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

Optimal. Leaf size=157 \[ \frac{2 B c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{a^{3/2} f}+\frac{(-B+i A) (c-i c \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac{2 B c \sqrt{c-i c \tan (e+f x)}}{a f \sqrt{a+i a \tan (e+f x)}} \]

[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))

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Rubi [A]  time = 0.265271, antiderivative size = 157, 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} \[ \frac{2 B c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{a^{3/2} f}+\frac{(-B+i A) (c-i c \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac{2 B c \sqrt{c-i c \tan (e+f x)}}{a f \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[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 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+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{(a c) \operatorname{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) \operatorname{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 ) \operatorname{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 ) \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 )}{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 ) \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 )}{a^2 f}\\ &=\frac{2 B c^{3/2} \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 )}{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]  time = 7.36481, size = 114, normalized size = 0.73 \[ \frac{\sqrt{2} c e^{-2 i (e+f x)} \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \left (i A+B \left (-1+6 e^{2 i (e+f x)}\right )+6 B e^{3 i (e+f x)} \tan ^{-1}\left (e^{i (e+f x)}\right )\right )}{3 a f \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[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]

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

________________________________________________________________________________________

Maple [B]  time = 0.119, size = 408, normalized size = 2.6 \begin{align*}{\frac{c}{3\,f{a}^{2} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{3}}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \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+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]

1/3/f*(-c*(-1+I*tan(f*x+e)))^(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+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)

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Maxima [A]  time = 2.84403, size = 228, normalized size = 1.45 \begin{align*} \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 ) +{\left (2 \, A + 2 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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 + 2*I*B)*c*sin(3*f*x + 3*e) - 1
2*I*B*c*sin(f*x + e))*sqrt(c)/(a^(3/2)*f)

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Fricas [B]  time = 1.6577, size = 1143, normalized size = 7.28 \begin{align*} -\frac{{\left (3 \, a^{2} f \sqrt{-\frac{B^{2} c^{3}}{a^{3} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac{2 \,{\left (B c e^{\left (2 i \, f x + 2 i \, e\right )} + B 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}} e^{\left (i \, f x + i \, e\right )} +{\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}}}}{2 \,{\left (B c e^{\left (2 i \, f x + 2 i \, e\right )} + B c\right )}}\right ) - 3 \, a^{2} f \sqrt{-\frac{B^{2} c^{3}}{a^{3} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac{2 \,{\left (B c e^{\left (2 i \, f x + 2 i \, e\right )} + B 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}} e^{\left (i \, f x + i \, e\right )} -{\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}}}}{2 \,{\left (B c e^{\left (2 i \, f x + 2 i \, e\right )} + B c\right )}}\right ) -{\left ({\left (-2 i \, A - 10 \, B\right )} c e^{\left (5 i \, f x + 5 i \, e\right )} + 12 \, B c e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-2 i \, A - 10 \, B\right )} c e^{\left (3 i \, f x + 3 i \, e\right )} +{\left (2 i \, A + 10 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (2 i \, A - 2 \, 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}} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{6 \, a^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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^(4*I*f*x + 4*I*e)*log(-1/2*(2*(B*c*e^(2*I*f*x + 2*I*e) + 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^(I*f*x + I*e) + (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^(4*I*f*x
+ 4*I*e)*log(-1/2*(2*(B*c*e^(2*I*f*x + 2*I*e) + 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^(I*f*x + I*e) - (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*I*A - 10*B)*c*e^(5*I*f*x + 5*I*e) + 12*B*c*e^(4*I*f*x + 4*I*e) + (-2*I*A - 10*B)*c*e^(3*I
*f*x + 3*I*e) + (2*I*A + 10*B)*c*e^(2*I*f*x + 2*I*e) + (2*I*A - 2*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^(I*f*x + I*e))*e^(-4*I*f*x - 4*I*e)/(a^2*f)

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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+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]

Exception raised: AttributeError

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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 \, 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)