3.836 \(\int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=287 \[ -\frac{5 c^{7/2} (-5 B+2 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{a^{3/2} f}-\frac{5 c^3 (-5 B+2 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a^2 f}-\frac{5 c^2 (-5 B+2 i A) \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^2 f}-\frac{2 c (-5 B+2 i A) (c-i c \tan (e+f x))^{5/2}}{3 a f \sqrt{a+i a \tan (e+f x)}}+\frac{(-B+i A) (c-i c \tan (e+f x))^{7/2}}{3 f (a+i a \tan (e+f x))^{3/2}} \]
[Out]
(-5*((2*I)*A - 5*B)*c^(7/2)*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) - (5*((2*I)*A - 5*B)*c^3*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*a^2*f) - (5*((
2*I)*A - 5*B)*c^2*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(3/2))/(6*a^2*f) - (2*((2*I)*A - 5*B)*c*(c
- I*c*Tan[e + f*x])^(5/2))/(3*a*f*Sqrt[a + I*a*Tan[e + f*x]]) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(7/2))/(3*f
*(a + I*a*Tan[e + f*x])^(3/2))
________________________________________________________________________________________
Rubi [A] time = 0.342676, antiderivative size = 287, 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} \[ -\frac{5 c^{7/2} (-5 B+2 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{a^{3/2} f}-\frac{5 c^3 (-5 B+2 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a^2 f}-\frac{5 c^2 (-5 B+2 i A) \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^2 f}-\frac{2 c (-5 B+2 i A) (c-i c \tan (e+f x))^{5/2}}{3 a f \sqrt{a+i a \tan (e+f x)}}+\frac{(-B+i A) (c-i c \tan (e+f x))^{7/2}}{3 f (a+i a \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(7/2))/(a + I*a*Tan[e + f*x])^(3/2),x]
[Out]
(-5*((2*I)*A - 5*B)*c^(7/2)*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) - (5*((2*I)*A - 5*B)*c^3*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*a^2*f) - (5*((
2*I)*A - 5*B)*c^2*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(3/2))/(6*a^2*f) - (2*((2*I)*A - 5*B)*c*(c
- I*c*Tan[e + f*x])^(5/2))/(3*a*f*Sqrt[a + I*a*Tan[e + f*x]]) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(7/2))/(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 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+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) (c-i c x)^{5/2}}{(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))^{7/2}}{3 f (a+i a \tan (e+f x))^{3/2}}-\frac{((2 A+5 i B) c) \operatorname{Subst}\left (\int \frac{(c-i c x)^{5/2}}{(a+i a x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 f}\\ &=-\frac{2 (2 i A-5 B) c (c-i c \tan (e+f x))^{5/2}}{3 a f \sqrt{a+i a \tan (e+f x)}}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac{\left (5 (2 A+5 i B) c^2\right ) \operatorname{Subst}\left (\int \frac{(c-i c x)^{3/2}}{\sqrt{a+i a x}} \, dx,x,\tan (e+f x)\right )}{3 a f}\\ &=-\frac{5 (2 i A-5 B) c^2 \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^2 f}-\frac{2 (2 i A-5 B) c (c-i c \tan (e+f x))^{5/2}}{3 a f \sqrt{a+i a \tan (e+f x)}}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac{\left (5 (2 A+5 i B) c^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c-i c x}}{\sqrt{a+i a x}} \, dx,x,\tan (e+f x)\right )}{2 a f}\\ &=-\frac{5 (2 i A-5 B) c^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a^2 f}-\frac{5 (2 i A-5 B) c^2 \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^2 f}-\frac{2 (2 i A-5 B) c (c-i c \tan (e+f x))^{5/2}}{3 a f \sqrt{a+i a \tan (e+f x)}}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac{\left (5 (2 A+5 i B) c^4\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 a f}\\ &=-\frac{5 (2 i A-5 B) c^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a^2 f}-\frac{5 (2 i A-5 B) c^2 \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^2 f}-\frac{2 (2 i A-5 B) c (c-i c \tan (e+f x))^{5/2}}{3 a f \sqrt{a+i a \tan (e+f x)}}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{3 f (a+i a \tan (e+f x))^{3/2}}-\frac{\left (5 (2 i A-5 B) c^4\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{5 (2 i A-5 B) c^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a^2 f}-\frac{5 (2 i A-5 B) c^2 \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^2 f}-\frac{2 (2 i A-5 B) c (c-i c \tan (e+f x))^{5/2}}{3 a f \sqrt{a+i a \tan (e+f x)}}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{3 f (a+i a \tan (e+f x))^{3/2}}-\frac{\left (5 (2 i A-5 B) c^4\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{5 (2 i A-5 B) c^{7/2} \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{5 (2 i A-5 B) c^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a^2 f}-\frac{5 (2 i A-5 B) c^2 \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^2 f}-\frac{2 (2 i A-5 B) c (c-i c \tan (e+f x))^{5/2}}{3 a f \sqrt{a+i a \tan (e+f x)}}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{3 f (a+i a \tan (e+f x))^{3/2}}\\ \end{align*}
Mathematica [A] time = 13.3648, size = 255, normalized size = 0.89 \[ \frac{\sqrt{\sec (e+f x)} (A+B \tan (e+f x)) \left (\frac{5 c^4 (5 B-2 i A) e^{i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt{\frac{c}{1+e^{2 i (e+f x)}}}}+\frac{1}{12} c^3 \sec ^{\frac{3}{2}}(e+f x) \sqrt{c-i c \tan (e+f x)} (33 (5 B-2 i A) \cos (e+f x)+(71 B-26 i A) \cos (3 (e+f x))+2 \sin (e+f x) ((34 A+79 i B) \cos (2 (e+f x))+34 A+82 i B))\right )}{f (a+i a \tan (e+f x))^{3/2} (A \cos (e+f x)+B \sin (e+f x))} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(7/2))/(a + I*a*Tan[e + f*x])^(3/2),x]
[Out]
(Sqrt[Sec[e + f*x]]*(A + B*Tan[e + f*x])*((5*((-2*I)*A + 5*B)*c^4*E^(I*(e + f*x))*Sqrt[E^(I*(e + f*x))/(1 + E^
((2*I)*(e + f*x)))]*ArcTan[E^(I*(e + f*x))])/Sqrt[c/(1 + E^((2*I)*(e + f*x)))] + (c^3*Sec[e + f*x]^(3/2)*(33*(
(-2*I)*A + 5*B)*Cos[e + f*x] + ((-26*I)*A + 71*B)*Cos[3*(e + f*x)] + 2*(34*A + (82*I)*B + (34*A + (79*I)*B)*Co
s[2*(e + f*x)])*Sin[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/12))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*T
an[e + f*x])^(3/2))
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Maple [B] time = 0.137, size = 733, 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+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e))^(3/2),x)
[Out]
1/6/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^3/a^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))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(a*c)^(1/2)/(-tan(f*x+e)+I)^3
________________________________________________________________________________________
Maxima [B] time = 5.96431, size = 1854, normalized size = 6.46 \begin{align*} \text{result too large to display} \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))^(7/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
-((1440*A + 3600*I*B)*c^3*cos(6*f*x + 6*e) + (2400*A + 6000*I*B)*c^3*cos(4*f*x + 4*e) + (768*A + 1920*I*B)*c^3
*cos(2*f*x + 2*e) + 720*(2*I*A - 5*B)*c^3*sin(6*f*x + 6*e) + 1200*(2*I*A - 5*B)*c^3*sin(4*f*x + 4*e) + 384*(2*
I*A - 5*B)*c^3*sin(2*f*x + 2*e) - (192*A + 192*I*B)*c^3 + ((720*A + 1800*I*B)*c^3*cos(7/2*arctan2(sin(2*f*x +
2*e), cos(2*f*x + 2*e))) + (1440*A + 3600*I*B)*c^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (720
*A + 1800*I*B)*c^3*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 360*(2*I*A - 5*B)*c^3*sin(7/2*arctan
2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 720*(2*I*A - 5*B)*c^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2
*e))) + 360*(2*I*A - 5*B)*c^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*arctan2(cos(1/2*arctan2(si
n(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) + ((720*A + 1800
*I*B)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (1440*A + 3600*I*B)*c^3*cos(5/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e))) + (720*A + 1800*I*B)*c^3*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
+ 360*(2*I*A - 5*B)*c^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 720*(2*I*A - 5*B)*c^3*sin(5/2*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 360*(2*I*A - 5*B)*c^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*
x + 2*e))))*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), c
os(2*f*x + 2*e))) + 1) + (180*(2*I*A - 5*B)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 360*(2*
I*A - 5*B)*c^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 180*(2*I*A - 5*B)*c^3*cos(3/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e))) - (360*A + 900*I*B)*c^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))
) - (720*A + 1800*I*B)*c^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (360*A + 900*I*B)*c^3*sin(3/
2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + s
in(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) + (180*(-2*I*A + 5*B)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 360*(-2*I*A + 5*B)*c^3*
cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 180*(-2*I*A + 5*B)*c^3*cos(3/2*arctan2(sin(2*f*x + 2*e)
, cos(2*f*x + 2*e))) + (360*A + 900*I*B)*c^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (720*A + 1
800*I*B)*c^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (360*A + 900*I*B)*c^3*sin(3/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e))))*log(cos(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))*sqrt(a)
*sqrt(c)/((-144*I*a^2*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 288*I*a^2*cos(5/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e))) - 144*I*a^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 144*a^2*sin(7
/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 288*a^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
+ 144*a^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*f)
________________________________________________________________________________________
Fricas [B] time = 1.75218, size = 1729, normalized size = 6.02 \begin{align*} \text{result too large to display} \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))^(7/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
1/12*(2*((46*I*A - 118*B)*c^3*e^(7*I*f*x + 7*I*e) + (-60*I*A + 150*B)*c^3*e^(6*I*f*x + 6*I*e) + (92*I*A - 236*
B)*c^3*e^(5*I*f*x + 5*I*e) + (-100*I*A + 250*B)*c^3*e^(4*I*f*x + 4*I*e) + (46*I*A - 118*B)*c^3*e^(3*I*f*x + 3*
I*e) + (-32*I*A + 80*B)*c^3*e^(2*I*f*x + 2*I*e) + (8*I*A - 8*B)*c^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*(a^2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))*sqrt((1
00*A^2 + 500*I*A*B - 625*B^2)*c^7/(a^3*f^2))*log((2*((10*I*A - 25*B)*c^3*e^(2*I*f*x + 2*I*e) + (10*I*A - 25*B)
*c^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) + (a^2*f*e^(2*I*f*x
+ 2*I*e) - a^2*f)*sqrt((100*A^2 + 500*I*A*B - 625*B^2)*c^7/(a^3*f^2)))/((-20*I*A + 50*B)*c^3*e^(2*I*f*x + 2*I*
e) + (-20*I*A + 50*B)*c^3)) - 3*(a^2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))*sqrt((100*A^2 + 500*I*
A*B - 625*B^2)*c^7/(a^3*f^2))*log((2*((10*I*A - 25*B)*c^3*e^(2*I*f*x + 2*I*e) + (10*I*A - 25*B)*c^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) - (a^2*f*e^(2*I*f*x + 2*I*e) - a^2*
f)*sqrt((100*A^2 + 500*I*A*B - 625*B^2)*c^7/(a^3*f^2)))/((-20*I*A + 50*B)*c^3*e^(2*I*f*x + 2*I*e) + (-20*I*A +
50*B)*c^3)))/(a^2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))
________________________________________________________________________________________
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+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(7/2)/(a+I*a*tan(f*x+e))**(3/2),x)
[Out]
Timed out
________________________________________________________________________________________
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{7}{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))^(7/2)/(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)^(7/2)/(I*a*tan(f*x + e) + a)^(3/2), x)