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

Optimal. Leaf size=679 \[ -\frac {d \left (A \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )+a^2 \left (-2 B c d+5 c^2 C+3 C d^2\right )-3 a b B \left (c^2+d^2\right )+2 b^2 c (c C-B d)\right )}{3 f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 (c+d \tan (e+f x))^{3/2}}-\frac {A b^2-a (b B-a C)}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {d \left (2 a^3 d^2 \left (B c^2-B d^2+2 c C d\right )+a^2 b \left (-6 B c^3 d-2 B c d^3+5 c^4 C+2 c^2 C d^2+C d^4\right )-A \left (4 a^3 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )+4 a b^2 c d^3-\left (b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )\right )-a b^2 \left (B c^4+3 B d^4-4 c C d^3\right )+2 b^3 c \left (-3 B c^2 d-B d^3+2 c^3 C\right )\right )}{f \left (a^2+b^2\right ) \left (c^2+d^2\right )^2 (b c-a d)^3 \sqrt {c+d \tan (e+f x)}}-\frac {b^{3/2} \left (-5 a^4 C d+7 a^3 b B d-a^2 b^2 (d (9 A+C)+2 B c)+a b^3 (4 A c+3 B d-4 c C)+b^4 (2 B c-5 A d)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{f \left (a^2+b^2\right )^2 (b c-a d)^{7/2}}-\frac {(i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (a-i b)^2 (c-i d)^{5/2}}-\frac {(B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (a+i b)^2 (c+i d)^{5/2}} \]

[Out]

-(I*A+B-I*C)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(a-I*b)^2/(c-I*d)^(5/2)/f-(B-I*(A-C))*arctanh((c+d*
tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(a+I*b)^2/(c+I*d)^(5/2)/f-b^(3/2)*(7*a^3*b*B*d-5*a^4*C*d+b^4*(-5*A*d+2*B*c)+a
*b^3*(4*A*c+3*B*d-4*C*c)-a^2*b^2*(2*B*c+(9*A+C)*d))*arctanh(b^(1/2)*(c+d*tan(f*x+e))^(1/2)/(-a*d+b*c)^(1/2))/(
a^2+b^2)^2/(-a*d+b*c)^(7/2)/f-d*(2*a^3*d^2*(B*c^2-B*d^2+2*C*c*d)+2*b^3*c*(-3*B*c^2*d-B*d^3+2*C*c^3)-a*b^2*(B*c
^4+3*B*d^4-4*C*c*d^3)+a^2*b*(-6*B*c^3*d-2*B*c*d^3+5*C*c^4+2*C*c^2*d^2+C*d^4)-A*(4*a^3*c*d^3+4*a*b^2*c*d^3-4*a^
2*b*d^2*(2*c^2+d^2)-b^3*(c^4+10*c^2*d^2+5*d^4)))/(a^2+b^2)/(-a*d+b*c)^3/(c^2+d^2)^2/f/(c+d*tan(f*x+e))^(1/2)-1
/3*d*(2*b^2*c*(-B*d+C*c)-3*a*b*B*(c^2+d^2)+a^2*(-2*B*c*d+5*C*c^2+3*C*d^2)+A*(2*a^2*d^2+b^2*(3*c^2+5*d^2)))/(a^
2+b^2)/(-a*d+b*c)^2/(c^2+d^2)/f/(c+d*tan(f*x+e))^(3/2)+(-A*b^2+a*(B*b-C*a))/(a^2+b^2)/(-a*d+b*c)/f/(a+b*tan(f*
x+e))/(c+d*tan(f*x+e))^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 5.06, antiderivative size = 678, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.149, Rules used = {3649, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac {d \left (-A \left (-4 a^2 b d^2 \left (2 c^2+d^2\right )+4 a^3 c d^3+4 a b^2 c d^3+b^3 \left (-\left (10 c^2 d^2+c^4+5 d^4\right )\right )\right )+a^2 b \left (-6 B c^3 d-2 B c d^3+2 c^2 C d^2+5 c^4 C+C d^4\right )+2 a^3 d^2 \left (B c^2-B d^2+2 c C d\right )-a b^2 \left (B c^4+3 B d^4-4 c C d^3\right )+2 b^3 c \left (-3 B c^2 d-B d^3+2 c^3 C\right )\right )}{f \left (a^2+b^2\right ) \left (c^2+d^2\right )^2 (b c-a d)^3 \sqrt {c+d \tan (e+f x)}}-\frac {d \left (2 a^2 A d^2+a^2 \left (-2 B c d+5 c^2 C+3 C d^2\right )-3 a b B \left (c^2+d^2\right )+A b^2 \left (3 c^2+5 d^2\right )+2 b^2 c (c C-B d)\right )}{3 f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 (c+d \tan (e+f x))^{3/2}}-\frac {A b^2-a (b B-a C)}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {b^{3/2} \left (-a^2 b^2 (d (9 A+C)+2 B c)+7 a^3 b B d-5 a^4 C d+a b^3 (4 A c+3 B d-4 c C)+b^4 (2 B c-5 A d)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{f \left (a^2+b^2\right )^2 (b c-a d)^{7/2}}-\frac {(i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (a-i b)^2 (c-i d)^{5/2}}-\frac {(B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (a+i b)^2 (c+i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((I*A + B - I*C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((a - I*b)^2*(c - I*d)^(5/2)*f)) - ((B - I
*(A - C))*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)^2*(c + I*d)^(5/2)*f) - (b^(3/2)*(7*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 + 3*B*d) - a^2*b^2*(2*B*c + (9*A + C)*d))*ArcTan
h[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/((a^2 + b^2)^2*(b*c - a*d)^(7/2)*f) - (d*(2*a^2*A*d^2 +
2*b^2*c*(c*C - B*d) - 3*a*b*B*(c^2 + d^2) + A*b^2*(3*c^2 + 5*d^2) + a^2*(5*c^2*C - 2*B*c*d + 3*C*d^2)))/(3*(a
^2 + b^2)*(b*c - a*d)^2*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) - (A*b^2 - a*(b*B - a*C))/((a^2 + b^2)*(b*c
- a*d)*f*(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(3/2)) - (d*(2*a^3*d^2*(B*c^2 + 2*c*C*d - B*d^2) + 2*b^3*c*
(2*c^3*C - 3*B*c^2*d - B*d^3) - a*b^2*(B*c^4 - 4*c*C*d^3 + 3*B*d^4) + a^2*b*(5*c^4*C - 6*B*c^3*d + 2*c^2*C*d^2
- 2*B*c*d^3 + C*d^4) - A*(4*a^3*c*d^3 + 4*a*b^2*c*d^3 - 4*a^2*b*d^2*(2*c^2 + d^2) - b^3*(c^4 + 10*c^2*d^2 + 5
*d^4))))/((a^2 + b^2)*(b*c - a*d)^3*(c^2 + d^2)^2*f*Sqrt[c + d*Tan[e + f*x]])

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 208

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

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^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 3539

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 3634

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 3649

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*T
an[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])))

Rule 3653

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 {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}} \, dx &=-\frac {A b^2-a (b B-a C)}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {\int \frac {\frac {1}{2} \left (5 A b^2 d-2 a A (b c-a d)-(b B-a C) (2 b c+3 a d)\right )+(A b-a B-b C) (b c-a d) \tan (e+f x)+\frac {5}{2} \left (A b^2-a (b B-a C)\right ) d \tan ^2(e+f x)}{(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 {d \left (2 a^2 A d^2+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+5 d^2\right )+a^2 \left (5 c^2 C-2 B c d+3 C d^2\right )\right )}{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 {A b^2-a (b B-a C)}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {2 \int \frac {-\frac {3}{4} \left (2 a^3 d^2 (A c-c C+B d)-a^2 b (4 A+C) d \left (c^2+d^2\right )+b^3 (2 B c-5 A d) \left (c^2+d^2\right )-a b^2 \left (2 c^3 C-B c^2 d+4 c C d^2-3 B d^3-2 A \left (c^3+2 c d^2\right )\right )\right )+\frac {3}{2} (b c-a d)^2 (A b c-a B c-b c C+a A d+b B d-a C d) \tan (e+f x)+\frac {3}{4} b d \left (2 a^2 A d^2+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+5 d^2\right )+a^2 \left (5 c^2 C-2 B c d+3 C d^2\right )\right ) \tan ^2(e+f x)}{(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 {d \left (2 a^2 A d^2+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+5 d^2\right )+a^2 \left (5 c^2 C-2 B c d+3 C d^2\right )\right )}{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 {A b^2-a (b B-a C)}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {d \left (2 a^3 d^2 \left (B c^2+2 c C d-B d^2\right )+2 b^3 c \left (2 c^3 C-3 B c^2 d-B d^3\right )-a b^2 \left (B c^4-4 c C d^3+3 B d^4\right )+a^2 b \left (5 c^4 C-6 B c^3 d+2 c^2 C d^2-2 B c d^3+C d^4\right )-A \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )\right )}{\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 {4 \int \frac {-\frac {3}{8} \left (b^4 (2 B c-5 A d) \left (c^2+d^2\right )^2+2 a^4 d^3 \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )-2 a^3 b d^2 \left (3 c^3 C-4 B c^2 d+c C d^2-2 B d^3-A c \left (3 c^2+d^2\right )\right )+a^2 b^2 d \left (c^4 C+4 c^2 C d^2-4 B c d^3-C d^4-2 A \left (3 c^4+7 c^2 d^2+2 d^4\right )\right )-a b^3 \left (2 c^5 C+B c^4 d+10 c^3 C d^2-6 B c^2 d^3+4 c C d^4-3 B d^5-2 A \left (c^5+5 c^3 d^2+2 c d^4\right )\right )\right )+\frac {3}{4} (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)+\frac {3}{8} b d \left (2 a^3 d^2 \left (B c^2+2 c C d-B d^2\right )+2 b^3 c \left (2 c^3 C-3 B c^2 d-B d^3\right )-a b^2 \left (B c^4-4 c C d^3+3 B d^4\right )+a^2 b \left (5 c^4 C-6 B c^3 d+2 c^2 C d^2-2 B c d^3+C d^4\right )-A \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )\right ) \tan ^2(e+f x)}{(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 {d \left (2 a^2 A d^2+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+5 d^2\right )+a^2 \left (5 c^2 C-2 B c d+3 C d^2\right )\right )}{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 {A b^2-a (b B-a C)}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {d \left (2 a^3 d^2 \left (B c^2+2 c C d-B d^2\right )+2 b^3 c \left (2 c^3 C-3 B c^2 d-B d^3\right )-a b^2 \left (B c^4-4 c C d^3+3 B d^4\right )+a^2 b \left (5 c^4 C-6 B c^3 d+2 c^2 C d^2-2 B c d^3+C d^4\right )-A \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )\right )}{\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 {4 \int \frac {\frac {3}{4} (b c-a d)^3 \left (a^2 \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )-b^2 \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )+2 a b \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )-\frac {3}{4} (b c-a d)^3 \left (2 a b \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )-a^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )+b^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \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 )^2}+\frac {\left (b^2 \left (7 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+3 B d)-a^2 b^2 (2 B c+(9 A+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 \left (a^2+b^2\right )^2 (b c-a d)^3}\\ &=-\frac {d \left (2 a^2 A d^2+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+5 d^2\right )+a^2 \left (5 c^2 C-2 B c d+3 C d^2\right )\right )}{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 {A b^2-a (b B-a C)}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {d \left (2 a^3 d^2 \left (B c^2+2 c C d-B d^2\right )+2 b^3 c \left (2 c^3 C-3 B c^2 d-B d^3\right )-a b^2 \left (B c^4-4 c C d^3+3 B d^4\right )+a^2 b \left (5 c^4 C-6 B c^3 d+2 c^2 C d^2-2 B c d^3+C d^4\right )-A \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )\right )}{\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 {c+d \tan (e+f x)}} \, dx}{2 (a-i b)^2 (c-i d)^2}+\frac {(A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (a+i b)^2 (c+i d)^2}+\frac {\left (b^2 \left (7 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+3 B d)-a^2 b^2 (2 B c+(9 A+C) d)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 \left (a^2+b^2\right )^2 (b c-a d)^3 f}\\ &=-\frac {d \left (2 a^2 A d^2+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+5 d^2\right )+a^2 \left (5 c^2 C-2 B c d+3 C d^2\right )\right )}{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 {A b^2-a (b B-a C)}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {d \left (2 a^3 d^2 \left (B c^2+2 c C d-B d^2\right )+2 b^3 c \left (2 c^3 C-3 B c^2 d-B d^3\right )-a b^2 \left (B c^4-4 c C d^3+3 B d^4\right )+a^2 b \left (5 c^4 C-6 B c^3 d+2 c^2 C d^2-2 B c d^3+C d^4\right )-A \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )\right )}{\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 {(i A+B-i C) \operatorname {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 (c-i d)^2 f}-\frac {(i (A+i B-C)) \operatorname {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 (c+i d)^2 f}+\frac {\left (b^2 \left (7 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+3 B d)-a^2 b^2 (2 B c+(9 A+C) d)\right )\right ) \operatorname {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 )}{\left (a^2+b^2\right )^2 d (b c-a d)^3 f}\\ &=-\frac {b^{3/2} \left (7 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+3 B d)-a^2 b^2 (2 B c+(9 A+C) d)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\left (a^2+b^2\right )^2 (b c-a d)^{7/2} f}-\frac {d \left (2 a^2 A d^2+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+5 d^2\right )+a^2 \left (5 c^2 C-2 B c d+3 C d^2\right )\right )}{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 {A b^2-a (b B-a C)}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {d \left (2 a^3 d^2 \left (B c^2+2 c C d-B d^2\right )+2 b^3 c \left (2 c^3 C-3 B c^2 d-B d^3\right )-a b^2 \left (B c^4-4 c C d^3+3 B d^4\right )+a^2 b \left (5 c^4 C-6 B c^3 d+2 c^2 C d^2-2 B c d^3+C d^4\right )-A \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )\right )}{\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 {(i (i A+B-i C)) \operatorname {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 (c-i d)^2 d f}-\frac {(A+i B-C) \operatorname {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 (c+i d)^2 d f}\\ &=-\frac {(i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(a-i b)^2 (c-i d)^{5/2} f}-\frac {(B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(a+i b)^2 (c+i d)^{5/2} f}-\frac {b^{3/2} \left (7 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+3 B d)-a^2 b^2 (2 B c+(9 A+C) d)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\left (a^2+b^2\right )^2 (b c-a d)^{7/2} f}-\frac {d \left (2 a^2 A d^2+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+5 d^2\right )+a^2 \left (5 c^2 C-2 B c d+3 C d^2\right )\right )}{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 {A b^2-a (b B-a C)}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {d \left (2 a^3 d^2 \left (B c^2+2 c C d-B d^2\right )+2 b^3 c \left (2 c^3 C-3 B c^2 d-B d^3\right )-a b^2 \left (B c^4-4 c C d^3+3 B d^4\right )+a^2 b \left (5 c^4 C-6 B c^3 d+2 c^2 C d^2-2 B c d^3+C d^4\right )-A \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )\right )}{\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 [B]  time = 6.43, size = 6052, normalized size = 8.91 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

[Out]

Timed out

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maple [B]  time = 0.85, size = 67570, normalized size = 99.51 \[ \text {output too large to display} \]

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

[Out]

result too large to display

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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))^2/(c+d*tan(f*x+e))^(5/2),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 details)Is a*d-b*c positive or negative?

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

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))^2*(c + d*tan(e + f*x))^(5/2)),x)

[Out]

\text{Hanged}

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

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

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