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

Optimal. Leaf size=365 \[ -\frac{2 b^{3/2} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{f \left (a^2+b^2\right ) (b c-a d)^{5/2}}+\frac{2 \left (b \left (c^2 d^2 (3 A-C)+A d^4-2 B c^3 d+c^4 C\right )-a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )\right )}{f \left (c^2+d^2\right )^2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}+\frac{2 \left (A d^2-B c d+c^2 C\right )}{3 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^{3/2}}+\frac{(A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (b+i a) (c-i d)^{5/2}}+\frac{(i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b) (c+i d)^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 2.46572, antiderivative size = 365, 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} \[ -\frac{2 b^{3/2} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{f \left (a^2+b^2\right ) (b c-a d)^{5/2}}+\frac{2 \left (b \left (c^2 d^2 (3 A-C)+A d^4-2 B c^3 d+c^4 C\right )-a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )\right )}{f \left (c^2+d^2\right )^2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}+\frac{2 \left (A d^2-B c d+c^2 C\right )}{3 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^{3/2}}+\frac{(A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (b+i a) (c-i d)^{5/2}}+\frac{(i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b) (c+i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A - I*B - C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((I*a + b)*(c - I*d)^(5/2)*f) + ((I*A - B - I*
C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)*(c + I*d)^(5/2)*f) - (2*b^(3/2)*(A*b^2 - a*(b*B
- a*C))*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/((a^2 + b^2)*(b*c - a*d)^(5/2)*f) + (2*(
c^2*C - B*c*d + A*d^2))/(3*(b*c - a*d)*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) + (2*(b*(c^4*C - 2*B*c^3*d +
c^2*(3*A - C)*d^2 + A*d^4) - a*d^2*(2*c*(A - C)*d - B*(c^2 - d^2))))/((b*c - a*d)^2*(c^2 + d^2)^2*f*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)) (c+d \tan (e+f x))^{5/2}} \, dx &=\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \int \frac{-\frac{3}{2} \left (a A c d-a d (c C-B d)-A b \left (c^2+d^2\right )\right )+\frac{3}{2} (b c-a d) (B c-(A-C) d) \tan (e+f x)+\frac{3}{2} b \left (c^2 C-B c d+A d^2\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx}{3 (b c-a d) \left (c^2+d^2\right )}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{4 \int \frac{-\frac{3}{4} \left (A \left (2 a b c^3 d-a^2 d^2 \left (c^2-d^2\right )-b^2 \left (c^2+d^2\right )^2\right )+a d \left (a d \left (c^2 C-2 B c d-C d^2\right )-b \left (2 c^3 C-3 B c^2 d-B d^3\right )\right )\right )-\frac{3}{4} (b c-a d)^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right ) \tan (e+f x)+\frac{3}{4} b \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^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 c-a d)^2 \left (c^2+d^2\right )^2}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{\left (b^2 \left (A b^2-a (b B-a C)\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 ) (b c-a d)^2}+\frac{4 \int \frac{-\frac{3}{4} (b c-a d)^2 \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}{4} (b c-a d)^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)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right )^2}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \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) (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) (c+i d)^2}+\frac{\left (b^2 \left (A b^2-a (b B-a C)\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 ) (b c-a d)^2 f}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \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) (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) (c+i d)^2 f}+\frac{\left (2 b^2 \left (A b^2-a (b B-a C)\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 ) d (b c-a d)^2 f}\\ &=-\frac{2 b^{3/2} \left (A b^2-a (b B-a C)\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 ) (b c-a d)^{5/2} f}+\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \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) (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) d (i c+d)^2 f}\\ &=\frac{(A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(i a+b) (c-i d)^{5/2} f}-\frac{(A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(i a-b) (c+i d)^{5/2} f}-\frac{2 b^{3/2} \left (A b^2-a (b B-a C)\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 ) (b c-a d)^{5/2} f}+\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 6.26429, size = 1948, normalized size = 5.34 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.25, size = 45119, normalized size = 123.6 \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))/(c+d*tan(f*x+e))^(5/2),x)

[Out]

result too large to display

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

[Out]

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