\(\int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{3/2}} \, dx\) [839]

Optimal result

Integrand size = 45, antiderivative size = 104 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A+2 B) \sqrt {c-i c \tan (e+f x)}}{3 a f \sqrt {a+i a \tan (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3669, 79, 37} \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {(2 B+i A) \sqrt {c-i c \tan (e+f x)}}{3 a f \sqrt {a+i a \tan (e+f x)}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 3.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.78 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {(2 A-i B+(i A+2 B) \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{3 a f (-i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i B \tan \left (f x +e \right )^{2}+3 i A \tan \left (f x +e \right )-A \tan \left (f x +e \right )^{2}-i B +3 B \tan \left (f x +e \right )+2 A \right )}{3 f \,a^{2} \left (i-\tan \left (f x +e \right )\right )^{3}}\)	\(103\)
default	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i B \tan \left (f x +e \right )^{2}+3 i A \tan \left (f x +e \right )-A \tan \left (f x +e \right )^{2}-i B +3 B \tan \left (f x +e \right )+2 A \right )}{3 f \,a^{2} \left (i-\tan \left (f x +e \right )\right )^{3}}\)	\(103\)
parts	\(\frac {A \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (3 i \tan \left (f x +e \right )-\tan \left (f x +e \right )^{2}+2\right )}{3 f \,a^{2} \left (i-\tan \left (f x +e \right )\right )^{3}}-\frac {i B \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (3 i \tan \left (f x +e \right )-2 \tan \left (f x +e \right )^{2}+1\right )}{3 f \,a^{2} \left (i-\tan \left (f x +e \right )\right )^{3}}\)	\(151\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{3/2}} \, dx=-\frac {{\left (3 \, {\left (-i \, A - B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-2 i \, A - B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-3 i \, f x - 3 i \, e\right )}}{6 \, a^{2} f} \]

[In]

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

[Out]

-1/6*(3*(-I*A - B)*e^(4*I*f*x + 4*I*e) + 2*(-2*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*f)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.69 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.88 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,3{}\mathrm {i}+3\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,4{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+2\,B\,\cos \left (2\,e+2\,f\,x\right )-B\,\cos \left (4\,e+4\,f\,x\right )+4\,A\,\sin \left (2\,e+2\,f\,x\right )+A\,\sin \left (4\,e+4\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,2{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}\right )}{12\,a^2\,f} \]

[In]

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