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

Optimal. Leaf size=155 \[ -\frac{2 a^{3/2} B \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{3/2} f}-\frac{(B+i A) (a+i a \tan (e+f x))^{3/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{2 a B \sqrt{a+i a \tan (e+f x)}}{c f \sqrt{c-i c \tan (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.264557, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3588, 78, 47, 63, 217, 203} \[ -\frac{2 a^{3/2} B \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{3/2} f}-\frac{(B+i A) (a+i a \tan (e+f x))^{3/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{2 a B \sqrt{a+i a \tan (e+f x)}}{c f \sqrt{c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3588

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]

Rule 78

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] || LtQ
[p, n]))))

Rule 47

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 63

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 217

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 203

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

Rubi steps

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

Mathematica [A]  time = 8.48806, size = 123, normalized size = 0.79 \[ -\frac{a e^{-i (e+f x)} \sqrt{a+i a \tan (e+f x)} \left (i A e^{3 i (e+f x)}+B e^{i (e+f x)} \left (-6+e^{2 i (e+f x)}\right )+6 B \tan ^{-1}\left (e^{i (e+f x)}\right )\right )}{3 \sqrt{2} c f \sqrt{\frac{c}{1+e^{2 i (e+f x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(I*A*E^((3*I)*(e + f*x)) + B*E^(I*(e + f*x))*(-6 + E^((2*I)*(e + f*x))) + 6*B*ArcTan[E^(I*(e + f*x))])*Sqr
t[a + I*a*Tan[e + f*x]])/(3*Sqrt[2]*c*E^(I*(e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*f)

________________________________________________________________________________________

Maple [B]  time = 0.116, size = 406, normalized size = 2.6 \begin{align*} -{\frac{a}{3\,f{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) } \left ( 3\,iB\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{3}ac-9\,iB\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) \tan \left ( fx+e \right ) ac-7\,iB\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \left ( \tan \left ( fx+e \right ) \right ) ^{2}-9\,B\ln \left ({\frac{ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}}{\sqrt{ac}}} \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{2}ac+A \left ( \tan \left ( fx+e \right ) \right ) ^{2}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+5\,iB\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+3\,B\ln \left ({\frac{ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}}{\sqrt{ac}}} \right ) ac+12\,B\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}\tan \left ( fx+e \right ) +A\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 2.20918, size = 232, normalized size = 1.5 \begin{align*} -\frac{{\left (6 \, B a \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 6 \, B a \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (-i \, A - B\right )} a \cos \left (3 \, f x + 3 \, e\right ) - 12 \, B a \cos \left (f x + e\right ) + 3 i \, B a \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 a \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 ) -{\left (2 \, A - 2 i \, B\right )} a \sin \left (3 \, f x + 3 \, e\right ) - 12 i \, B a \sin \left (f x + e\right )\right )} \sqrt{a}}{6 \, c^{\frac{3}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.59669, size = 946, normalized size = 6.1 \begin{align*} \frac{3 \, c^{2} f \sqrt{-\frac{B^{2} a^{3}}{c^{3} f^{2}}} \log \left (\frac{4 \,{\left (2 \,{\left (B a e^{\left (2 i \, f x + 2 i \, e\right )} + B a\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 )} +{\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{2} f\right )} \sqrt{-\frac{B^{2} a^{3}}{c^{3} f^{2}}}\right )}}{B a e^{\left (2 i \, f x + 2 i \, e\right )} + B a}\right ) - 3 \, c^{2} f \sqrt{-\frac{B^{2} a^{3}}{c^{3} f^{2}}} \log \left (\frac{4 \,{\left (2 \,{\left (B a e^{\left (2 i \, f x + 2 i \, e\right )} + B a\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 )} -{\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{2} f\right )} \sqrt{-\frac{B^{2} a^{3}}{c^{3} f^{2}}}\right )}}{B a e^{\left (2 i \, f x + 2 i \, e\right )} + B a}\right ) +{\left ({\left (-2 i \, A - 2 \, B\right )} a e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-2 i \, A + 10 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + 12 \, B a\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 )}}{6 \, c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*c^2*f*sqrt(-B^2*a^3/(c^3*f^2))*log(4*(2*(B*a*e^(2*I*f*x + 2*I*e) + B*a)*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) + (c^2*f*e^(2*I*f*x + 2*I*e) - c^2*f)*sqrt(-B^2*a^3/(c^3*
f^2)))/(B*a*e^(2*I*f*x + 2*I*e) + B*a)) - 3*c^2*f*sqrt(-B^2*a^3/(c^3*f^2))*log(4*(2*(B*a*e^(2*I*f*x + 2*I*e) +
B*a)*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) - (c^2*f*e^(2*I*f*x
+ 2*I*e) - c^2*f)*sqrt(-B^2*a^3/(c^3*f^2)))/(B*a*e^(2*I*f*x + 2*I*e) + B*a)) + ((-2*I*A - 2*B)*a*e^(4*I*f*x +
4*I*e) + (-2*I*A + 10*B)*a*e^(2*I*f*x + 2*I*e) + 12*B*a)*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))/(c^2*f)

________________________________________________________________________________________

Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (f x + e\right ) + A\right )}{\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{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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