3.2.8 \(\int \frac {(c+d \tan (e+f x))^{5/2} (A+B \tan (e+f x)+C \tan ^2(e+f x))}{(a+b \tan (e+f x))^2} \, dx\) [108]

Optimal. Leaf size=473 \[ -\frac {(i A+B-i C) (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(a-i b)^2 f}-\frac {(B-i (A-C)) (c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(a+i b)^2 f}+\frac {(b c-a d)^{3/2} \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{7/2} \left (a^2+b^2\right )^2 f}-\frac {d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))} \]

[Out]

-(I*A+B-I*C)*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(a-I*b)^2/f-(B-I*(A-C))*(c+I*d)^(5/2)
*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(a+I*b)^2/f+(-a*d+b*c)^(3/2)*(3*a^3*b*B*d-5*a^4*C*d-b^4*(5*A*d+
2*B*c)-a*b^3*(4*A*c-7*B*d-4*C*c)+a^2*b^2*(2*B*c-(A+9*C)*d))*arctanh(b^(1/2)*(c+d*tan(f*x+e))^(1/2)/(-a*d+b*c)^
(1/2))/b^(7/2)/(a^2+b^2)^2/f-d*(5*a^3*C*d-A*b^2*(-a*d+b*c)-2*b^3*(B*d+2*C*c)-a^2*b*(3*B*d+5*C*c)+a*b^2*(B*c+4*
C*d))*(c+d*tan(f*x+e))^(1/2)/b^3/(a^2+b^2)/f+1/3*(3*A*b^2-3*B*a*b+5*C*a^2+2*C*b^2)*d*(c+d*tan(f*x+e))^(3/2)/b^
2/(a^2+b^2)/f-(A*b^2-a*(B*b-C*a))*(c+d*tan(f*x+e))^(5/2)/b/(a^2+b^2)/f/(a+b*tan(f*x+e))

________________________________________________________________________________________

Rubi [A]
time = 2.64, antiderivative size = 473, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {3726, 3728, 3734, 3620, 3618, 65, 214, 3715} \begin {gather*} -\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac {d \left (5 a^2 C-3 a b B+3 A b^2+2 b^2 C\right ) (c+d \tan (e+f x))^{3/2}}{3 b^2 f \left (a^2+b^2\right )}-\frac {d \sqrt {c+d \tan (e+f x)} \left (5 a^3 C d-a^2 b (3 B d+5 c C)-A b^2 (b c-a d)+a b^2 (B c+4 C d)-2 b^3 (B d+2 c C)\right )}{b^3 f \left (a^2+b^2\right )}+\frac {(b c-a d)^{3/2} \left (-5 a^4 C d+3 a^3 b B d+a^2 b^2 (2 B c-d (A+9 C))-a b^3 (4 A c-7 B d-4 c C)-b^4 (5 A d+2 B c)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{7/2} f \left (a^2+b^2\right )^2}-\frac {(c-i d)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (a-i b)^2}-\frac {(c+i d)^{5/2} (B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (a+i b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((c + d*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2))/(a + b*Tan[e + f*x])^2,x]

[Out]

-(((I*A + B - I*C)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((a - I*b)^2*f)) - ((B - I
*(A - C))*(c + I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)^2*f) + ((b*c - a*d)^(3/2
)*(3*a^3*b*B*d - 5*a^4*C*d - b^4*(2*B*c + 5*A*d) - a*b^3*(4*A*c - 4*c*C - 7*B*d) + a^2*b^2*(2*B*c - (A + 9*C)*
d))*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/(b^(7/2)*(a^2 + b^2)^2*f) - (d*(5*a^3*C*d - A
*b^2*(b*c - a*d) - 2*b^3*(2*c*C + B*d) - a^2*b*(5*c*C + 3*B*d) + a*b^2*(B*c + 4*C*d))*Sqrt[c + d*Tan[e + f*x]]
)/(b^3*(a^2 + b^2)*f) + ((3*A*b^2 - 3*a*b*B + 5*a^2*C + 2*b^2*C)*d*(c + d*Tan[e + f*x])^(3/2))/(3*b^2*(a^2 + b
^2)*f) - ((A*b^2 - a*(b*B - a*C))*(c + d*Tan[e + f*x])^(5/2))/(b*(a^2 + b^2)*f*(a + b*Tan[e + f*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 214

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

Rule 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Rule 3620

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3726

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Ta
n[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3728

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d
*Tan[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3734

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[(c + d*Tan[e + f*x])^n*((1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rubi steps

\begin {align*} \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^2} \, dx &=-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {\int \frac {(c+d \tan (e+f x))^{3/2} \left (\frac {1}{2} \left (2 (b B-a C) \left (b c-\frac {5 a d}{2}\right )+2 A b \left (a c+\frac {5 b d}{2}\right )\right )-b ((A-C) (b c-a d)-B (a c+b d)) \tan (e+f x)+\frac {1}{2} \left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx}{b \left (a^2+b^2\right )}\\ &=\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {2 \int \frac {\sqrt {c+d \tan (e+f x)} \left (-\frac {3}{4} \left (a \left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d^2-b c ((b B-a C) (2 b c-5 a d)+A b (2 a c+5 b d))\right )+\frac {3}{2} b^2 \left (2 a A c d-2 a c C d-A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )+b \left (c^2 C+2 B c d-C d^2\right )\right ) \tan (e+f x)-\frac {3}{4} d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx}{3 b^2 \left (a^2+b^2\right )}\\ &=-\frac {d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {4 \int \frac {\frac {3}{8} \left (5 a^4 C d^3+b^4 c^2 (2 B c+5 A d)-a^3 b d^2 (10 c C+3 B d)+a^2 b^2 d \left (5 c^2 C+4 B c d+(A+4 C) d^2\right )+a b^3 \left (2 A c^3-2 c^3 C-5 B c^2 d-4 A c d^2-6 c C d^2-2 B d^3\right )\right )+\frac {3}{4} b^3 \left (a A d \left (3 c^2-d^2\right )-A b \left (c^3-3 c d^2\right )+b \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3\right )-a \left (C d \left (3 c^2-d^2\right )-B \left (c^3-3 c d^2\right )\right )\right ) \tan (e+f x)+\frac {3}{8} d \left (5 a^4 C d^2-a^3 b d (10 c C+3 B d)+a b^3 \left (B c^2-12 c C d-4 B d^2\right )+2 b^4 \left (3 c^2 C+3 B c d-C d^2\right )+a^2 b^2 \left (5 c^2 C+4 B c d+4 C d^2\right )+A b^2 \left (2 a b c d+a^2 d^2-b^2 \left (c^2-2 d^2\right )\right )\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{3 b^3 \left (a^2+b^2\right )}\\ &=-\frac {d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {4 \int \frac {-\frac {3}{4} b^3 \left (b^2 \left (A c^3-c^3 C-3 B c^2 d-3 A c d^2+3 c C d^2+B d^3\right )+a^2 \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3-A \left (c^3-3 c d^2\right )\right )-2 a b \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right )-\frac {3}{4} b^3 \left (2 a b \left (A c^3-c^3 C-3 B c^2 d-3 A c d^2+3 c C d^2+B d^3\right )-a^2 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )+b^2 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{3 b^3 \left (a^2+b^2\right )^2}-\frac {\left ((b c-a d)^2 \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right )\right ) \int \frac {1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{2 b^3 \left (a^2+b^2\right )^2}\\ &=-\frac {d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {\left ((A-i B-C) (c-i d)^3\right ) \int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (a-i b)^2}+\frac {\left ((A+i B-C) (c+i d)^3\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (a+i b)^2}-\frac {\left ((b c-a d)^2 \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right )\right ) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 b^3 \left (a^2+b^2\right )^2 f}\\ &=-\frac {d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {\left ((i A+B-i C) (c-i d)^3\right ) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (a-i b)^2 f}-\frac {\left (i (A+i B-C) (c+i d)^3\right ) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (a+i b)^2 f}-\frac {\left ((b c-a d)^2 \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right )\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{b^3 \left (a^2+b^2\right )^2 d f}\\ &=\frac {(b c-a d)^{3/2} \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{7/2} \left (a^2+b^2\right )^2 f}-\frac {d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac {\left ((A-i B-C) (c-i d)^3\right ) \text {Subst}\left (\int \frac {1}{-1-\frac {i c}{d}+\frac {i x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{(a-i b)^2 d f}-\frac {\left ((A+i B-C) (c+i d)^3\right ) \text {Subst}\left (\int \frac {1}{-1+\frac {i c}{d}-\frac {i x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{(a+i b)^2 d f}\\ &=-\frac {(i A+B-i C) (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(a-i b)^2 f}-\frac {(B-i (A-C)) (c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(a+i b)^2 f}+\frac {(b c-a d)^{3/2} \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{7/2} \left (a^2+b^2\right )^2 f}-\frac {d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(6112\) vs. \(2(473)=946\).
time = 6.35, size = 6112, normalized size = 12.92 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((c + d*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2))/(a + b*Tan[e + f*x])^2,x]

[Out]

Result too large to show

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(5213\) vs. \(2(434)=868\).
time = 0.69, size = 5214, normalized size = 11.02

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(5214\)
default	\(\text {Expression too large to display}\)	\(5214\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(a+b*tan(f*x+e))**2,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c + d*tan(e + f*x))^(5/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)^2))/(a + b*tan(e + f*x))^2,x)

[Out]

\text{Hanged}

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