3.844 \(\int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=284 \[ \frac{2 c^{7/2} (-6 B+i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{a^{5/2} f}+\frac{c^3 (-6 B+i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a^3 f}+\frac{2 c^2 (-6 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 c (-6 B+i A) (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(-B+i A) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]
[Out]
(2*(I*A - 6*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^(
5/2)*f) + ((I*A - 6*B)*c^3*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*f) + (2*(I*A - 6*B)*c^2
*(c - I*c*Tan[e + f*x])^(3/2))/(3*a^2*f*Sqrt[a + I*a*Tan[e + f*x]]) - (2*(I*A - 6*B)*c*(c - I*c*Tan[e + f*x])^
(5/2))/(15*a*f*(a + I*a*Tan[e + f*x])^(3/2)) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(7/2))/(5*f*(a + I*a*Tan[e +
f*x])^(5/2))
________________________________________________________________________________________
Rubi [A] time = 0.342894, antiderivative size = 284, 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} \[ \frac{2 c^{7/2} (-6 B+i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{a^{5/2} f}+\frac{c^3 (-6 B+i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a^3 f}+\frac{2 c^2 (-6 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 c (-6 B+i A) (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(-B+i A) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]
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])^(5/2),x]
[Out]
(2*(I*A - 6*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^(
5/2)*f) + ((I*A - 6*B)*c^3*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*f) + (2*(I*A - 6*B)*c^2
*(c - I*c*Tan[e + f*x])^(3/2))/(3*a^2*f*Sqrt[a + I*a*Tan[e + f*x]]) - (2*(I*A - 6*B)*c*(c - I*c*Tan[e + f*x])^
(5/2))/(15*a*f*(a + I*a*Tan[e + f*x])^(3/2)) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(7/2))/(5*f*(a + I*a*Tan[e +
f*x])^(5/2))
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))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) (c-i c x)^{5/2}}{(a+i a x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac{((A+6 i B) c) \operatorname{Subst}\left (\int \frac{(c-i c x)^{5/2}}{(a+i a x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac{2 (i A-6 B) c (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{\left ((A+6 i B) c^2\right ) \operatorname{Subst}\left (\int \frac{(c-i c x)^{3/2}}{(a+i a x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 a f}\\ &=\frac{2 (i A-6 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 (i A-6 B) c (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac{\left ((A+6 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 )}{a^2 f}\\ &=\frac{(i A-6 B) c^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a^3 f}+\frac{2 (i A-6 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 (i A-6 B) c (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac{\left ((A+6 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 )}{a^2 f}\\ &=\frac{(i A-6 B) c^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a^3 f}+\frac{2 (i A-6 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 (i A-6 B) c (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{\left (2 (i A-6 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^3 f}\\ &=\frac{(i A-6 B) c^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a^3 f}+\frac{2 (i A-6 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 (i A-6 B) c (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{\left (2 (i A-6 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^3 f}\\ &=\frac{2 (i A-6 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^{5/2} f}+\frac{(i A-6 B) c^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a^3 f}+\frac{2 (i A-6 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 (i A-6 B) c (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}}\\ \end{align*}
Mathematica [A] time = 14.2969, size = 205, normalized size = 0.72 \[ \frac{2 \sqrt{2} c^2 e^{-4 i (e+f x)} \left (\frac{c}{1+e^{2 i (e+f x)}}\right )^{3/2} \left (15 i (A+6 i B) e^{5 i (e+f x)} \left (1+e^{2 i (e+f x)}\right ) \tan ^{-1}\left (e^{i (e+f x)}\right )+i A \left (-2 e^{2 i (e+f x)}+10 e^{4 i (e+f x)}+15 e^{6 i (e+f x)}+3\right )-3 B \left (-4 e^{2 i (e+f x)}+20 e^{4 i (e+f x)}+30 e^{6 i (e+f x)}+1\right )\right )}{15 a^2 f \sqrt{a+i a \tan (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(7/2))/(a + I*a*Tan[e + f*x])^(5/2),x]
[Out]
(2*Sqrt[2]*c^2*(c/(1 + E^((2*I)*(e + f*x))))^(3/2)*(I*A*(3 - 2*E^((2*I)*(e + f*x)) + 10*E^((4*I)*(e + f*x)) +
15*E^((6*I)*(e + f*x))) - 3*B*(1 - 4*E^((2*I)*(e + f*x)) + 20*E^((4*I)*(e + f*x)) + 30*E^((6*I)*(e + f*x))) +
(15*I)*(A + (6*I)*B)*E^((5*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))*ArcTan[E^(I*(e + f*x))]))/(15*a^2*E^((4*I)*
(e + f*x))*f*Sqrt[a + I*a*Tan[e + f*x]])
________________________________________________________________________________________
Maple [B] time = 0.121, size = 835, 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+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e))^(5/2),x)
[Out]
1/15/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^3/a^3*(-474*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*
(a*c)^(1/2)*tan(f*x+e)+540*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-15*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)^4*a*c+24
6*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^3+60*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)^3*a*c-360*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-15*B*tan(f*x+e)^4*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)-90*I*B*ln((a*c*t
an(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)^4*a*c-60*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+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)^2*a*c+46*A*tan(f*x+e)^3*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/
2)-90*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+26*I*A*(a*c*(1+tan(f*x
+e)^2))^(1/2)*(a*c)^(1/2)+360*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+564*B*tan(f*x+e)^2*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)-94*I*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*
c)^(1/2)*tan(f*x+e)^2-15*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-74*A*
(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)-141*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c*(1+ta
n(f*x+e)^2))^(1/2)/(-tan(f*x+e)+I)^4/(a*c)^(1/2)
________________________________________________________________________________________
Maxima [B] time = 3.9302, size = 1416, normalized size = 4.99 \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))^(5/2),x, algorithm="maxima")
[Out]
((900*A + 5400*I*B)*c^3*cos(6*f*x + 6*e) + (600*A + 3600*I*B)*c^3*cos(4*f*x + 4*e) - (120*A + 720*I*B)*c^3*cos
(2*f*x + 2*e) + 900*(I*A - 6*B)*c^3*sin(6*f*x + 6*e) + 600*(I*A - 6*B)*c^3*sin(4*f*x + 4*e) + 120*(-I*A + 6*B)
*c^3*sin(2*f*x + 2*e) + (180*A + 180*I*B)*c^3 + ((450*A + 2700*I*B)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(
2*f*x + 2*e))) + (450*A + 2700*I*B)*c^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 450*(I*A - 6*B)
*c^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 450*(I*A - 6*B)*c^3*sin(5/2*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), cos(2*f*x + 2*e))) + 1) + ((450*A + 2700*I*B)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
)) + (450*A + 2700*I*B)*c^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 450*(I*A - 6*B)*c^3*sin(7/2
*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 450*(I*A - 6*B)*c^3*sin(5/2*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) + (225*(I*A - 6*B)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 225*(I*A
- 6*B)*c^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (225*A + 1350*I*B)*c^3*sin(7/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e))) - (225*A + 1350*I*B)*c^3*sin(5/2*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) + (225*(-I*A + 6*B)*c^3*cos(7/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) + 225*(-I*A + 6*B)*c^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))
) + (225*A + 1350*I*B)*c^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (225*A + 1350*I*B)*c^3*sin(5
/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)/((-450*I*a^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 450*I*a^3*cos(5/2*
arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 450*a^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
450*a^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*f)
________________________________________________________________________________________
Fricas [B] time = 1.73902, size = 1531, normalized size = 5.39 \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))^(5/2),x, algorithm="fricas")
[Out]
-1/60*(15*a^3*sqrt((4*A^2 + 48*I*A*B - 144*B^2)*c^7/(a^5*f^2))*f*e^(6*I*f*x + 6*I*e)*log((2*((2*I*A - 12*B)*c^
3*e^(2*I*f*x + 2*I*e) + (2*I*A - 12*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^3*f*e^(2*I*f*x + 2*I*e) - a^3*f)*sqrt((4*A^2 + 48*I*A*B - 144*B^2)*c^7/(a^5*f^2)))/((-4
*I*A + 24*B)*c^3*e^(2*I*f*x + 2*I*e) + (-4*I*A + 24*B)*c^3)) - 15*a^3*sqrt((4*A^2 + 48*I*A*B - 144*B^2)*c^7/(a
^5*f^2))*f*e^(6*I*f*x + 6*I*e)*log((2*((2*I*A - 12*B)*c^3*e^(2*I*f*x + 2*I*e) + (2*I*A - 12*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^3*f*e^(2*I*f*x + 2*I*e) - a^3*f
)*sqrt((4*A^2 + 48*I*A*B - 144*B^2)*c^7/(a^5*f^2)))/((-4*I*A + 24*B)*c^3*e^(2*I*f*x + 2*I*e) + (-4*I*A + 24*B)
*c^3)) - 2*((-52*I*A + 282*B)*c^3*e^(7*I*f*x + 7*I*e) + (60*I*A - 360*B)*c^3*e^(6*I*f*x + 6*I*e) + (-52*I*A +
282*B)*c^3*e^(5*I*f*x + 5*I*e) + (40*I*A - 240*B)*c^3*e^(4*I*f*x + 4*I*e) + (-8*I*A + 48*B)*c^3*e^(2*I*f*x + 2
*I*e) + (12*I*A - 12*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))*e^(-6*I*f*x - 6*I*e)/(a^3*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+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(7/2)/(a+I*a*tan(f*x+e))**(5/2),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{5}{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))^(5/2),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)^(5/2), x)