3.9.24 \(\int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\) [824]

Optimal. Leaf size=251 \[ -\frac {2 a^{7/2} B \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{7/2} f}-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^3 B \sqrt {a+i a \tan (e+f x)}}{c^3 f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

-2*a^(7/2)*B*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x+e))^(1/2))/c^(7/2)/f+2*a^3*B*(a+I*
a*tan(f*x+e))^(1/2)/c^3/f/(c-I*c*tan(f*x+e))^(1/2)-1/7*(I*A+B)*(a+I*a*tan(f*x+e))^(7/2)/f/(c-I*c*tan(f*x+e))^(
7/2)+2/5*a*B*(a+I*a*tan(f*x+e))^(5/2)/c/f/(c-I*c*tan(f*x+e))^(5/2)-2/3*a^2*B*(a+I*a*tan(f*x+e))^(3/2)/c^2/f/(c
-I*c*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.21, antiderivative size = 251, normalized size of antiderivative = 1.00, 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 = {3669, 79, 49, 65, 223, 209} \begin {gather*} -\frac {2 a^{7/2} B \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{7/2} f}+\frac {2 a^3 B \sqrt {a+i a \tan (e+f x)}}{c^3 f \sqrt {c-i c \tan (e+f x)}}-\frac {2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 + 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 65

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 79

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] || L
tQ[p, n]))))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

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 3669

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]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(a+i a x)^{5/2} (A+B x)}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {(i a B) \text {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {\left (i a^2 B\right ) \text {Subst}\left (\int \frac {(a+i a x)^{3/2}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{c f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {\left (i a^3 B\right ) \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{c^2 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^3 B \sqrt {a+i a \tan (e+f x)}}{c^3 f \sqrt {c-i c \tan (e+f x)}}-\frac {\left (i a^4 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{c^3 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^3 B \sqrt {a+i a \tan (e+f x)}}{c^3 f \sqrt {c-i c \tan (e+f x)}}-\frac {\left (2 a^3 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{c^3 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^3 B \sqrt {a+i a \tan (e+f x)}}{c^3 f \sqrt {c-i c \tan (e+f x)}}-\frac {\left (2 a^3 B\right ) \text {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 )}{c^3 f}\\ &=-\frac {2 a^{7/2} 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 )}{c^{7/2} f}-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^3 B \sqrt {a+i a \tan (e+f x)}}{c^3 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 13.77, size = 226, normalized size = 0.90 \begin {gather*} \frac {a^3 \cos ^3(e+f x) \left (\cos \left (\frac {1}{2} (e-4 f x)\right )-i \sin \left (\frac {1}{2} (e-4 f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e-4 f x)\right )+i \sin \left (\frac {1}{2} (e-4 f x)\right )\right ) \left (28 i B \cos (e+f x)-15 A \cos (3 (e+f x))-195 i B \cos (3 (e+f x))+112 B \sin (e+f x)-15 i A \sin (3 (e+f x))-225 B \sin (3 (e+f x))+210 B \text {ArcTan}\left (e^{i (e+f x)}\right ) (i \cos (4 (e+f x))+\sin (4 (e+f x)))\right ) (-i+\tan (e+f x))^3 \sqrt {a+i a \tan (e+f x)}}{105 c^3 f \sqrt {c-i c \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Cos[e + f*x]^3*(Cos[(e - 4*f*x)/2] - I*Sin[(e - 4*f*x)/2])*(Cos[(e - 4*f*x)/2] + I*Sin[(e - 4*f*x)/2])*((
28*I)*B*Cos[e + f*x] - 15*A*Cos[3*(e + f*x)] - (195*I)*B*Cos[3*(e + f*x)] + 112*B*Sin[e + f*x] - (15*I)*A*Sin[
3*(e + f*x)] - 225*B*Sin[3*(e + f*x)] + 210*B*ArcTan[E^(I*(e + f*x))]*(I*Cos[4*(e + f*x)] + Sin[4*(e + f*x)]))
*(-I + Tan[e + f*x])^3*Sqrt[a + I*a*Tan[e + f*x]])/(105*c^3*f*Sqrt[c - I*c*Tan[e + f*x]])

________________________________________________________________________________________

Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 637 vs. \(2 (206 ) = 412\).
time = 0.41, size = 638, normalized size = 2.54

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (-105 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{5}\left (f x +e \right )\right )+1050 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{3}\left (f x +e \right )\right )+337 i B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{4}\left (f x +e \right )\right )+525 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{4}\left (f x +e \right )\right )+30 i A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{3}\left (f x +e \right )\right )-15 A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{4}\left (f x +e \right )\right )-525 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-1176 i B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )-1050 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-950 B \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )+30 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )+167 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}+105 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c +730 B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )+15 A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{105 f \,c^{4} \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (i+\tan \left (f x +e \right )\right )^{5} \sqrt {a c}}\)	\(638\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (-105 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{5}\left (f x +e \right )\right )+1050 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{3}\left (f x +e \right )\right )+337 i B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{4}\left (f x +e \right )\right )+525 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{4}\left (f x +e \right )\right )+30 i A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{3}\left (f x +e \right )\right )-15 A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{4}\left (f x +e \right )\right )-525 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-1176 i B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )-1050 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-950 B \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )+30 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )+167 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}+105 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c +730 B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )+15 A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{105 f \,c^{4} \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (i+\tan \left (f x +e \right )\right )^{5} \sqrt {a c}}\)	\(638\)

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))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/105/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^3/c^4*(-105*I*B*ln((a*c*tan(f*x+e)+(a*c)^(1/2
)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)^5+1050*I*B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+
tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)^3+337*I*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e
)^4+525*B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)^4+30*I*A*(a
*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^3-15*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^4
-525*I*B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)-1176*I*B*(a*
c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^2-1050*B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2)
)^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)^2-950*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^3+30*I*A*(a*c
)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)+167*I*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)+105*B*ln((a*c
*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c+730*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c
)^(1/2)*tan(f*x+e)+15*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(I+tan(f*x+e))^
5/(a*c)^(1/2)

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Maxima [A]
time = 0.62, size = 264, normalized size = 1.05 \begin {gather*} -\frac {{\left (210 \, B a^{3} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 210 \, B a^{3} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - 30 \, {\left (-i \, A - B\right )} a^{3} \cos \left (7 \, f x + 7 \, e\right ) - 84 \, B a^{3} \cos \left (5 \, f x + 5 \, e\right ) + 140 \, B a^{3} \cos \left (3 \, f x + 3 \, e\right ) - 420 \, B a^{3} \cos \left (f x + e\right ) + 105 i \, B a^{3} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - 105 i \, B a^{3} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 30 \, {\left (A - i \, B\right )} a^{3} \sin \left (7 \, f x + 7 \, e\right ) - 84 i \, B a^{3} \sin \left (5 \, f x + 5 \, e\right ) + 140 i \, B a^{3} \sin \left (3 \, f x + 3 \, e\right ) - 420 i \, B a^{3} \sin \left (f x + e\right )\right )} \sqrt {a}}{210 \, c^{\frac {7}{2}} f} \end {gather*}

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))^(7/2),x, algorithm="maxima")

[Out]

-1/210*(210*B*a^3*arctan2(cos(f*x + e), sin(f*x + e) + 1) + 210*B*a^3*arctan2(cos(f*x + e), -sin(f*x + e) + 1)
- 30*(-I*A - B)*a^3*cos(7*f*x + 7*e) - 84*B*a^3*cos(5*f*x + 5*e) + 140*B*a^3*cos(3*f*x + 3*e) - 420*B*a^3*cos
(f*x + e) + 105*I*B*a^3*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - 105*I*B*a^3*log(cos(f*x +
e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) - 30*(A - I*B)*a^3*sin(7*f*x + 7*e) - 84*I*B*a^3*sin(5*f*x + 5*e)
+ 140*I*B*a^3*sin(3*f*x + 3*e) - 420*I*B*a^3*sin(f*x + e))*sqrt(a)/(c^(7/2)*f)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 453 vs. \(2 (207) = 414\).
time = 1.78, size = 453, normalized size = 1.80 \begin {gather*} \frac {105 \, c^{4} f \sqrt {-\frac {B^{2} a^{7}}{c^{7} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (B a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + B a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (c^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{4} f\right )} \sqrt {-\frac {B^{2} a^{7}}{c^{7} f^{2}}}\right )}}{B a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{3}}\right ) - 105 \, c^{4} f \sqrt {-\frac {B^{2} a^{7}}{c^{7} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (B a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + B a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (c^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{4} f\right )} \sqrt {-\frac {B^{2} a^{7}}{c^{7} f^{2}}}\right )}}{B a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{3}}\right ) - 2 \, {\left (15 \, {\left (i \, A + B\right )} a^{3} e^{\left (9 i \, f x + 9 i \, e\right )} + 3 \, {\left (5 i \, A - 9 \, B\right )} a^{3} e^{\left (7 i \, f x + 7 i \, e\right )} + 28 \, B a^{3} e^{\left (5 i \, f x + 5 i \, e\right )} - 140 \, B a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} - 210 \, B a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{210 \, c^{4} f} \end {gather*}

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))^(7/2),x, algorithm="fricas")

[Out]

1/210*(105*c^4*f*sqrt(-B^2*a^7/(c^7*f^2))*log(4*(2*(B*a^3*e^(3*I*f*x + 3*I*e) + B*a^3*e^(I*f*x + I*e))*sqrt(a/
(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) + (c^4*f*e^(2*I*f*x + 2*I*e) - c^4*f)*sqrt(-B^2*a
^7/(c^7*f^2)))/(B*a^3*e^(2*I*f*x + 2*I*e) + B*a^3)) - 105*c^4*f*sqrt(-B^2*a^7/(c^7*f^2))*log(4*(2*(B*a^3*e^(3*
I*f*x + 3*I*e) + B*a^3*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) -
(c^4*f*e^(2*I*f*x + 2*I*e) - c^4*f)*sqrt(-B^2*a^7/(c^7*f^2)))/(B*a^3*e^(2*I*f*x + 2*I*e) + B*a^3)) - 2*(15*(I*
A + B)*a^3*e^(9*I*f*x + 9*I*e) + 3*(5*I*A - 9*B)*a^3*e^(7*I*f*x + 7*I*e) + 28*B*a^3*e^(5*I*f*x + 5*I*e) - 140*
B*a^3*e^(3*I*f*x + 3*I*e) - 210*B*a^3*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x +
2*I*e) + 1)))/(c^4*f)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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))**(7/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3879 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^(7/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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