3.2.58 \(\int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \, dx\) [158]

Optimal. Leaf size=598 \[ -\frac {(i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{5/2} (c-i d)^{3/2} f}-\frac {(B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{5/2} (c+i d)^{3/2} f}-\frac {2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+2 (5 A-C) d)\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (8 a^3 b B d \left (c^2+d^2\right )+2 a b^3 (3 A c-3 c C+B d) \left (c^2+d^2\right )-a^4 d \left (8 c^2 C-3 B c d+(3 A+5 C) d^2\right )-a^2 b^2 \left (3 B c^3+11 A c^2 d+5 c^2 C d-3 B c d^2+17 A d^3-C d^3\right )-b^4 \left (d \left (5 A c^2+3 c^2 C+8 A d^2\right )-3 B \left (c^3+2 c d^2\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}} \]

[Out]

-(I*A+B-I*C)*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a-I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))/(a-I*b)^(5/2)/
(c-I*d)^(3/2)/f-(B-I*(A-C))*arctanh((c+I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))
/(a+I*b)^(5/2)/(c+I*d)^(3/2)/f-2/3*(7*a^3*b*B*d-4*a^4*C*d+b^4*(-4*A*d+3*B*c)+a*b^3*(6*A*c+B*d-6*C*c)-a^2*b^2*(
3*B*c+2*(5*A-C)*d))/(a^2+b^2)^2/(-a*d+b*c)^2/f/(a+b*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(1/2)-2/3*d*(8*a^3*b*B*
d*(c^2+d^2)+2*a*b^3*(3*A*c+B*d-3*C*c)*(c^2+d^2)-a^4*d*(8*c^2*C-3*B*c*d+(3*A+5*C)*d^2)-a^2*b^2*(11*A*c^2*d+17*A
*d^3+3*B*c^3-3*B*c*d^2+5*C*c^2*d-C*d^3)-b^4*(d*(5*A*c^2+8*A*d^2+3*C*c^2)-3*B*(c^3+2*c*d^2)))*(a+b*tan(f*x+e))^
(1/2)/(a^2+b^2)^2/(-a*d+b*c)^3/(c^2+d^2)/f/(c+d*tan(f*x+e))^(1/2)-2/3*(A*b^2-a*(B*b-C*a))/(a^2+b^2)/(-a*d+b*c)
/f/(c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^(3/2)

________________________________________________________________________________________

Rubi [A]
time = 2.36, antiderivative size = 598, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {3730, 3697, 3696, 95, 214} \begin {gather*} -\frac {2 \left (A b^2-a (b B-a C)\right )}{3 f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \sqrt {a+b \tan (e+f x)} \left (a^4 (-d) \left (d^2 (3 A+5 C)-3 B c d+8 c^2 C\right )+8 a^3 b B d \left (c^2+d^2\right )-a^2 b^2 \left (11 A c^2 d+17 A d^3+3 B c^3-3 B c d^2+5 c^2 C d-C d^3\right )+2 a b^3 \left (c^2+d^2\right ) (3 A c+B d-3 c C)-b^4 \left (d \left (5 A c^2+8 A d^2+3 c^2 C\right )-3 B \left (c^3+2 c d^2\right )\right )\right )}{3 f \left (a^2+b^2\right )^2 \left (c^2+d^2\right ) (b c-a d)^3 \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (-4 a^4 C d+7 a^3 b B d-a^2 b^2 (2 d (5 A-C)+3 B c)+a b^3 (6 A c+B d-6 c C)+b^4 (3 B c-4 A d)\right )}{3 f \left (a^2+b^2\right )^2 (b c-a d)^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {(i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{5/2} (c-i d)^{3/2}}-\frac {(B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{5/2} (c+i d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2)/((a + b*Tan[e + f*x])^(5/2)*(c + d*Tan[e + f*x])^(3/2)),x]

[Out]

-(((I*A + B - I*C)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]])])
/((a - I*b)^(5/2)*(c - I*d)^(3/2)*f)) - ((B - I*(A - C))*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqr
t[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/((a + I*b)^(5/2)*(c + I*d)^(3/2)*f) - (2*(A*b^2 - a*(b*B - a*C)))/(3*(a
^2 + b^2)*(b*c - a*d)*f*(a + b*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]]) - (2*(7*a^3*b*B*d - 4*a^4*C*d + b
^4*(3*B*c - 4*A*d) + a*b^3*(6*A*c - 6*c*C + B*d) - a^2*b^2*(3*B*c + 2*(5*A - C)*d)))/(3*(a^2 + b^2)^2*(b*c - a
*d)^2*f*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]) - (2*d*(8*a^3*b*B*d*(c^2 + d^2) + 2*a*b^3*(3*A*c -
3*c*C + B*d)*(c^2 + d^2) - a^4*d*(8*c^2*C - 3*B*c*d + (3*A + 5*C)*d^2) - a^2*b^2*(3*B*c^3 + 11*A*c^2*d + 5*c^2
*C*d - 3*B*c*d^2 + 17*A*d^3 - C*d^3) - b^4*(d*(5*A*c^2 + 3*c^2*C + 8*A*d^2) - 3*B*(c^3 + 2*c*d^2)))*Sqrt[a + b
*Tan[e + f*x]])/(3*(a^2 + b^2)^2*(b*c - a*d)^3*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]])

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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 3696

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^2/f, Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e
+ f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 +
B^2, 0]

Rule 3697

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 + I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 -
I*Tan[e + f*x]), x], x] + Dist[(A - I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 + I*Tan[e +
f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2
+ B^2, 0]

Rule 3730

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*b^2 - a*(b*B - a*C))*(a + b*Ta
n[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \, dx &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {2 \int \frac {\frac {1}{2} \left (4 A b^2 d-3 a A (b c-a d)-(b B-a C) (3 b c+a d)\right )+\frac {3}{2} (A b-a B-b C) (b c-a d) \tan (e+f x)+2 \left (A b^2-a (b B-a C)\right ) d \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx}{3 \left (a^2+b^2\right ) (b c-a d)}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+2 (5 A-C) d)\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {4 \int \frac {\frac {1}{4} \left (-\left (a b c-a^2 d-2 b^2 d\right ) \left (a^2 (3 A+C) d-b^2 (3 B c-4 A d)-a b (3 A c-3 c C+B d)\right )+(b c+a d) \left (3 b^3 c C-7 a^2 b B d+4 a^3 C d-A b^2 (3 b c-7 a d)+3 a b^2 (B c-C d)\right )\right )+\frac {3}{4} \left (a^2 B-b^2 B-2 a b (A-C)\right ) (b c-a d)^2 \tan (e+f x)-\frac {1}{2} d \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+10 A d-2 C d)\right ) \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx}{3 \left (a^2+b^2\right )^2 (b c-a d)^2}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+2 (5 A-C) d)\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (8 a^3 b B d \left (c^2+d^2\right )+2 a b^3 (3 A c-3 c C+B d) \left (c^2+d^2\right )-a^4 d \left (8 c^2 C-3 B c d+(3 A+5 C) d^2\right )-a^2 b^2 \left (3 B c^3+11 A c^2 d+5 c^2 C d-3 B c d^2+17 A d^3-C d^3\right )-b^4 \left (d \left (5 A c^2+3 c^2 C+8 A d^2\right )-3 B \left (c^3+2 c d^2\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {8 \int \frac {\frac {3}{8} (b c-a d)^3 \left (a^2 (A c-c C+B d)-b^2 (A c-c C+B d)+2 a b (B c-(A-C) d)\right )-\frac {3}{8} (b c-a d)^3 \left (2 a b (A c-c C+B d)-a^2 (B c-(A-C) d)+b^2 (B c-(A-C) d)\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right )}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+2 (5 A-C) d)\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (8 a^3 b B d \left (c^2+d^2\right )+2 a b^3 (3 A c-3 c C+B d) \left (c^2+d^2\right )-a^4 d \left (8 c^2 C-3 B c d+(3 A+5 C) d^2\right )-a^2 b^2 \left (3 B c^3+11 A c^2 d+5 c^2 C d-3 B c d^2+17 A d^3-C d^3\right )-b^4 \left (d \left (5 A c^2+3 c^2 C+8 A d^2\right )-3 B \left (c^3+2 c d^2\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {(A-i B-C) \int \frac {1+i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (a-i b)^2 (c-i d)}+\frac {(A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (a+i b)^2 (c+i d)}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+2 (5 A-C) d)\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (8 a^3 b B d \left (c^2+d^2\right )+2 a b^3 (3 A c-3 c C+B d) \left (c^2+d^2\right )-a^4 d \left (8 c^2 C-3 B c d+(3 A+5 C) d^2\right )-a^2 b^2 \left (3 B c^3+11 A c^2 d+5 c^2 C d-3 B c d^2+17 A d^3-C d^3\right )-b^4 \left (d \left (5 A c^2+3 c^2 C+8 A d^2\right )-3 B \left (c^3+2 c d^2\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {(A-i B-C) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a-i b)^2 (c-i d) f}+\frac {(A+i B-C) \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a+i b)^2 (c+i d) f}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+2 (5 A-C) d)\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (8 a^3 b B d \left (c^2+d^2\right )+2 a b^3 (3 A c-3 c C+B d) \left (c^2+d^2\right )-a^4 d \left (8 c^2 C-3 B c d+(3 A+5 C) d^2\right )-a^2 b^2 \left (3 B c^3+11 A c^2 d+5 c^2 C d-3 B c d^2+17 A d^3-C d^3\right )-b^4 \left (d \left (5 A c^2+3 c^2 C+8 A d^2\right )-3 B \left (c^3+2 c d^2\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {(A-i B-C) \text {Subst}\left (\int \frac {1}{i a+b-(i c+d) x^2} \, dx,x,\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^2 (c-i d) f}+\frac {(A+i B-C) \text {Subst}\left (\int \frac {1}{-i a+b-(-i c+d) x^2} \, dx,x,\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^2 (c+i d) f}\\ &=-\frac {(i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{5/2} (c-i d)^{3/2} f}-\frac {(B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{5/2} (c+i d)^{3/2} f}-\frac {2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+2 (5 A-C) d)\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (8 a^3 b B d \left (c^2+d^2\right )+2 a b^3 (3 A c-3 c C+B d) \left (c^2+d^2\right )-a^4 d \left (8 c^2 C-3 B c d+(3 A+5 C) d^2\right )-a^2 b^2 \left (3 B c^3+11 A c^2 d+5 c^2 C d-3 B c d^2+17 A d^3-C d^3\right )-b^4 \left (d \left (5 A c^2+3 c^2 C+8 A d^2\right )-3 B \left (c^3+2 c d^2\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 6.62, size = 902, normalized size = 1.51 \begin {gather*} -\frac {2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (-\frac {2 \left (-a \left (-2 a \left (A b^2-a (b B-a C)\right ) d+\frac {3}{2} b (A b-a B-b C) (b c-a d)\right )+\frac {1}{2} b^2 \left (4 A b^2 d-3 a A (b c-a d)-(b B-a C) (3 b c+a d)\right )\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (-\frac {3 (b c-a d)^3 \left (\frac {(a+i b)^2 (i A+B-i C) (c+i d) \tanh ^{-1}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+i b} \sqrt {-c+i d}}+\frac {(a-i b)^2 (A+i B-C) (i c+d) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} \sqrt {c+i d}}\right )}{4 (-b c+a d) \left (c^2+d^2\right ) f}-\frac {2 \left (d^2 \left (\left (-\frac {b c}{2}-\frac {a d}{2}\right ) \left (-2 a \left (A b^2-a (b B-a C)\right ) d+\frac {3}{2} b (A b-a B-b C) (b c-a d)\right )+\frac {1}{2} \left (b^2 d-\frac {1}{2} a (b c-a d)\right ) \left (4 A b^2 d-3 a A (b c-a d)-(b B-a C) (3 b c+a d)\right )\right )-c \left (\frac {1}{2} d (b c-a d) \left (-2 b \left (A b^2-a (b B-a C)\right ) d-\frac {3}{2} a (A b-a B-b C) (b c-a d)+\frac {1}{2} b \left (4 A b^2 d-3 a A (b c-a d)-(b B-a C) (3 b c+a d)\right )\right )-c d \left (-a \left (-2 a \left (A b^2-a (b B-a C)\right ) d+\frac {3}{2} b (A b-a B-b C) (b c-a d)\right )+\frac {1}{2} b^2 \left (4 A b^2 d-3 a A (b c-a d)-(b B-a C) (3 b c+a d)\right )\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{(-b c+a d) \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}\right )}{\left (a^2+b^2\right ) (b c-a d)}\right )}{3 \left (a^2+b^2\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2)/((a + b*Tan[e + f*x])^(5/2)*(c + d*Tan[e + f*x])^(3/2)),x]

[Out]

(-2*(A*b^2 - a*(b*B - a*C)))/(3*(a^2 + b^2)*(b*c - a*d)*f*(a + b*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]])
- (2*((-2*(-(a*(-2*a*(A*b^2 - a*(b*B - a*C))*d + (3*b*(A*b - a*B - b*C)*(b*c - a*d))/2)) + (b^2*(4*A*b^2*d -
3*a*A*(b*c - a*d) - (b*B - a*C)*(3*b*c + a*d)))/2))/((a^2 + b^2)*(b*c - a*d)*f*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c
+ d*Tan[e + f*x]]) - (2*((-3*(b*c - a*d)^3*(((a + I*b)^2*(I*A + B - I*C)*(c + I*d)*ArcTanh[(Sqrt[-c + I*d]*Sq
rt[a + b*Tan[e + f*x]])/(Sqrt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[-a + I*b]*Sqrt[-c + I*d]) + ((a - I*
b)^2*(A + I*B - C)*(I*c + d)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e
+ f*x]])])/(Sqrt[a + I*b]*Sqrt[c + I*d])))/(4*(-(b*c) + a*d)*(c^2 + d^2)*f) - (2*(d^2*((-1/2*(b*c) - (a*d)/2)*
(-2*a*(A*b^2 - a*(b*B - a*C))*d + (3*b*(A*b - a*B - b*C)*(b*c - a*d))/2) + ((b^2*d - (a*(b*c - a*d))/2)*(4*A*b
^2*d - 3*a*A*(b*c - a*d) - (b*B - a*C)*(3*b*c + a*d)))/2) - c*((d*(b*c - a*d)*(-2*b*(A*b^2 - a*(b*B - a*C))*d
- (3*a*(A*b - a*B - b*C)*(b*c - a*d))/2 + (b*(4*A*b^2*d - 3*a*A*(b*c - a*d) - (b*B - a*C)*(3*b*c + a*d)))/2))/
2 - c*d*(-(a*(-2*a*(A*b^2 - a*(b*B - a*C))*d + (3*b*(A*b - a*B - b*C)*(b*c - a*d))/2)) + (b^2*(4*A*b^2*d - 3*a
*A*(b*c - a*d) - (b*B - a*C)*(3*b*c + a*d)))/2)))*Sqrt[a + b*Tan[e + f*x]])/((-(b*c) + a*d)*(c^2 + d^2)*f*Sqrt
[c + d*Tan[e + f*x]])))/((a^2 + b^2)*(b*c - a*d))))/(3*(a^2 + b^2)*(b*c - a*d))

________________________________________________________________________________________

Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {A +B \tan \left (f x +e \right )+C \left (\tan ^{2}\left (f x +e \right )\right )}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {5}{2}} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x)

[Out]

int((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x)

________________________________________________________________________________________

Maxima [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((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(a+b*tan(f*x+e))**(5/2)/(c+d*tan(f*x+e))**(3/2),x)

[Out]

Integral((A + B*tan(e + f*x) + C*tan(e + f*x)**2)/((a + b*tan(e + f*x))**(5/2)*(c + d*tan(e + f*x))**(3/2)), x
)

________________________________________________________________________________________

Giac [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((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

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((A + B*tan(e + f*x) + C*tan(e + f*x)^2)/((a + b*tan(e + f*x))^(5/2)*(c + d*tan(e + f*x))^(3/2)),x)

[Out]

\text{Hanged}

________________________________________________________________________________________