3.822 \(\int \frac{(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=285 \[ -\frac{5 a^{7/2} (5 B+2 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{3/2} f}+\frac{5 a^3 (5 B+2 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^2 f}+\frac{5 a^2 (5 B+2 i A) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c^2 f}+\frac{2 a (5 B+2 i A) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \]
[Out]
(-5*a^(7/2)*((2*I)*A + 5*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])^(7/2))/(3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (2*a*((2*I)*A + 5
*B)*(a + I*a*Tan[e + f*x])^(5/2))/(3*c*f*Sqrt[c - I*c*Tan[e + f*x]]) + (5*a^3*((2*I)*A + 5*B)*Sqrt[a + I*a*Tan
[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*c^2*f) + (5*a^2*((2*I)*A + 5*B)*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c
- I*c*Tan[e + f*x]])/(6*c^2*f)
________________________________________________________________________________________
Rubi [A] time = 0.34433, antiderivative size = 285, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 7, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used =
{3588, 78, 47, 50, 63, 217, 203} \[ -\frac{5 a^{7/2} (5 B+2 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{3/2} f}+\frac{5 a^3 (5 B+2 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^2 f}+\frac{5 a^2 (5 B+2 i A) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c^2 f}+\frac{2 a (5 B+2 i A) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(3/2),x]
[Out]
(-5*a^(7/2)*((2*I)*A + 5*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])^(7/2))/(3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (2*a*((2*I)*A + 5
*B)*(a + I*a*Tan[e + f*x])^(5/2))/(3*c*f*Sqrt[c - I*c*Tan[e + f*x]]) + (5*a^3*((2*I)*A + 5*B)*Sqrt[a + I*a*Tan
[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*c^2*f) + (5*a^2*((2*I)*A + 5*B)*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c
- I*c*Tan[e + f*x]])/(6*c^2*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 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 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))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2} (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))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}-\frac{(a (2 A-5 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{2 a (2 i A+5 B) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{\left (5 a^2 (2 A-5 i B)\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{3 c f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{2 a (2 i A+5 B) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{5 a^2 (2 i A+5 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c^2 f}+\frac{\left (5 a^3 (2 A-5 i B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 c f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{2 a (2 i A+5 B) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{5 a^3 (2 i A+5 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^2 f}+\frac{5 a^2 (2 i A+5 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c^2 f}+\frac{\left (5 a^4 (2 A-5 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 )}{2 c f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{2 a (2 i A+5 B) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{5 a^3 (2 i A+5 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^2 f}+\frac{5 a^2 (2 i A+5 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c^2 f}-\frac{\left (5 a^3 (2 i A+5 B)\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 )}{c f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{2 a (2 i A+5 B) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{5 a^3 (2 i A+5 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^2 f}+\frac{5 a^2 (2 i A+5 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c^2 f}-\frac{\left (5 a^3 (2 i A+5 B)\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 )}{c f}\\ &=-\frac{5 a^{7/2} (2 i A+5 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))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{2 a (2 i A+5 B) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{5 a^3 (2 i A+5 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^2 f}+\frac{5 a^2 (2 i A+5 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c^2 f}\\ \end{align*}
Mathematica [A] time = 17.4703, size = 517, normalized size = 1.81 \[ \frac{\cos ^4(e+f x) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left ((11 B+5 i A) \cos (2 f x) \left (\frac{2 \cos (e)}{3 c^2}-\frac{2 i \sin (e)}{3 c^2}\right )+(A-i B) \cos (4 f x) \left (\frac{2 \sin (e)}{3 c^2}-\frac{2 i \cos (e)}{3 c^2}\right )+(5 A-11 i B) \sin (2 f x) \left (-\frac{2 \cos (e)}{3 c^2}+\frac{2 i \sin (e)}{3 c^2}\right )+(A-i B) \sin (4 f x) \left (\frac{2 \cos (e)}{3 c^2}+\frac{2 i \sin (e)}{3 c^2}\right )+\sec (e) \left (\frac{\cos (3 e)}{2 c^2}-\frac{i \sin (3 e)}{2 c^2}\right ) (10 i A \cos (e)+i B \sin (e)+26 B \cos (e))+i B \sec (e) \sin (f x) \left (\frac{\cos (3 e)}{2 c^2}-\frac{i \sin (3 e)}{2 c^2}\right ) \sec (e+f x)\right )}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}-\frac{5 i (2 A-5 i B) \sqrt{e^{i f x}} e^{-i (4 e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{c f \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \sec ^{\frac{9}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{7/2} (A \cos (e+f x)+B \sin (e+f x))} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(3/2),x]
[Out]
((-5*I)*(2*A - (5*I)*B)*Sqrt[E^(I*f*x)]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*ArcTan[E^(I*(e + f*x))
]*(a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c*E^(I*(4*e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*f*
Sec[e + f*x]^(9/2)*(Cos[f*x] + I*Sin[f*x])^(7/2)*(A*Cos[e + f*x] + B*Sin[e + f*x])) + (Cos[e + f*x]^4*(((5*I)*
A + 11*B)*Cos[2*f*x]*((2*Cos[e])/(3*c^2) - (((2*I)/3)*Sin[e])/c^2) + (A - I*B)*Cos[4*f*x]*((((-2*I)/3)*Cos[e])
/c^2 + (2*Sin[e])/(3*c^2)) + Sec[e]*((10*I)*A*Cos[e] + 26*B*Cos[e] + I*B*Sin[e])*(Cos[3*e]/(2*c^2) - ((I/2)*Si
n[3*e])/c^2) + I*B*Sec[e]*Sec[e + f*x]*(Cos[3*e]/(2*c^2) - ((I/2)*Sin[3*e])/c^2)*Sin[f*x] + (5*A - (11*I)*B)*(
(-2*Cos[e])/(3*c^2) + (((2*I)/3)*Sin[e])/c^2)*Sin[2*f*x] + (A - I*B)*((2*Cos[e])/(3*c^2) + (((2*I)/3)*Sin[e])/
c^2)*Sin[4*f*x])*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(a + I*a*Tan[e + f*x])^(7/2)*(A + B*Ta
n[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x]))
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Maple [B] time = 0.115, size = 731, normalized size = 2.6 \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))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x)
[Out]
1/6/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^3/c^2*(6*I*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c
)^(1/2)*tan(f*x+e)^3-114*I*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)-75*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-118*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c
)^(1/2)+30*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)^3*a*c+3*I*B*
(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^4+185*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x
+e)^2+225*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+21*B*(a
*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^3+225*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-30*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))*a*c-90*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-74
*A*tan(f*x+e)^2*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+90*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)^2*a*c-75*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-279*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)+46*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*
(a*c)^(1/2))/(tan(f*x+e)+I)^3/(a*c)^(1/2)/(a*c*(1+tan(f*x+e)^2))^(1/2)
________________________________________________________________________________________
Maxima [B] time = 3.24362, size = 1598, normalized size = 5.61 \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))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
-(((120*A - 300*I*B)*a^3*cos(4*f*x + 4*e) + (240*A - 600*I*B)*a^3*cos(2*f*x + 2*e) - 60*(-2*I*A - 5*B)*a^3*sin
(4*f*x + 4*e) - 120*(-2*I*A - 5*B)*a^3*sin(2*f*x + 2*e) + (120*A - 300*I*B)*a^3)*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) + ((120*A - 300*I*B
)*a^3*cos(4*f*x + 4*e) + (240*A - 600*I*B)*a^3*cos(2*f*x + 2*e) - 60*(-2*I*A - 5*B)*a^3*sin(4*f*x + 4*e) - 120
*(-2*I*A - 5*B)*a^3*sin(2*f*x + 2*e) + (120*A - 300*I*B)*a^3)*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) + ((32*A - 32*I*B)*a^3*cos(4*f*x + 4*
e) + (64*A - 64*I*B)*a^3*cos(2*f*x + 2*e) - 32*(-I*A - B)*a^3*sin(4*f*x + 4*e) - 64*(-I*A - B)*a^3*sin(2*f*x +
2*e) - (16*A - 232*I*B)*a^3)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - ((192*A - 384*I*B)*a^3*co
s(4*f*x + 4*e) + (384*A - 768*I*B)*a^3*cos(2*f*x + 2*e) + 192*(I*A + 2*B)*a^3*sin(4*f*x + 4*e) + 384*(I*A + 2*
B)*a^3*sin(2*f*x + 2*e) + (240*A - 600*I*B)*a^3)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (30*(-
2*I*A - 5*B)*a^3*cos(4*f*x + 4*e) + 60*(-2*I*A - 5*B)*a^3*cos(2*f*x + 2*e) + (60*A - 150*I*B)*a^3*sin(4*f*x +
4*e) + (120*A - 300*I*B)*a^3*sin(2*f*x + 2*e) + 30*(-2*I*A - 5*B)*a^3)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), c
os(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) - (30*(2*I*A + 5*B)*a^3*cos(4*f*x + 4*e) + 60*(2*I*A + 5*B)*a^3*cos(2*f*x + 2*e)
- (60*A - 150*I*B)*a^3*sin(4*f*x + 4*e) - (120*A - 300*I*B)*a^3*sin(2*f*x + 2*e) + 30*(2*I*A + 5*B)*a^3)*log(c
os(1/2*arctan2(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) - (32*(-I*A - B)*a^3*cos(4*f*x + 4*e) + 64*(-I*
A - B)*a^3*cos(2*f*x + 2*e) + (32*A - 32*I*B)*a^3*sin(4*f*x + 4*e) + (64*A - 64*I*B)*a^3*sin(2*f*x + 2*e) + 8*
(2*I*A + 29*B)*a^3)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (192*(I*A + 2*B)*a^3*cos(4*f*x + 4*
e) + 384*(I*A + 2*B)*a^3*cos(2*f*x + 2*e) - (192*A - 384*I*B)*a^3*sin(4*f*x + 4*e) - (384*A - 768*I*B)*a^3*sin
(2*f*x + 2*e) + 120*(2*I*A + 5*B)*a^3)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/(
(-24*I*c^2*cos(4*f*x + 4*e) - 48*I*c^2*cos(2*f*x + 2*e) + 24*c^2*sin(4*f*x + 4*e) + 48*c^2*sin(2*f*x + 2*e) -
24*I*c^2)*f)
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Fricas [B] time = 1.65747, size = 1477, normalized size = 5.18 \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))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
1/12*(2*((-8*I*A - 8*B)*a^3*e^(6*I*f*x + 6*I*e) + (32*I*A + 80*B)*a^3*e^(4*I*f*x + 4*I*e) + (100*I*A + 250*B)*
a^3*e^(2*I*f*x + 2*I*e) + (60*I*A + 150*B)*a^3)*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) + 3*(c^2*f*e^(2*I*f*x + 2*I*e) + c^2*f)*sqrt((100*A^2 - 500*I*A*B - 625*B^2)*a^7/(c^3*f^
2))*log(2*(((40*I*A + 100*B)*a^3*e^(2*I*f*x + 2*I*e) + (40*I*A + 100*B)*a^3)*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) + 2*(c^2*f*e^(2*I*f*x + 2*I*e) - c^2*f)*sqrt((100*A^2 - 500
*I*A*B - 625*B^2)*a^7/(c^3*f^2)))/((10*I*A + 25*B)*a^3*e^(2*I*f*x + 2*I*e) + (10*I*A + 25*B)*a^3)) - 3*(c^2*f*
e^(2*I*f*x + 2*I*e) + c^2*f)*sqrt((100*A^2 - 500*I*A*B - 625*B^2)*a^7/(c^3*f^2))*log(2*(((40*I*A + 100*B)*a^3*
e^(2*I*f*x + 2*I*e) + (40*I*A + 100*B)*a^3)*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) - 2*(c^2*f*e^(2*I*f*x + 2*I*e) - c^2*f)*sqrt((100*A^2 - 500*I*A*B - 625*B^2)*a^7/(c^3*f^2)))
/((10*I*A + 25*B)*a^3*e^(2*I*f*x + 2*I*e) + (10*I*A + 25*B)*a^3)))/(c^2*f*e^(2*I*f*x + 2*I*e) + c^2*f)
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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))**(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(3/2),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{7}{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))^(7/2)*(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)^(7/2)/(-I*c*tan(f*x + e) + c)^(3/2), x)