3.799 \(\int \frac{(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{\sqrt{c-i c \tan (e+f x)}} \, dx\)
Optimal. Leaf size=169 \[ \frac{2 a^{3/2} (2 B+i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{c} f}-\frac{a (2 B+i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c f}-\frac{(B+i A) (a+i a \tan (e+f x))^{3/2}}{f \sqrt{c-i c \tan (e+f x)}} \]
[Out]
(2*a^(3/2)*(I*A + 2*B)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(Sqr
t[c]*f) - ((I*A + B)*(a + I*a*Tan[e + f*x])^(3/2))/(f*Sqrt[c - I*c*Tan[e + f*x]]) - (a*(I*A + 2*B)*Sqrt[a + I*
a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(c*f)
________________________________________________________________________________________
Rubi [A] time = 0.269004, antiderivative size = 169, 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, 50, 63, 217, 203} \[ \frac{2 a^{3/2} (2 B+i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{c} f}-\frac{a (2 B+i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c f}-\frac{(B+i A) (a+i a \tan (e+f x))^{3/2}}{f \sqrt{c-i c \tan (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[((a + I*a*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x]))/Sqrt[c - I*c*Tan[e + f*x]],x]
[Out]
(2*a^(3/2)*(I*A + 2*B)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(Sqr
t[c]*f) - ((I*A + B)*(a + I*a*Tan[e + f*x])^(3/2))/(f*Sqrt[c - I*c*Tan[e + f*x]]) - (a*(I*A + 2*B)*Sqrt[a + I*
a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(c*f)
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 50
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 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))}{\sqrt{c-i c \tan (e+f x)}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x} (A+B 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}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{(a (A-2 i B)) \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{3/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{a (i A+2 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c f}-\frac{\left (a^2 (A-2 i 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 )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{3/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{a (i A+2 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c f}+\frac{(2 a (i A+2 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 )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{3/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{a (i A+2 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c f}+\frac{(2 a (i A+2 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 )}{f}\\ &=\frac{2 a^{3/2} (i A+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 )}{\sqrt{c} f}-\frac{(i A+B) (a+i a \tan (e+f x))^{3/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{a (i A+2 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c f}\\ \end{align*}
Mathematica [A] time = 6.42171, size = 190, normalized size = 1.12 \[ \frac{2 a e^{-2 i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} (\tan (e+f x)-i) \sqrt{a+i a \tan (e+f x)} \left (e^{i (e+f x)} \left (A \left (1+e^{2 i (e+f x)}\right )-i B \left (2+e^{2 i (e+f x)}\right )\right )-(A-2 i B) \left (1+e^{2 i (e+f x)}\right ) \tan ^{-1}\left (e^{i (e+f x)}\right )\right )}{c f \sec ^{\frac{3}{2}}(e+f x)} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + I*a*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x]))/Sqrt[c - I*c*Tan[e + f*x]],x]
[Out]
(2*a*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*(E^(I*(e + f*x))*(A*(1
+ E^((2*I)*(e + f*x))) - I*B*(2 + E^((2*I)*(e + f*x)))) - (A - (2*I)*B)*(1 + E^((2*I)*(e + f*x)))*ArcTan[E^(I*
(e + f*x))])*(-I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]])/(c*E^((2*I)*(e + f*x))*f*Sec[e + f*x]^(3/2))
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Maple [B] time = 0.181, size = 497, normalized size = 2.9 \begin{align*} \text{result too large to display} \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))^(1/2),x)
[Out]
1/f*(2*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)^2*a*c-2*I*A*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-A*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-2*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))*a*c-4*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)-4*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-B*tan(f*x+e)^2*(a*c*(1+ta
n(f*x+e)^2))^(1/2)*(a*c)^(1/2)+2*I*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)+A*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+2*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)+3*B*(a
*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a/c/(a*c*(1+ta
n(f*x+e)^2))^(1/2)/(tan(f*x+e)+I)^2/(a*c)^(1/2)
________________________________________________________________________________________
Maxima [B] time = 2.13346, size = 829, normalized size = 4.91 \begin{align*} \text{result too large to display} \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))^(1/2),x, algorithm="maxima")
[Out]
(((2*A - 4*I*B)*a*cos(2*f*x + 2*e) - 2*(-I*A - 2*B)*a*sin(2*f*x + 2*e) + (2*A - 4*I*B)*a)*arctan2(cos(1/2*arct
an2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + ((2*A -
4*I*B)*a*cos(2*f*x + 2*e) - 2*(-I*A - 2*B)*a*sin(2*f*x + 2*e) + (2*A - 4*I*B)*a)*arctan2(cos(1/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - ((4*A - 4*I*B)*a
*cos(2*f*x + 2*e) + 4*(I*A + B)*a*sin(2*f*x + 2*e) + (4*A - 8*I*B)*a)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*
f*x + 2*e))) + ((I*A + 2*B)*a*cos(2*f*x + 2*e) - (A - 2*I*B)*a*sin(2*f*x + 2*e) + (I*A + 2*B)*a)*log(cos(1/2*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin
(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + ((-I*A - 2*B)*a*cos(2*f*x + 2*e) + (A - 2*I*B)*a*sin(
2*f*x + 2*e) + (-I*A - 2*B)*a)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - (4*(I*A +
B)*a*cos(2*f*x + 2*e) - (4*A - 4*I*B)*a*sin(2*f*x + 2*e) + 4*(I*A + 2*B)*a)*sin(1/2*arctan2(sin(2*f*x + 2*e),
cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((-2*I*c*cos(2*f*x + 2*e) + 2*c*sin(2*f*x + 2*e) - 2*I*c)*f)
________________________________________________________________________________________
Fricas [B] time = 1.6406, size = 1112, normalized size = 6.58 \begin{align*} -\frac{c \sqrt{\frac{{\left (4 \, A^{2} - 16 i \, A B - 16 \, B^{2}\right )} a^{3}}{c f^{2}}} f \log \left (\frac{2 \,{\left ({\left ({\left (4 i \, A + 8 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (4 i \, A + 8 \, B\right )} 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 f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )} \sqrt{\frac{{\left (4 \, A^{2} - 16 i \, A B - 16 \, B^{2}\right )} a^{3}}{c f^{2}}}\right )}}{{\left (i \, A + 2 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, A + 2 \, B\right )} a}\right ) - c \sqrt{\frac{{\left (4 \, A^{2} - 16 i \, A B - 16 \, B^{2}\right )} a^{3}}{c f^{2}}} f \log \left (\frac{2 \,{\left ({\left ({\left (4 i \, A + 8 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (4 i \, A + 8 \, B\right )} 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 f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )} \sqrt{\frac{{\left (4 \, A^{2} - 16 i \, A B - 16 \, B^{2}\right )} a^{3}}{c f^{2}}}\right )}}{{\left (i \, A + 2 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, A + 2 \, B\right )} a}\right ) - 2 \,{\left ({\left (-4 i \, A - 4 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-4 i \, A - 8 \, B\right )} 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 )}}{4 \, c 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))^(1/2),x, algorithm="fricas")
[Out]
-1/4*(c*sqrt((4*A^2 - 16*I*A*B - 16*B^2)*a^3/(c*f^2))*f*log(2*(((4*I*A + 8*B)*a*e^(2*I*f*x + 2*I*e) + (4*I*A +
8*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*f*e^(2*I*f*x
+ 2*I*e) - c*f)*sqrt((4*A^2 - 16*I*A*B - 16*B^2)*a^3/(c*f^2)))/((I*A + 2*B)*a*e^(2*I*f*x + 2*I*e) + (I*A + 2*
B)*a)) - c*sqrt((4*A^2 - 16*I*A*B - 16*B^2)*a^3/(c*f^2))*f*log(2*(((4*I*A + 8*B)*a*e^(2*I*f*x + 2*I*e) + (4*I*
A + 8*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*f*e^(2*I*
f*x + 2*I*e) - c*f)*sqrt((4*A^2 - 16*I*A*B - 16*B^2)*a^3/(c*f^2)))/((I*A + 2*B)*a*e^(2*I*f*x + 2*I*e) + (I*A +
2*B)*a)) - 2*((-4*I*A - 4*B)*a*e^(2*I*f*x + 2*I*e) + (-4*I*A - 8*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*f)
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**(1/2),x)
[Out]
Timed out
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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 \, a \tan \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}{\sqrt{-i \, c \tan \left (f x + e\right ) + c}}\,{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))^(1/2),x, algorithm="giac")
[Out]
integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(3/2)/sqrt(-I*c*tan(f*x + e) + c), x)