3.795 \(\int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx\)

Optimal. Leaf size=279 \[ -\frac{a^{3/2} c^{7/2} (-2 B+5 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{4 f}+\frac{a c^3 (5 A+2 i B) \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}-\frac{c^2 (-2 B+5 i A) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}-\frac{c (-2 B+5 i A) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac{B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f} \]

[Out]

-(a^(3/2)*((5*I)*A - 2*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) + (a*(5*A + (2*I)*B)*c^3*Tan[e + f*x]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(8*f
) - (((5*I)*A - 2*B)*c^2*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2))/(12*f) - (((5*I)*A - 2*B)*
c*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2))/(20*f) + (B*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c
*Tan[e + f*x])^(7/2))/(5*f)

________________________________________________________________________________________

Rubi [A]  time = 0.336548, antiderivative size = 279, 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, 80, 49, 38, 63, 217, 203} \[ -\frac{a^{3/2} c^{7/2} (-2 B+5 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{4 f}+\frac{a c^3 (5 A+2 i B) \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}-\frac{c^2 (-2 B+5 i A) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}-\frac{c (-2 B+5 i A) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac{B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(7/2),x]

[Out]

-(a^(3/2)*((5*I)*A - 2*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) + (a*(5*A + (2*I)*B)*c^3*Tan[e + f*x]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(8*f
) - (((5*I)*A - 2*B)*c^2*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2))/(12*f) - (((5*I)*A - 2*B)*
c*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2))/(20*f) + (B*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c
*Tan[e + f*x])^(7/2))/(5*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 49

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[(2*c*n)/(m + n + 1), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x]
&& EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0] && IGtQ[n + 1/2, 0] && LtQ[m, n]

Rule 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(x*(a + b*x)^m*(c + d*x)^m)/(2*m + 1)
, x] + Dist[(2*a*c*m)/(2*m + 1), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0]

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))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \sqrt{a+i a x} (A+B x) (c-i c x)^{5/2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f}+\frac{(a (5 A+2 i B) c) \operatorname{Subst}\left (\int \sqrt{a+i a x} (c-i c x)^{5/2} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac{(5 i A-2 B) c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac{B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f}+\frac{\left (a (5 A+2 i B) c^2\right ) \operatorname{Subst}\left (\int \sqrt{a+i a x} (c-i c x)^{3/2} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=-\frac{(5 i A-2 B) c^2 (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}-\frac{(5 i A-2 B) c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac{B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f}+\frac{\left (a (5 A+2 i B) c^3\right ) \operatorname{Subst}\left (\int \sqrt{a+i a x} \sqrt{c-i c x} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac{a (5 A+2 i B) c^3 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}-\frac{(5 i A-2 B) c^2 (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}-\frac{(5 i A-2 B) c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac{B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f}+\frac{\left (a^2 (5 A+2 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{a (5 A+2 i B) c^3 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}-\frac{(5 i A-2 B) c^2 (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}-\frac{(5 i A-2 B) c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac{B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f}-\frac{\left (a (5 i A-2 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{a (5 A+2 i B) c^3 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}-\frac{(5 i A-2 B) c^2 (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}-\frac{(5 i A-2 B) c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac{B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f}-\frac{\left (a (5 i A-2 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{a^{3/2} (5 i A-2 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{a (5 A+2 i B) c^3 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}-\frac{(5 i A-2 B) c^2 (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}-\frac{(5 i A-2 B) c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac{B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f}\\ \end{align*}

Mathematica [A]  time = 13.189, size = 257, normalized size = 0.92 \[ \frac{(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \left (\frac{c^4 (2 B-5 i A) e^{-2 i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt{\frac{c}{1+e^{2 i (e+f x)}}}}-\frac{1}{240} c^3 (\tan (e+f x)+i) \sec ^{\frac{7}{2}}(e+f x) \sqrt{c-i c \tan (e+f x)} (30 (6 B+i A) \sin (2 (e+f x))+(5 A+2 i B) (64+15 i \sin (4 (e+f x)))+320 (A+i B) \cos (2 (e+f x)))\right )}{4 f \sec ^{\frac{5}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(7/2),x]

[Out]

((a + I*a*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x])*((((-5*I)*A + 2*B)*c^4*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(
e + f*x)))]*ArcTan[E^(I*(e + f*x))])/(E^((2*I)*(e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]) - (c^3*Sec[e + f*
x]^(7/2)*(320*(A + I*B)*Cos[2*(e + f*x)] + 30*(I*A + 6*B)*Sin[2*(e + f*x)] + (5*A + (2*I)*B)*(64 + (15*I)*Sin[
4*(e + f*x)]))*(I + Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/240))/(4*f*Sec[e + f*x]^(5/2)*(A*Cos[e + f*x] +
B*Sin[e + f*x]))

________________________________________________________________________________________

Maple [A]  time = 0.115, size = 412, normalized size = 1.5 \begin{align*} -{\frac{a{c}^{3}}{120\,f}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) } \left ( 60\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{3}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+24\,B \left ( \tan \left ( fx+e \right ) \right ) ^{4}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+80\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{2}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+30\,A \left ( \tan \left ( fx+e \right ) \right ) ^{3}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}-30\,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+30\,iB\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}\tan \left ( fx+e \right ) -32\,B \left ( \tan \left ( fx+e \right ) \right ) ^{2}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+80\,iA\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}-75\,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-45\,A\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\tan \left ( fx+e \right ) -56\,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))^(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x)

[Out]

-1/120/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*c^3*a*(60*I*B*tan(f*x+e)^3*(a*c*(1+tan(f*x+e)
^2))^(1/2)*(a*c)^(1/2)+24*B*tan(f*x+e)^4*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+80*I*A*tan(f*x+e)^2*(a*c*(1+
tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+30*A*tan(f*x+e)^3*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)-30*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+30*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(
1/2)*tan(f*x+e)-32*B*tan(f*x+e)^2*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+80*I*A*(a*c*(1+tan(f*x+e)^2))^(1/2)
*(a*c)^(1/2)-75*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-45*A*(a*c)^(1/
2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)-56*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)/(a*c)^(1/2)

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Maxima [B]  time = 13.9878, size = 2215, normalized size = 7.94 \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))^(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

-((144000*A + 57600*I*B)*a*c^3*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (672000*A + 268800*I*B)*
a*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (1228800*A + 491520*I*B)*a*c^3*cos(5/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e))) + (556800*A + 960000*I*B)*a*c^3*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) - (144000*A + 57600*I*B)*a*c^3*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 28800*(5*I*A -
2*B)*a*c^3*sin(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 134400*(5*I*A - 2*B)*a*c^3*sin(7/2*arctan2(
sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 245760*(5*I*A - 2*B)*a*c^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) + 19200*(29*I*A - 50*B)*a*c^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 28800*(-5*I*A +
2*B)*a*c^3*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + ((72000*A + 28800*I*B)*a*c^3*cos(10*f*x + 10
*e) + (360000*A + 144000*I*B)*a*c^3*cos(8*f*x + 8*e) + (720000*A + 288000*I*B)*a*c^3*cos(6*f*x + 6*e) + (72000
0*A + 288000*I*B)*a*c^3*cos(4*f*x + 4*e) + (360000*A + 144000*I*B)*a*c^3*cos(2*f*x + 2*e) + 14400*(5*I*A - 2*B
)*a*c^3*sin(10*f*x + 10*e) + 72000*(5*I*A - 2*B)*a*c^3*sin(8*f*x + 8*e) + 144000*(5*I*A - 2*B)*a*c^3*sin(6*f*x
+ 6*e) + 144000*(5*I*A - 2*B)*a*c^3*sin(4*f*x + 4*e) + 72000*(5*I*A - 2*B)*a*c^3*sin(2*f*x + 2*e) + (72000*A
+ 28800*I*B)*a*c^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) + ((72000*A + 28800*I*B)*a*c^3*cos(10*f*x + 10*e) + (360000*A + 144000*I*B)*a*c^
3*cos(8*f*x + 8*e) + (720000*A + 288000*I*B)*a*c^3*cos(6*f*x + 6*e) + (720000*A + 288000*I*B)*a*c^3*cos(4*f*x
+ 4*e) + (360000*A + 144000*I*B)*a*c^3*cos(2*f*x + 2*e) + 14400*(5*I*A - 2*B)*a*c^3*sin(10*f*x + 10*e) + 72000
*(5*I*A - 2*B)*a*c^3*sin(8*f*x + 8*e) + 144000*(5*I*A - 2*B)*a*c^3*sin(6*f*x + 6*e) + 144000*(5*I*A - 2*B)*a*c
^3*sin(4*f*x + 4*e) + 72000*(5*I*A - 2*B)*a*c^3*sin(2*f*x + 2*e) + (72000*A + 28800*I*B)*a*c^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) + (
7200*(5*I*A - 2*B)*a*c^3*cos(10*f*x + 10*e) + 36000*(5*I*A - 2*B)*a*c^3*cos(8*f*x + 8*e) + 72000*(5*I*A - 2*B)
*a*c^3*cos(6*f*x + 6*e) + 72000*(5*I*A - 2*B)*a*c^3*cos(4*f*x + 4*e) + 36000*(5*I*A - 2*B)*a*c^3*cos(2*f*x + 2
*e) - (36000*A + 14400*I*B)*a*c^3*sin(10*f*x + 10*e) - (180000*A + 72000*I*B)*a*c^3*sin(8*f*x + 8*e) - (360000
*A + 144000*I*B)*a*c^3*sin(6*f*x + 6*e) - (360000*A + 144000*I*B)*a*c^3*sin(4*f*x + 4*e) - (180000*A + 72000*I
*B)*a*c^3*sin(2*f*x + 2*e) + 7200*(5*I*A - 2*B)*a*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) + (7200*(-5*I*A + 2*B)*a*c^3*cos(10*f*x + 10*e) + 36000*(-5*I*A + 2*B)*a*c^3*cos(8*f*x + 8*e) + 7
2000*(-5*I*A + 2*B)*a*c^3*cos(6*f*x + 6*e) + 72000*(-5*I*A + 2*B)*a*c^3*cos(4*f*x + 4*e) + 36000*(-5*I*A + 2*B
)*a*c^3*cos(2*f*x + 2*e) + (36000*A + 14400*I*B)*a*c^3*sin(10*f*x + 10*e) + (180000*A + 72000*I*B)*a*c^3*sin(8
*f*x + 8*e) + (360000*A + 144000*I*B)*a*c^3*sin(6*f*x + 6*e) + (360000*A + 144000*I*B)*a*c^3*sin(4*f*x + 4*e)
+ (180000*A + 72000*I*B)*a*c^3*sin(2*f*x + 2*e) + 7200*(-5*I*A + 2*B)*a*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))*sqrt(a)*sqrt(c)/(f*(-115200*I*cos(10*f*x + 10*e) - 576000*I*cos(8*f*x + 8*
e) - 1152000*I*cos(6*f*x + 6*e) - 1152000*I*cos(4*f*x + 4*e) - 576000*I*cos(2*f*x + 2*e) + 115200*sin(10*f*x +
10*e) + 576000*sin(8*f*x + 8*e) + 1152000*sin(6*f*x + 6*e) + 1152000*sin(4*f*x + 4*e) + 576000*sin(2*f*x + 2*
e) - 115200*I))

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Fricas [B]  time = 1.65431, size = 1802, 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+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/240*(4*((-75*I*A + 30*B)*a*c^3*e^(8*I*f*x + 8*I*e) + (-350*I*A + 140*B)*a*c^3*e^(6*I*f*x + 6*I*e) + (-640*I*
A + 256*B)*a*c^3*e^(4*I*f*x + 4*I*e) + (-290*I*A + 500*B)*a*c^3*e^(2*I*f*x + 2*I*e) + (75*I*A - 30*B)*a*c^3)*s
qrt(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) - 15*sqrt((25*A^2 + 20*I*A*
B - 4*B^2)*a^3*c^7/f^2)*(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*
I*f*x + 2*I*e) + f)*log(2*(((-20*I*A + 8*B)*a*c^3*e^(2*I*f*x + 2*I*e) + (-20*I*A + 8*B)*a*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((25*A^2 + 20*I*A*B - 4*B^2)*a^3*
c^7/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-5*I*A + 2*B)*a*c^3*e^(2*I*f*x + 2*I*e) + (-5*I*A + 2*B)*a*c^3)) + 15*
sqrt((25*A^2 + 20*I*A*B - 4*B^2)*a^3*c^7/f^2)*(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*
x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)*log(2*(((-20*I*A + 8*B)*a*c^3*e^(2*I*f*x + 2*I*e) + (-20*I*A + 8*B)*
a*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((25*A^2 +
20*I*A*B - 4*B^2)*a^3*c^7/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-5*I*A + 2*B)*a*c^3*e^(2*I*f*x + 2*I*e) + (-5*I*
A + 2*B)*a*c^3)))/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x
+ 2*I*e) + f)

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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))**(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(7/2),x)

[Out]

Timed out

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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{3}{2}}{\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))^(3/2)*(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)*(I*a*tan(f*x + e) + a)^(3/2)*(-I*c*tan(f*x + e) + c)^(7/2), x)