\(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx\) [838]

Optimal result

Integrand size = 45, antiderivative size = 157 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {2 B c^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{3/2} f}+\frac {2 B c \sqrt {c-i c \tan (e+f x)}}{a f \sqrt {a+i a \tan (e+f x)}}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3669, 79, 49, 65, 223, 209} \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {2 B c^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{3/2} f}+\frac {(-B+i A) (c-i c \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {2 B c \sqrt {c-i c \tan (e+f x)}}{a f \sqrt {a+i a \tan (e+f x)}} \]

[In]

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

[Out]

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

Rule 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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

Mathematica [A] (verified)

Time = 5.23 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.87 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {c \sqrt {c-i c \tan (e+f x)} \left (\frac {6 B \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {1-i \tan (e+f x)}}+\frac {\sqrt {a} (A-5 i B+(-i A+7 B) \tan (e+f x))}{(-i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}}\right )}{3 a^{3/2} f} \]

[In]

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

[Out]

(c*Sqrt[c - I*c*Tan[e + f*x]]*((6*B*ArcSin[Sqrt[a + I*a*Tan[e + f*x]]/(Sqrt[2]*Sqrt[a])])/Sqrt[1 - I*Tan[e + f
*x]] + (Sqrt[a]*(A - (5*I)*B + ((-I)*A + 7*B)*Tan[e + f*x]))/((-I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]]))
)/(3*a^(3/2)*f)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 407 vs. \(2 (128 ) = 256\).

Time = 0.34 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.60

method	result	size
derivativedivides	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (-3 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{3}+9 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+7 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}-9 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}+A \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )^{2}-5 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+3 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +12 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{3 f \,a^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i-\tan \left (f x +e \right )\right )^{3} \sqrt {a c}}\)	\(408\)
default	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (-3 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{3}+9 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+7 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}-9 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}+A \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )^{2}-5 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+3 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +12 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{3 f \,a^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i-\tan \left (f x +e \right )\right )^{3} \sqrt {a c}}\)	\(408\)
parts	\(\frac {A \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan \left (f x +e \right )^{2}\right )}{3 f \,a^{2} \left (i-\tan \left (f x +e \right )\right )^{3}}+\frac {i B \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (9 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right )^{2} a c -3 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right )^{3} a c -3 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c -12 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+9 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right ) a c +7 \tan \left (f x +e \right )^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}-5 \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{3 f \,a^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i-\tan \left (f x +e \right )\right )^{3} \sqrt {a c}}\)	\(415\)

[In]

int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(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*c*(-3*I*B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c
*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)^3+9*I*B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e
)^2))^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)+7*I*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^2-9*B*ln((a
*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)^2+A*(a*c*(1+tan(f*x+e)^2))
^(1/2)*(a*c)^(1/2)*tan(f*x+e)^2-5*I*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)+3*B*ln((a*c*tan(f*x+e)+(a*c)^(1
/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c+12*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)+A*
(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(I-tan(f*x+e))^3/(a*c)^(1/2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 400 vs. \(2 (119) = 238\).

Time = 0.28 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.55 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=-\frac {{\left (3 \, a^{2} f \sqrt {-\frac {B^{2} c^{3}}{a^{3} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \left (\frac {4 \, {\left (2 \, {\left (B c e^{\left (3 i \, f x + 3 i \, e\right )} + B c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{2} f\right )} \sqrt {-\frac {B^{2} c^{3}}{a^{3} f^{2}}}\right )}}{B c e^{\left (2 i \, f x + 2 i \, e\right )} + B c}\right ) - 3 \, a^{2} f \sqrt {-\frac {B^{2} c^{3}}{a^{3} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \left (\frac {4 \, {\left (2 \, {\left (B c e^{\left (3 i \, f x + 3 i \, e\right )} + B c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{2} f\right )} \sqrt {-\frac {B^{2} c^{3}}{a^{3} f^{2}}}\right )}}{B c e^{\left (2 i \, f x + 2 i \, e\right )} + B c}\right ) - 2 \, {\left (6 \, B c e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (-i \, A - 5 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (-i \, A + B\right )} c\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{6 \, a^{2} f} \]

[In]

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

[Out]

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

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {{\left (6 \, B c \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 6 \, B c \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (-i \, A + B\right )} c \cos \left (3 \, f x + 3 \, e\right ) + 12 \, B c \cos \left (f x + e\right ) + 3 i \, B c \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - 3 i \, B c \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (A + i \, B\right )} c \sin \left (3 \, f x + 3 \, e\right ) - 12 i \, B c \sin \left (f x + e\right )\right )} \sqrt {c}}{6 \, a^{\frac {3}{2}} f} \]

[In]

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

[Out]

1/6*(6*B*c*arctan2(cos(f*x + e), sin(f*x + e) + 1) + 6*B*c*arctan2(cos(f*x + e), -sin(f*x + e) + 1) - 2*(-I*A
+ B)*c*cos(3*f*x + 3*e) + 12*B*c*cos(f*x + e) + 3*I*B*c*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) +
1) - 3*I*B*c*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) + 2*(A + I*B)*c*sin(3*f*x + 3*e) - 12*
I*B*c*sin(f*x + e))*sqrt(c)/(a^(3/2)*f)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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