3.820 \(\int (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) \sqrt{c-i c \tan (e+f x)} \, dx\)

Optimal. Leaf size=272 \[ -\frac{5 a^{7/2} \sqrt{c} (3 B+4 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{4 f}+\frac{5 a^3 (3 B+4 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}+\frac{5 a^2 (3 B+4 i A) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{24 f}+\frac{a (3 B+4 i A) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{12 f}+\frac{B (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}}{4 f} \]

[Out]

(-5*a^(7/2)*((4*I)*A + 3*B)*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e +
f*x]])])/(4*f) + (5*a^3*((4*I)*A + 3*B)*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(8*f) + (5*a^2*
((4*I)*A + 3*B)*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]])/(24*f) + (a*((4*I)*A + 3*B)*(a + I*a*
Tan[e + f*x])^(5/2)*Sqrt[c - I*c*Tan[e + f*x]])/(12*f) + (B*(a + I*a*Tan[e + f*x])^(7/2)*Sqrt[c - I*c*Tan[e +
f*x]])/(4*f)

________________________________________________________________________________________

Rubi [A]  time = 0.320036, antiderivative size = 272, 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, 80, 50, 63, 217, 203} \[ -\frac{5 a^{7/2} \sqrt{c} (3 B+4 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{4 f}+\frac{5 a^3 (3 B+4 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}+\frac{5 a^2 (3 B+4 i A) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{24 f}+\frac{a (3 B+4 i A) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{12 f}+\frac{B (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}}{4 f} \]

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)*((4*I)*A + 3*B)*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e +
f*x]])])/(4*f) + (5*a^3*((4*I)*A + 3*B)*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(8*f) + (5*a^2*
((4*I)*A + 3*B)*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]])/(24*f) + (a*((4*I)*A + 3*B)*(a + I*a*
Tan[e + f*x])^(5/2)*Sqrt[c - I*c*Tan[e + f*x]])/(12*f) + (B*(a + I*a*Tan[e + f*x])^(7/2)*Sqrt[c - I*c*Tan[e +
f*x]])/(4*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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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 (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)}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{B (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}}{4 f}+\frac{(a (4 A-3 i B) c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac{a (4 i A+3 B) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{12 f}+\frac{B (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}}{4 f}+\frac{\left (5 a^2 (4 A-3 i B) c\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{12 f}\\ &=\frac{5 a^2 (4 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{24 f}+\frac{a (4 i A+3 B) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{12 f}+\frac{B (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}}{4 f}+\frac{\left (5 a^3 (4 A-3 i B) c\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac{5 a^3 (4 i A+3 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}+\frac{5 a^2 (4 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{24 f}+\frac{a (4 i A+3 B) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{12 f}+\frac{B (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}}{4 f}+\frac{\left (5 a^4 (4 A-3 i B) c\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac{5 a^3 (4 i A+3 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}+\frac{5 a^2 (4 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{24 f}+\frac{a (4 i A+3 B) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{12 f}+\frac{B (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}}{4 f}-\frac{\left (5 a^3 (4 i A+3 B) c\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 )}{4 f}\\ &=\frac{5 a^3 (4 i A+3 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}+\frac{5 a^2 (4 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{24 f}+\frac{a (4 i A+3 B) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{12 f}+\frac{B (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}}{4 f}-\frac{\left (5 a^3 (4 i A+3 B) c\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 )}{4 f}\\ &=-\frac{5 a^{7/2} (4 i A+3 B) \sqrt{c} \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 )}{4 f}+\frac{5 a^3 (4 i A+3 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}+\frac{5 a^2 (4 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{24 f}+\frac{a (4 i A+3 B) (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{12 f}+\frac{B (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}}{4 f}\\ \end{align*}

Mathematica [A]  time = 11.427, size = 465, normalized size = 1.71 \[ \frac{\cos ^4(e+f x) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left (\sec (e) \left (-\frac{1}{12} \sin (3 e)-\frac{1}{12} i \cos (3 e)\right ) \sec ^2(e+f x) (4 A \cos (e)+3 B \sin (e)-12 i B \cos (e))+\sec (e) \left (\frac{1}{8} \cos (3 e)-\frac{1}{8} i \sin (3 e)\right ) \sec (e+f x) (-12 A \sin (f x)+17 i B \sin (f x))+\sec (e) \left (\frac{1}{8} \cos (3 e)-\frac{1}{8} i \sin (3 e)\right ) (-12 A \sin (e)+32 i A \cos (e)+17 i B \sin (e)+32 B \cos (e))-i B \sec (e) \left (\frac{1}{4} \cos (3 e)-\frac{1}{4} i \sin (3 e)\right ) \sin (f x) \sec ^3(e+f x)\right )}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}-\frac{5 i c (4 A-3 i B) \sqrt{e^{i f x}} e^{-i (4 e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{4 f \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \sec ^{\frac{9}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{7/2} (A \cos (e+f x)+B \sin (e+f x))} \]

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*I)/4)*(4*A - (3*I)*B)*c*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*(Sec
[e]*Sec[e + f*x]^2*(4*A*Cos[e] - (12*I)*B*Cos[e] + 3*B*Sin[e])*((-I/12)*Cos[3*e] - Sin[3*e]/12) + Sec[e]*((32*
I)*A*Cos[e] + 32*B*Cos[e] - 12*A*Sin[e] + (17*I)*B*Sin[e])*(Cos[3*e]/8 - (I/8)*Sin[3*e]) - I*B*Sec[e]*Sec[e +
f*x]^3*(Cos[3*e]/4 - (I/4)*Sin[3*e])*Sin[f*x] + Sec[e]*Sec[e + f*x]*(Cos[3*e]/8 - (I/8)*Sin[3*e])*(-12*A*Sin[f
*x] + (17*I)*B*Sin[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 [A]  time = 0.102, size = 349, normalized size = 1.3 \begin{align*} -{\frac{{a}^{3}}{24\,f}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) } \left ( 6\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{3}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+8\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{2}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+45\,iB\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) ac-45\,iB\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\tan \left ( fx+e \right ) +24\,B \left ( \tan \left ( fx+e \right ) \right ) ^{2}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}-88\,iA\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }-60\,A\ln \left ({\frac{ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}}{\sqrt{ac}}} \right ) ac+36\,A\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\tan \left ( fx+e \right ) -72\,B\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) } \right ){\frac{1}{\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e)),x)

[Out]

-1/24/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*a^3*(6*I*B*tan(f*x+e)^3*(a*c*(1+tan(f*x+e)^2))
^(1/2)*(a*c)^(1/2)+8*I*A*tan(f*x+e)^2*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+45*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-45*I*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+
e)+24*B*tan(f*x+e)^2*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)-88*I*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)-
60*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+36*A*(a*c)^(1/2)*(a*c*(1+ta
n(f*x+e)^2))^(1/2)*tan(f*x+e)-72*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2)/(a*c*(1+tan(f*x+e)^2)
)^(1/2)

________________________________________________________________________________________

Maxima [B]  time = 6.57883, size = 1800, normalized size = 6.62 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e)),x, algorithm="maxima")

[Out]

((50688*A - 56448*I*B)*a^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (112128*A - 84096*I*B)*a^3*c
os(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (84480*A - 63360*I*B)*a^3*cos(3/2*arctan2(sin(2*f*x + 2*
e), cos(2*f*x + 2*e))) + (23040*A - 17280*I*B)*a^3*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1152
*(44*I*A + 49*B)*a^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 28032*(4*I*A + 3*B)*a^3*sin(5/2*ar
ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 21120*(4*I*A + 3*B)*a^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f
*x + 2*e))) + 5760*(4*I*A + 3*B)*a^3*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - ((11520*A - 8640*I
*B)*a^3*cos(8*f*x + 8*e) + (46080*A - 34560*I*B)*a^3*cos(6*f*x + 6*e) + (69120*A - 51840*I*B)*a^3*cos(4*f*x +
4*e) + (46080*A - 34560*I*B)*a^3*cos(2*f*x + 2*e) - 2880*(-4*I*A - 3*B)*a^3*sin(8*f*x + 8*e) - 11520*(-4*I*A -
3*B)*a^3*sin(6*f*x + 6*e) - 17280*(-4*I*A - 3*B)*a^3*sin(4*f*x + 4*e) - 11520*(-4*I*A - 3*B)*a^3*sin(2*f*x +
2*e) + (11520*A - 8640*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) - ((11520*A - 8640*I*B)*a^3*cos(8*f*x + 8*e) + (46080*A - 34560*I*B
)*a^3*cos(6*f*x + 6*e) + (69120*A - 51840*I*B)*a^3*cos(4*f*x + 4*e) + (46080*A - 34560*I*B)*a^3*cos(2*f*x + 2*
e) - 2880*(-4*I*A - 3*B)*a^3*sin(8*f*x + 8*e) - 11520*(-4*I*A - 3*B)*a^3*sin(6*f*x + 6*e) - 17280*(-4*I*A - 3*
B)*a^3*sin(4*f*x + 4*e) - 11520*(-4*I*A - 3*B)*a^3*sin(2*f*x + 2*e) + (11520*A - 8640*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) + (
1440*(-4*I*A - 3*B)*a^3*cos(8*f*x + 8*e) + 5760*(-4*I*A - 3*B)*a^3*cos(6*f*x + 6*e) + 8640*(-4*I*A - 3*B)*a^3*
cos(4*f*x + 4*e) + 5760*(-4*I*A - 3*B)*a^3*cos(2*f*x + 2*e) + (5760*A - 4320*I*B)*a^3*sin(8*f*x + 8*e) + (2304
0*A - 17280*I*B)*a^3*sin(6*f*x + 6*e) + (34560*A - 25920*I*B)*a^3*sin(4*f*x + 4*e) + (23040*A - 17280*I*B)*a^3
*sin(2*f*x + 2*e) + 1440*(-4*I*A - 3*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) + (1440*(4*I*A + 3*B)*a^3*cos(8*f*x + 8*e) + 5760*(4*I*A + 3*B)*a^3*cos(6*f*x + 6*e) + 8640*(4*I*A + 3*B)*a
^3*cos(4*f*x + 4*e) + 5760*(4*I*A + 3*B)*a^3*cos(2*f*x + 2*e) - (5760*A - 4320*I*B)*a^3*sin(8*f*x + 8*e) - (23
040*A - 17280*I*B)*a^3*sin(6*f*x + 6*e) - (34560*A - 25920*I*B)*a^3*sin(4*f*x + 4*e) - (23040*A - 17280*I*B)*a
^3*sin(2*f*x + 2*e) + 1440*(4*I*A + 3*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))*sqrt(a)*sqrt(c)/(f*(-4608*I*cos(8*f*x + 8*e) - 18432*I*cos(6*f*x + 6*e) - 27648*I*cos(4*f*x + 4*e) - 1843
2*I*cos(2*f*x + 2*e) + 4608*sin(8*f*x + 8*e) + 18432*sin(6*f*x + 6*e) + 27648*sin(4*f*x + 4*e) + 18432*sin(2*f
*x + 2*e) - 4608*I))

________________________________________________________________________________________

Fricas [B]  time = 1.73749, size = 1612, normalized size = 5.93 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/48*(4*((132*I*A + 147*B)*a^3*e^(6*I*f*x + 6*I*e) + (292*I*A + 219*B)*a^3*e^(4*I*f*x + 4*I*e) + (220*I*A + 16
5*B)*a^3*e^(2*I*f*x + 2*I*e) + (60*I*A + 45*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((400*A^2 - 600*I*A*B - 225*B^2)*a^7*c/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(
4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)*log(2*(((80*I*A + 60*B)*a^3*e^(2*I*f*x + 2*I*e) + (80*I*A + 60
*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((400*A^2
- 600*I*A*B - 225*B^2)*a^7*c/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((20*I*A + 15*B)*a^3*e^(2*I*f*x + 2*I*e) + (20
*I*A + 15*B)*a^3)) - 3*sqrt((400*A^2 - 600*I*A*B - 225*B^2)*a^7*c/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x
+ 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)*log(2*(((80*I*A + 60*B)*a^3*e^(2*I*f*x + 2*I*e) + (80*I*A + 60*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((400*A^2 - 600*
I*A*B - 225*B^2)*a^7*c/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((20*I*A + 15*B)*a^3*e^(2*I*f*x + 2*I*e) + (20*I*A +
15*B)*a^3)))/(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + 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((c-I*c*tan(f*x+e))**(1/2)*(a+I*a*tan(f*x+e))**(7/2)*(A+B*tan(f*x+e)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e)),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)