-i \, A + B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, {\left (i \, A - B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} - {\left (7 i \, A - B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-3 i \, f x - 3 i \, e\right )}}{12 \, a^{2} c f} \]

[In]

integrate((A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/12*(3*(I*A + B)*e^(6*I*f*x + 6*I*e) + 4*(I*A - B)*e^(5*I*f*x + 5*I*e) + 3*(-I*A + B)*e^(4*I*f*x + 4*I*e) +
4*(I*A - B)*e^(3*I*f*x + 3*I*e) - (7*I*A - 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*f)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

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

Mupad [B] (verification not implemented)

Time = 9.36 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.12 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (6\,A\,\sin \left (2\,e+2\,f\,x\right )-3\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-B\,\cos \left (4\,e+4\,f\,x\right )-A\,3{}\mathrm {i}+A\,\sin \left (4\,e+4\,f\,x\right )+B\,\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}\right )}{12\,a^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

[In]

int((A + B*tan(e + f*x))/((a + a*tan(e + f*x)*1i)^(3/2)*(c - c*tan(e + f*x)*1i)^(1/2)),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)*6i - 3*B
- A*3i + A*cos(4*e + 4*f*x)*1i - B*cos(4*e + 4*f*x) + 6*A*sin(2*e + 2*f*x) + A*sin(4*e + 4*f*x) + B*sin(4*e +
4*f*x)*1i))/(12*a^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))