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

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Rubi [A]  time = 0.26, antiderivative size = 155, normalized size of antiderivative = 1.00, 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 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 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 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])

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 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]

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*}

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Mathematica [A]  time = 9.20, 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]

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

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fricas [B]  time = 1.77, size = 381, normalized size = 2.46 \[ \frac {3 \, c^{2} f \sqrt {-\frac {B^{2} a^{3}}{c^{3} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (B a e^{\left (3 i \, f x + 3 i \, e\right )} + B a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{2} f\right )} \sqrt {-\frac {B^{2} a^{3}}{c^{3} f^{2}}}\right )}}{B a e^{\left (2 i \, f x + 2 i \, e\right )} + B a}\right ) - 3 \, c^{2} f \sqrt {-\frac {B^{2} a^{3}}{c^{3} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (B a e^{\left (3 i \, f x + 3 i \, e\right )} + B a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{2} f\right )} \sqrt {-\frac {B^{2} a^{3}}{c^{3} f^{2}}}\right )}}{B a e^{\left (2 i \, f x + 2 i \, e\right )} + B a}\right ) + {\left ({\left (-2 i \, A - 2 \, B\right )} a e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (-2 i \, A + 10 \, B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + 12 \, B a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{6 \, c^{2} f} \]

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^(3*I*f*x + 3*I*e) + B*a*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)) + (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^(3*I*f*x + 3*I*e) +
B*a*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)) - (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^(5*I*f*x +
5*I*e) + (-2*I*A + 10*B)*a*e^(3*I*f*x + 3*I*e) + 12*B*a*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqr
t(c/(e^(2*I*f*x + 2*I*e) + 1)))/(c^2*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [B]  time = 0.51, size = 406, normalized size = 2.62 \[ -\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a \left (3 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{3}\left (f x +e \right )\right ) a c -9 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c -7 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \left (\tan ^{2}\left (f x +e \right )\right )-9 B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c +A \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+5 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+3 B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +12 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{3 f \,c^{2} \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan \left (f x +e \right )+i\right )^{3} \sqrt {c a}} \]

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

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maxima [A]  time = 0.85, size = 169, normalized size = 1.09 \[ -\frac {{\left (6 \, B a \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 6 \, B a \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) + {\left (2 i \, A + 2 \, B\right )} a \cos \left (3 \, f x + 3 \, e\right ) - 12 \, B a \cos \left (f x + e\right ) + 3 i \, B a \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - 3 i \, B a \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (A - i \, B\right )} a \sin \left (3 \, f x + 3 \, e\right ) - 12 i \, B a \sin \left (f x + e\right )\right )} \sqrt {a}}{6 \, c^{\frac {3}{2}} f} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (e + f x \right )}\right )}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}}}\, dx \]

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]

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

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