3.821 \(\int \frac{(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{\sqrt{c-i c \tan (e+f x)}} \, dx\)
Optimal. Leaf size=283 \[ \frac{5 a^{7/2} (4 B+3 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{c} f}-\frac{5 a^3 (4 B+3 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{5 a^2 (4 B+3 i A) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c f}-\frac{a (4 B+3 i A) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{3 c f}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{f \sqrt{c-i c \tan (e+f x)}} \]
[Out]
(5*a^(7/2)*((3*I)*A + 4*B)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/
(Sqrt[c]*f) - ((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(f*Sqrt[c - I*c*Tan[e + f*x]]) - (5*a^3*((3*I)*A + 4*B)
*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*c*f) - (5*a^2*((3*I)*A + 4*B)*(a + I*a*Tan[e + f*x]
)^(3/2)*Sqrt[c - I*c*Tan[e + f*x]])/(6*c*f) - (a*((3*I)*A + 4*B)*(a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c - I*c*Tan
[e + f*x]])/(3*c*f)
________________________________________________________________________________________
Rubi [A] time = 0.335987, antiderivative size = 283, normalized size of antiderivative = 1.,
number of steps used = 8, 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} \[ \frac{5 a^{7/2} (4 B+3 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{c} f}-\frac{5 a^3 (4 B+3 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{5 a^2 (4 B+3 i A) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c f}-\frac{a (4 B+3 i A) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{3 c f}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{f \sqrt{c-i c \tan (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/Sqrt[c - I*c*Tan[e + f*x]],x]
[Out]
(5*a^(7/2)*((3*I)*A + 4*B)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/
(Sqrt[c]*f) - ((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(f*Sqrt[c - I*c*Tan[e + f*x]]) - (5*a^3*((3*I)*A + 4*B)
*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*c*f) - (5*a^2*((3*I)*A + 4*B)*(a + I*a*Tan[e + f*x]
)^(3/2)*Sqrt[c - I*c*Tan[e + f*x]])/(6*c*f) - (a*((3*I)*A + 4*B)*(a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c - I*c*Tan
[e + f*x]])/(3*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))^{7/2} (A+B \tan (e+f x))}{\sqrt{c-i c \tan (e+f x)}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2} (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))^{7/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{(a (3 A-4 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{a (3 i A+4 B) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{3 c f}-\frac{\left (5 a^2 (3 A-4 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 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{5 a^2 (3 i A+4 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c f}-\frac{a (3 i A+4 B) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{3 c f}-\frac{\left (5 a^3 (3 A-4 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 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{5 a^3 (3 i A+4 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{5 a^2 (3 i A+4 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c f}-\frac{a (3 i A+4 B) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{3 c f}-\frac{\left (5 a^4 (3 A-4 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 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{5 a^3 (3 i A+4 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{5 a^2 (3 i A+4 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c f}-\frac{a (3 i A+4 B) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{3 c f}+\frac{\left (5 a^3 (3 i A+4 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 )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{5 a^3 (3 i A+4 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{5 a^2 (3 i A+4 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c f}-\frac{a (3 i A+4 B) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{3 c f}+\frac{\left (5 a^3 (3 i A+4 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 )}{f}\\ &=\frac{5 a^{7/2} (3 i A+4 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))^{7/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{5 a^3 (3 i A+4 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{5 a^2 (3 i A+4 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c f}-\frac{a (3 i A+4 B) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{3 c f}\\ \end{align*}
Mathematica [A] time = 13.4676, size = 481, normalized size = 1.7 \[ \frac{\cos ^4(e+f x) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left ((A-i B) \cos (2 f x) \left (-\frac{4 \sin (e)}{c}-\frac{4 i \cos (e)}{c}\right )+(A-i B) \sin (2 f x) \left (\frac{4 \cos (e)}{c}-\frac{4 i \sin (e)}{c}\right )+\sec (e) \left (\frac{\cos (3 e)}{2 c}-\frac{i \sin (3 e)}{2 c}\right ) \sec (e+f x) (A \sin (f x)-4 i B \sin (f x))+\sec (e) \left (-\frac{\sin (3 e)}{2 c}-\frac{i \cos (3 e)}{2 c}\right ) (i A \sin (e)+16 A \cos (e)+4 B \sin (e)-24 i B \cos (e))+\sec ^2(e+f x) \left (\frac{B \cos (3 e)}{3 c}-\frac{i B \sin (3 e)}{3 c}\right )\right )}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac{5 (4 B+3 i A) \sqrt{e^{i f x}} e^{-i (4 e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{f \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \sec ^{\frac{9}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{7/2} (A \cos (e+f x)+B \sin (e+f x))} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/Sqrt[c - I*c*Tan[e + f*x]],x]
[Out]
(5*((3*I)*A + 4*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]))/(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*((A - I*B)*Cos[
2*f*x]*(((-4*I)*Cos[e])/c - (4*Sin[e])/c) + Sec[e]*(16*A*Cos[e] - (24*I)*B*Cos[e] + I*A*Sin[e] + 4*B*Sin[e])*(
((-I/2)*Cos[3*e])/c - Sin[3*e]/(2*c)) + Sec[e + f*x]^2*((B*Cos[3*e])/(3*c) - ((I/3)*B*Sin[3*e])/c) + Sec[e]*Se
c[e + f*x]*(Cos[3*e]/(2*c) - ((I/2)*Sin[3*e])/c)*(A*Sin[f*x] - (4*I)*B*Sin[f*x]) + (A - I*B)*((4*Cos[e])/c - (
(4*I)*Sin[e])/c)*Sin[2*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*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x]))
________________________________________________________________________________________
Maple [B] time = 0.175, size = 627, normalized size = 2.2 \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))^(1/2),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*(-60*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+8*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(
f*x+e)^3-2*B*tan(f*x+e)^4*(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)*a*c+18*I*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e
)^2+45*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-3*A*tan(f*
x+e)^3*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+60*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+128*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)+120*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+24*B*tan(f*x+e)^2*(a*c*(1+tan(f*x+e)^2))^(
1/2)*(a*c)^(1/2)-72*I*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)-45*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-93*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)-94*B*(a*c)^(1/2
)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(tan(f*x+e)+I)^2/(a*c)^(1/2)
________________________________________________________________________________________
Maxima [B] time = 4.64865, size = 1794, normalized size = 6.34 \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))^(1/2),x, algorithm="maxima")
[Out]
-((648*A - 1440*I*B)*a^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (1152*A - 2112*I*B)*a^3*cos(3/
2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 72*(9*I*A + 20*B)*a^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2
*f*x + 2*e))) + 192*(6*I*A + 11*B)*a^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - ((540*A - 720*I*
B)*a^3*cos(6*f*x + 6*e) + (1620*A - 2160*I*B)*a^3*cos(4*f*x + 4*e) + (1620*A - 2160*I*B)*a^3*cos(2*f*x + 2*e)
- 180*(-3*I*A - 4*B)*a^3*sin(6*f*x + 6*e) - 540*(-3*I*A - 4*B)*a^3*sin(4*f*x + 4*e) - 540*(-3*I*A - 4*B)*a^3*s
in(2*f*x + 2*e) + (540*A - 720*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) - ((540*A - 720*I*B)*a^3*cos(6*f*x + 6*e) + (1620*A - 2160*
I*B)*a^3*cos(4*f*x + 4*e) + (1620*A - 2160*I*B)*a^3*cos(2*f*x + 2*e) - 180*(-3*I*A - 4*B)*a^3*sin(6*f*x + 6*e)
- 540*(-3*I*A - 4*B)*a^3*sin(4*f*x + 4*e) - 540*(-3*I*A - 4*B)*a^3*sin(2*f*x + 2*e) + (540*A - 720*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) + ((576*A - 576*I*B)*a^3*cos(6*f*x + 6*e) + (1728*A - 1728*I*B)*a^3*cos(4*f*x + 4*e) + (1728*A - 17
28*I*B)*a^3*cos(2*f*x + 2*e) + 576*(I*A + B)*a^3*sin(6*f*x + 6*e) + 1728*(I*A + B)*a^3*sin(4*f*x + 4*e) + 1728
*(I*A + B)*a^3*sin(2*f*x + 2*e) + (1080*A - 1440*I*B)*a^3)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))
) + (90*(-3*I*A - 4*B)*a^3*cos(6*f*x + 6*e) + 270*(-3*I*A - 4*B)*a^3*cos(4*f*x + 4*e) + 270*(-3*I*A - 4*B)*a^3
*cos(2*f*x + 2*e) + (270*A - 360*I*B)*a^3*sin(6*f*x + 6*e) + (810*A - 1080*I*B)*a^3*sin(4*f*x + 4*e) + (810*A
- 1080*I*B)*a^3*sin(2*f*x + 2*e) + 90*(-3*I*A - 4*B)*a^3)*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) + (90*(3*I*A + 4*B)*a^3*cos(6*f*x + 6*e) + 270*(3*I*A + 4*B)*a^3*cos(4*f*x + 4*e) + 270*(3*I*A
+ 4*B)*a^3*cos(2*f*x + 2*e) - (270*A - 360*I*B)*a^3*sin(6*f*x + 6*e) - (810*A - 1080*I*B)*a^3*sin(4*f*x + 4*e
) - (810*A - 1080*I*B)*a^3*sin(2*f*x + 2*e) + 90*(3*I*A + 4*B)*a^3)*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) + (576*(I*A + B)*a^3*cos(6*f*x + 6*e) + 1728*(I*A + B)*a^3*cos(4*f*x + 4*e) + 1728*(
I*A + B)*a^3*cos(2*f*x + 2*e) - (576*A - 576*I*B)*a^3*sin(6*f*x + 6*e) - (1728*A - 1728*I*B)*a^3*sin(4*f*x + 4
*e) - (1728*A - 1728*I*B)*a^3*sin(2*f*x + 2*e) + 360*(3*I*A + 4*B)*a^3)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(
2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((-72*I*c*cos(6*f*x + 6*e) - 216*I*c*cos(4*f*x + 4*e) - 216*I*c*cos(2*f*x + 2*
e) + 72*c*sin(6*f*x + 6*e) + 216*c*sin(4*f*x + 4*e) + 216*c*sin(2*f*x + 2*e) - 72*I*c)*f)
________________________________________________________________________________________
Fricas [B] time = 1.5423, size = 1557, normalized size = 5.5 \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))^(1/2),x, algorithm="fricas")
[Out]
1/12*(2*((-48*I*A - 48*B)*a^3*e^(6*I*f*x + 6*I*e) + (-198*I*A - 264*B)*a^3*e^(4*I*f*x + 4*I*e) + (-240*I*A - 3
20*B)*a^3*e^(2*I*f*x + 2*I*e) + (-90*I*A - 120*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*sqrt((225*A^2 - 600*I*A*B - 400*B^2)*a^7/(c*f^2))*(c*f*e^(4*I*f*x + 4*I*e) +
2*c*f*e^(2*I*f*x + 2*I*e) + c*f)*log(2*(((60*I*A + 80*B)*a^3*e^(2*I*f*x + 2*I*e) + (60*I*A + 80*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*sqrt((225*A^2 - 600*I*A*B -
400*B^2)*a^7/(c*f^2))*(c*f*e^(2*I*f*x + 2*I*e) - c*f))/((15*I*A + 20*B)*a^3*e^(2*I*f*x + 2*I*e) + (15*I*A + 20
*B)*a^3)) + 3*sqrt((225*A^2 - 600*I*A*B - 400*B^2)*a^7/(c*f^2))*(c*f*e^(4*I*f*x + 4*I*e) + 2*c*f*e^(2*I*f*x +
2*I*e) + c*f)*log(2*(((60*I*A + 80*B)*a^3*e^(2*I*f*x + 2*I*e) + (60*I*A + 80*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*sqrt((225*A^2 - 600*I*A*B - 400*B^2)*a^7/(c*f^2
))*(c*f*e^(2*I*f*x + 2*I*e) - c*f))/((15*I*A + 20*B)*a^3*e^(2*I*f*x + 2*I*e) + (15*I*A + 20*B)*a^3)))/(c*f*e^(
4*I*f*x + 4*I*e) + 2*c*f*e^(2*I*f*x + 2*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))**(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(1/2),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 \, a \tan \left (f x + e\right ) + a\right )}^{\frac{7}{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))^(7/2)*(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)^(7/2)/sqrt(-I*c*tan(f*x + e) + c), x)