3.2.63 \(\int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx\) [163]

Optimal. Leaf size=651 \[ -\frac {(i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{3/2} (c-i d)^{5/2} f}-\frac {(B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2} (c+i d)^{5/2} f}-\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (b^2 c (c C-B d)-3 a b B \left (c^2+d^2\right )+a^2 \left (4 c^2 C-B c d+3 C d^2\right )+A \left (a^2 d^2+b^2 \left (3 c^2+4 d^2\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (b^3 c \left (5 c^3 C-8 B c^2 d-c C d^2-2 B d^3\right )+a^2 b \left (8 c^4 C-8 B c^3 d+5 c^2 C d^2-2 B c d^3+3 C d^4\right )+3 a^3 d^2 \left (2 c C d+B \left (c^2-d^2\right )\right )+3 a b^2 \left (2 c C d^3-B \left (c^4+c^2 d^2+2 d^4\right )\right )-A \left (6 a^3 c d^3+6 a b^2 c d^3-a^2 b d^2 \left (11 c^2+5 d^2\right )-b^3 \left (3 c^4+17 c^2 d^2+8 d^4\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \]

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

________________________________________________________________________________________

Rubi [A]
time = 2.40, antiderivative size = 650, 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 d \sqrt {a+b \tan (e+f x)} \left (a^2 A d^2+a^2 \left (-B c d+4 c^2 C+3 C d^2\right )-3 a b B \left (c^2+d^2\right )+A b^2 \left (3 c^2+4 d^2\right )+b^2 c (c C-B d)\right )}{3 f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (A b^2-a (b B-a C)\right )}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \sqrt {a+b \tan (e+f x)} \left (3 a^3 d^2 \left (B \left (c^2-d^2\right )+2 c C d\right )+a^2 b \left (-8 B c^3 d-2 B c d^3+8 c^4 C+5 c^2 C d^2+3 C d^4\right )-A \left (6 a^3 c d^3-a^2 b d^2 \left (11 c^2+5 d^2\right )+6 a b^2 c d^3-\left (b^3 \left (3 c^4+17 c^2 d^2+8 d^4\right )\right )\right )+3 a b^2 \left (2 c C d^3-B \left (c^4+c^2 d^2+2 d^4\right )\right )+b^3 c \left (-8 B c^2 d-2 B d^3+5 c^3 C-c C d^2\right )\right )}{3 f \left (a^2+b^2\right ) \left (c^2+d^2\right )^2 (b c-a d)^3 \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)^{3/2} (c-i d)^{5/2}}-\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)^{3/2} (c+i d)^{5/2}} \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])^(3/2)*(c + d*Tan[e + f*x])^(5/2)),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)^(3/2)*(c - I*d)^(5/2)*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)^(3/2)*(c + I*d)^(5/2)*f) - (2*(A*b^2 - a*(b*B - a*C)))/((a^2
+ b^2)*(b*c - a*d)*f*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2)) - (2*d*(a^2*A*d^2 + b^2*c*(c*C - B*
d) - 3*a*b*B*(c^2 + d^2) + A*b^2*(3*c^2 + 4*d^2) + a^2*(4*c^2*C - B*c*d + 3*C*d^2))*Sqrt[a + b*Tan[e + f*x]])/
(3*(a^2 + b^2)*(b*c - a*d)^2*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) - (2*d*(b^3*c*(5*c^3*C - 8*B*c^2*d - c*
C*d^2 - 2*B*d^3) + a^2*b*(8*c^4*C - 8*B*c^3*d + 5*c^2*C*d^2 - 2*B*c*d^3 + 3*C*d^4) + 3*a^3*d^2*(2*c*C*d + B*(c
^2 - d^2)) + 3*a*b^2*(2*c*C*d^3 - B*(c^4 + c^2*d^2 + 2*d^4)) - A*(6*a^3*c*d^3 + 6*a*b^2*c*d^3 - a^2*b*d^2*(11*
c^2 + 5*d^2) - b^3*(3*c^4 + 17*c^2*d^2 + 8*d^4)))*Sqrt[a + b*Tan[e + f*x]])/(3*(a^2 + b^2)*(b*c - a*d)^3*(c^2
+ d^2)^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))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 \int \frac {\frac {1}{2} \left (4 A b^2 d-a A (b c-a d)-(b B-a C) (b c+3 a d)\right )+\frac {1}{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)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx}{\left (a^2+b^2\right ) (b c-a d)}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (a^2 A d^2+b^2 c (c C-B d)-3 a b B \left (c^2+d^2\right )+A b^2 \left (3 c^2+4 d^2\right )+a^2 \left (4 c^2 C-B c d+3 C d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {4 \int \frac {\frac {1}{4} \left (\left (3 b c^2-3 a c d+2 b d^2\right ) \left (a^2 (A+3 C) d-b^2 (B c-4 A d)-a b (A c-c C+3 B d)\right )-d (b c-3 a d) \left (b^2 c C+A b (3 b c+a d)+a^2 (4 c C-B d)-a b (3 B c+C d)\right )\right )-\frac {3}{4} (b c-a d)^2 (b c C-b B d-A (b c+a d)+a (B c+C d)) \tan (e+f x)+\frac {1}{2} b d \left (a^2 A d^2+b^2 c (c C-B d)-3 a b B \left (c^2+d^2\right )+A b^2 \left (3 c^2+4 d^2\right )+a^2 \left (4 c^2 C-B c d+3 C d^2\right )\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 ) (b c-a d)^2 \left (c^2+d^2\right )}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (a^2 A d^2+b^2 c (c C-B d)-3 a b B \left (c^2+d^2\right )+A b^2 \left (3 c^2+4 d^2\right )+a^2 \left (4 c^2 C-B c d+3 C d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (b^3 c \left (5 c^3 C-8 B c^2 d-c C d^2-2 B d^3\right )+a^2 b \left (8 c^4 C-8 B c^3 d+5 c^2 C d^2-2 B c d^3+3 C d^4\right )+3 a^3 d^2 \left (2 c C d+B \left (c^2-d^2\right )\right )+3 a b^2 \left (2 c C d^3-B \left (c^4+c^2 d^2+2 d^4\right )\right )-A \left (6 a^3 c d^3+6 a b^2 c d^3-a^2 b d^2 \left (11 c^2+5 d^2\right )-b^3 \left (3 c^4+17 c^2 d^2+8 d^4\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}-\frac {8 \int \frac {\frac {3}{8} (b c-a d)^3 \left (a \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )+b \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )+\frac {3}{8} (b c-a d)^3 \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)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{3 \left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (a^2 A d^2+b^2 c (c C-B d)-3 a b B \left (c^2+d^2\right )+A b^2 \left (3 c^2+4 d^2\right )+a^2 \left (4 c^2 C-B c d+3 C d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (b^3 c \left (5 c^3 C-8 B c^2 d-c C d^2-2 B d^3\right )+a^2 b \left (8 c^4 C-8 B c^3 d+5 c^2 C d^2-2 B c d^3+3 C d^4\right )+3 a^3 d^2 \left (2 c C d+B \left (c^2-d^2\right )\right )+3 a b^2 \left (2 c C d^3-B \left (c^4+c^2 d^2+2 d^4\right )\right )-A \left (6 a^3 c d^3+6 a b^2 c d^3-a^2 b d^2 \left (11 c^2+5 d^2\right )-b^3 \left (3 c^4+17 c^2 d^2+8 d^4\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 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) (c-i d)^2}+\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) (c+i d)^2}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (a^2 A d^2+b^2 c (c C-B d)-3 a b B \left (c^2+d^2\right )+A b^2 \left (3 c^2+4 d^2\right )+a^2 \left (4 c^2 C-B c d+3 C d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (b^3 c \left (5 c^3 C-8 B c^2 d-c C d^2-2 B d^3\right )+a^2 b \left (8 c^4 C-8 B c^3 d+5 c^2 C d^2-2 B c d^3+3 C d^4\right )+3 a^3 d^2 \left (2 c C d+B \left (c^2-d^2\right )\right )+3 a b^2 \left (2 c C d^3-B \left (c^4+c^2 d^2+2 d^4\right )\right )-A \left (6 a^3 c d^3+6 a b^2 c d^3-a^2 b d^2 \left (11 c^2+5 d^2\right )-b^3 \left (3 c^4+17 c^2 d^2+8 d^4\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 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) (c-i d)^2 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) (c+i d)^2 f}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (a^2 A d^2+b^2 c (c C-B d)-3 a b B \left (c^2+d^2\right )+A b^2 \left (3 c^2+4 d^2\right )+a^2 \left (4 c^2 C-B c d+3 C d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (b^3 c \left (5 c^3 C-8 B c^2 d-c C d^2-2 B d^3\right )+a^2 b \left (8 c^4 C-8 B c^3 d+5 c^2 C d^2-2 B c d^3+3 C d^4\right )+3 a^3 d^2 \left (2 c C d+B \left (c^2-d^2\right )\right )+3 a b^2 \left (2 c C d^3-B \left (c^4+c^2 d^2+2 d^4\right )\right )-A \left (6 a^3 c d^3+6 a b^2 c d^3-a^2 b d^2 \left (11 c^2+5 d^2\right )-b^3 \left (3 c^4+17 c^2 d^2+8 d^4\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 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) (c-i d)^2 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) (c+i d)^2 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)^{3/2} (c-i d)^{5/2} 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)^{3/2} (c+i d)^{5/2} f}-\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (a^2 A d^2+b^2 c (c C-B d)-3 a b B \left (c^2+d^2\right )+A b^2 \left (3 c^2+4 d^2\right )+a^2 \left (4 c^2 C-B c d+3 C d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (b^3 c \left (5 c^3 C-8 B c^2 d-c C d^2-2 B d^3\right )+a^2 b \left (8 c^4 C-8 B c^3 d+5 c^2 C d^2-2 B c d^3+3 C d^4\right )+3 a^3 d^2 \left (2 c C d+B \left (c^2-d^2\right )\right )+3 a b^2 \left (2 c C d^3-B \left (c^4+c^2 d^2+2 d^4\right )\right )-A \left (6 a^3 c d^3+6 a b^2 c d^3-a^2 b d^2 \left (11 c^2+5 d^2\right )-b^3 \left (3 c^4+17 c^2 d^2+8 d^4\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 6.65, size = 903, normalized size = 1.39 \begin {gather*} -\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (-\frac {2 \left (-c \left (-2 c \left (A b^2-a (b B-a C)\right ) d+\frac {1}{2} (A b-a B-b C) d (b c-a d)\right )+\frac {1}{2} d^2 \left (4 A b^2 d-a A (b c-a d)-(b B-a C) (b c+3 a d)\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 (-b c+a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (\frac {3 (b c-a d)^3 \left (\frac {(a+i b) (i A+B-i C) (c+i d)^2 \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 {(i a+b) (A+i B-C) (c-i d)^2 \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 {3 a d}{2}\right ) \left (-2 c \left (A b^2-a (b B-a C)\right ) d+\frac {1}{2} (A b-a B-b C) d (b c-a d)\right )+\frac {1}{2} \left (b d^2-\frac {3}{2} c (-b c+a d)\right ) \left (4 A b^2 d-a A (b c-a d)-(b B-a C) (b c+3 a d)\right )\right )-c \left (\frac {3}{2} d (-b c+a d) \left (-2 \left (A b^2-a (b B-a C)\right ) d^2-\frac {1}{2} c (A b-a B-b C) (b c-a d)+\frac {1}{2} d \left (4 A b^2 d-a A (b c-a d)-(b B-a C) (b c+3 a d)\right )\right )-b c \left (-c \left (-2 c \left (A b^2-a (b B-a C)\right ) d+\frac {1}{2} (A b-a B-b C) d (b c-a d)\right )+\frac {1}{2} d^2 \left (4 A b^2 d-a A (b c-a d)-(b B-a C) (b c+3 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 )}{3 (-b c+a d) \left (c^2+d^2\right )}\right )}{\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])^(3/2)*(c + d*Tan[e + f*x])^(5/2)),x]

[Out]

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

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Maple [F(-1)] grade_annotation
time = 180.00, size = 0, normalized size = 0.00 hanged

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))^(3/2)/(c+d*tan(f*x+e))^(5/2),x)

[Out]

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

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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))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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

[Out]

Timed out

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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 {3}{2}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{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))**(3/2)/(c+d*tan(f*x+e))**(5/2),x)

[Out]

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

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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))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out

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

[Out]

\text{Hanged}

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