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

Optimal. Leaf size=447 \[ -\frac{d \left (A \left (2 a^2 d^2+b^2 \left (c^2+3 d^2\right )\right )+a^2 \left (-2 B c d+3 c^2 C+C d^2\right )-a b B \left (c^2+d^2\right )+2 b^2 c (c C-B d)\right )}{f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}-\frac{A b^2-a (b B-a C)}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{\sqrt{b} \left (-a^2 b^2 (d (7 A-C)+2 B c)+5 a^3 b B d-3 a^4 C d+a b^3 (4 A c+B d-4 c C)+b^4 (2 B c-3 A d)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{f \left (a^2+b^2\right )^2 (b c-a d)^{5/2}}-\frac{(i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)^2 (c-i d)^{3/2}}-\frac{(B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)^2 (c+i d)^{3/2}} \]

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

________________________________________________________________________________________

Rubi [A]  time = 2.88148, antiderivative size = 446, normalized size of antiderivative = 1., number of steps used = 13, 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} \[ -\frac{d \left (2 a^2 A d^2+a^2 \left (-2 B c d+3 c^2 C+C d^2\right )-a b B \left (c^2+d^2\right )+A b^2 \left (c^2+3 d^2\right )+2 b^2 c (c C-B d)\right )}{f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}-\frac{A b^2-a (b B-a C)}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{\sqrt{b} \left (-a^2 b^2 (d (7 A-C)+2 B c)+5 a^3 b B d-3 a^4 C d+a b^3 (4 A c+B d-4 c C)+b^4 (2 B c-3 A d)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{f \left (a^2+b^2\right )^2 (b c-a d)^{5/2}}-\frac{(i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)^2 (c-i d)^{3/2}}-\frac{(B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)^2 (c+i d)^{3/2}} \]

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

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]

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 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 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 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]

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

Mathematica [B]  time = 6.25202, size = 2078, normalized size = 4.65 \[ \text{Result too large to show} \]

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])^(3/2)),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.263, size = 40619, normalized size = 90.9 \begin{align*} \text{output too large to display} \end{align*}

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))^(3/2),x)

[Out]

result too large to display

________________________________________________________________________________________

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

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))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

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))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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

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))**(3/2),x)

[Out]

Timed out

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

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))^(3/2),x, algorithm="giac")

[Out]

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