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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.329006, 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} \[ -\frac{5 \sqrt{a} c^{7/2} (-3 B+4 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{4 f}-\frac{5 c^3 (-3 B+4 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}-\frac{5 c^2 (-3 B+4 i A) \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac{c (-3 B+4 i A) \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{12 f}+\frac{B \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{7/2}}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + I*a*Tan[e + f*x]]*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(7/2),x]

[Out]

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

Mathematica [A]  time = 9.30935, size = 257, normalized size = 0.94 \[ \frac{\sqrt{a+i a \tan (e+f x)} (A+B \tan (e+f x)) \left (\frac{5 c^4 (3 B-4 i A) e^{-i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt{\frac{c}{1+e^{2 i (e+f x)}}}}+\frac{1}{24} c^3 \sec ^{\frac{7}{2}}(e+f x) \sqrt{c-i c \tan (e+f x)} (64 (3 B-4 i A) \cos (e+f x)+96 (B-i A) \cos (3 (e+f x))-6 \sin (e+f x) ((12 A+17 i B) \cos (2 (e+f x))+12 A+13 i B))\right )}{4 f \sec ^{\frac{3}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + I*a*Tan[e + f*x]]*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(7/2),x]

[Out]

(Sqrt[a + I*a*Tan[e + f*x]]*(A + B*Tan[e + f*x])*((5*((-4*I)*A + 3*B)*c^4*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(
e + f*x)))]*ArcTan[E^(I*(e + f*x))])/(E^(I*(e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]) + (c^3*Sec[e + f*x]^(
7/2)*(64*((-4*I)*A + 3*B)*Cos[e + f*x] + 96*((-I)*A + B)*Cos[3*(e + f*x)] - 6*(12*A + (13*I)*B + (12*A + (17*I
)*B)*Cos[2*(e + f*x)])*Sin[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/24))/(4*f*Sec[e + f*x]^(3/2)*(A*Cos[e + f*x]
+ B*Sin[e + f*x]))

________________________________________________________________________________________

Maple [A]  time = 0.173, size = 349, normalized size = 1.3 \begin{align*}{\frac{{c}^{3}}{24\,f}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \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((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x)

[Out]

1/24/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*c^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)+6
0*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+tan
(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 = 7.75012, size = 1798, normalized size = 6.61 \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))^(1/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

-((23040*A + 17280*I*B)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (84480*A + 63360*I*B)*c^3*c
os(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (112128*A + 84096*I*B)*c^3*cos(3/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e))) + (50688*A + 56448*I*B)*c^3*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 576
0*(4*I*A - 3*B)*c^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 21120*(4*I*A - 3*B)*c^3*sin(5/2*arc
tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 28032*(4*I*A - 3*B)*c^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*
x + 2*e))) + 1152*(44*I*A - 49*B)*c^3*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + ((11520*A + 8640*
I*B)*c^3*cos(8*f*x + 8*e) + (46080*A + 34560*I*B)*c^3*cos(6*f*x + 6*e) + (69120*A + 51840*I*B)*c^3*cos(4*f*x +
4*e) + (46080*A + 34560*I*B)*c^3*cos(2*f*x + 2*e) + 2880*(4*I*A - 3*B)*c^3*sin(8*f*x + 8*e) + 11520*(4*I*A -
3*B)*c^3*sin(6*f*x + 6*e) + 17280*(4*I*A - 3*B)*c^3*sin(4*f*x + 4*e) + 11520*(4*I*A - 3*B)*c^3*sin(2*f*x + 2*e
) + (11520*A + 8640*I*B)*c^3)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + ((11520*A + 8640*I*B)*c^3*cos(8*f*x + 8*e) + (46080*A + 34560*I*B)*c
^3*cos(6*f*x + 6*e) + (69120*A + 51840*I*B)*c^3*cos(4*f*x + 4*e) + (46080*A + 34560*I*B)*c^3*cos(2*f*x + 2*e)
+ 2880*(4*I*A - 3*B)*c^3*sin(8*f*x + 8*e) + 11520*(4*I*A - 3*B)*c^3*sin(6*f*x + 6*e) + 17280*(4*I*A - 3*B)*c^3
*sin(4*f*x + 4*e) + 11520*(4*I*A - 3*B)*c^3*sin(2*f*x + 2*e) + (11520*A + 8640*I*B)*c^3)*arctan2(cos(1/2*arcta
n2(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)*c^3*cos(8*f*x + 8*e) + 5760*(4*I*A - 3*B)*c^3*cos(6*f*x + 6*e) + 8640*(4*I*A - 3*B)*c^3*cos(4*f*x
+ 4*e) + 5760*(4*I*A - 3*B)*c^3*cos(2*f*x + 2*e) - (5760*A + 4320*I*B)*c^3*sin(8*f*x + 8*e) - (23040*A + 17280
*I*B)*c^3*sin(6*f*x + 6*e) - (34560*A + 25920*I*B)*c^3*sin(4*f*x + 4*e) - (23040*A + 17280*I*B)*c^3*sin(2*f*x
+ 2*e) + 1440*(4*I*A - 3*B)*c^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)*c^3*cos(8*f*x + 8*e) + 5760*(-4*I*A + 3*B)*c^3*cos(6*f*x + 6*e) + 8640*(-4*I*A + 3*B)*c^3*cos(4*
f*x + 4*e) + 5760*(-4*I*A + 3*B)*c^3*cos(2*f*x + 2*e) + (5760*A + 4320*I*B)*c^3*sin(8*f*x + 8*e) + (23040*A +
17280*I*B)*c^3*sin(6*f*x + 6*e) + (34560*A + 25920*I*B)*c^3*sin(4*f*x + 4*e) + (23040*A + 17280*I*B)*c^3*sin(2
*f*x + 2*e) + 1440*(-4*I*A + 3*B)*c^3)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*ar
ctan2(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))*sq
rt(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) - 18432*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 = 1.61683, size = 1628, normalized size = 5.99 \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))^(1/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/48*(4*((-60*I*A + 45*B)*c^3*e^(6*I*f*x + 6*I*e) + (-220*I*A + 165*B)*c^3*e^(4*I*f*x + 4*I*e) + (-292*I*A + 2
19*B)*c^3*e^(2*I*f*x + 2*I*e) + (-132*I*A + 147*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*sqrt((400*A^2 + 600*I*A*B - 225*B^2)*a*c^7/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)*c^3*e^(2*I*f*x + 2*I*e) + (-80*I*
A + 60*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) + 2*sqrt((4
00*A^2 + 600*I*A*B - 225*B^2)*a*c^7/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-20*I*A + 15*B)*c^3*e^(2*I*f*x + 2*I*e
) + (-20*I*A + 15*B)*c^3)) + 3*sqrt((400*A^2 + 600*I*A*B - 225*B^2)*a*c^7/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)*c^3*e^(2*I*f*x + 2*I*e) + (-80*I*A +
60*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) - 2*sqrt((400*
A^2 + 600*I*A*B - 225*B^2)*a*c^7/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-20*I*A + 15*B)*c^3*e^(2*I*f*x + 2*I*e) +
(-20*I*A + 15*B)*c^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((a+I*a*tan(f*x+e))**(1/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(7/2),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 )} \sqrt{i \, a \tan \left (f x + e\right ) + a}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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