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

Optimal. Leaf size=643 \[ \frac{\sqrt{b c-a d} \left (2 a^3 b^3 \left (4 c d (A-C)+B \left (4 c^2+3 d^2\right )\right )-3 a^2 b^4 \left (8 A c^2-6 A d^2-16 B c d-8 c^2 C+21 C d^2\right )+a^4 b^2 d (d (A-46 C)+4 B c)+3 a^5 b B d^2-15 a^6 C d^2-a b^5 \left (56 c d (A-C)+B \left (24 c^2-35 d^2\right )\right )-b^6 \left (4 c (5 B d+2 c C)-A \left (8 c^2-15 d^2\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{4 b^{7/2} f \left (a^2+b^2\right )^3}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{2 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^2}+\frac{(c+d \tan (e+f x))^{3/2} \left (a^2 b^2 (3 A d+4 B c-13 C d)+a^3 b B d-5 a^4 C d-a b^3 (8 A c-9 B d-8 c C)-b^4 (5 A d+4 B c)\right )}{4 b^2 f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))}-\frac{d \sqrt{c+d \tan (e+f x)} \left (a^2 b^2 (d (A-31 C)+4 B c)+3 a^3 b B d-15 a^4 C d-a b^3 (8 A c-11 B d-8 c C)-b^4 (7 A d+4 B c+8 C d)\right )}{4 b^3 f \left (a^2+b^2\right )^2}-\frac{(c-i d)^{5/2} (A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (b+i a)^3}+\frac{(c+i d)^{5/2} (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (-b+i a)^3} \]

[Out]

-(((A - I*B - C)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((I*a + b)^3*f)) + ((A + I*B
- C)*(c + I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((I*a - b)^3*f) + (Sqrt[b*c - a*d]*(3*a
^5*b*B*d^2 - 15*a^6*C*d^2 + a^4*b^2*d*(4*B*c + (A - 46*C)*d) - 3*a^2*b^4*(8*A*c^2 - 8*c^2*C - 16*B*c*d - 6*A*d
^2 + 21*C*d^2) - a*b^5*(56*c*(A - C)*d + B*(24*c^2 - 35*d^2)) - b^6*(4*c*(2*c*C + 5*B*d) - A*(8*c^2 - 15*d^2))
+ 2*a^3*b^3*(4*c*(A - C)*d + B*(4*c^2 + 3*d^2)))*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])
/(4*b^(7/2)*(a^2 + b^2)^3*f) - (d*(3*a^3*b*B*d - 15*a^4*C*d - a*b^3*(8*A*c - 8*c*C - 11*B*d) + a^2*b^2*(4*B*c
+ (A - 31*C)*d) - b^4*(4*B*c + 7*A*d + 8*C*d))*Sqrt[c + d*Tan[e + f*x]])/(4*b^3*(a^2 + b^2)^2*f) + ((a^3*b*B*d
- 5*a^4*C*d - b^4*(4*B*c + 5*A*d) - a*b^3*(8*A*c - 8*c*C - 9*B*d) + a^2*b^2*(4*B*c + 3*A*d - 13*C*d))*(c + d*
Tan[e + f*x])^(3/2))/(4*b^2*(a^2 + b^2)^2*f*(a + b*Tan[e + f*x])) - ((A*b^2 - a*(b*B - a*C))*(c + d*Tan[e + f*
x])^(5/2))/(2*b*(a^2 + b^2)*f*(a + b*Tan[e + f*x])^2)

________________________________________________________________________________________

Rubi [A]  time = 6.06519, antiderivative size = 643, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.17, Rules used = {3645, 3647, 3653, 3539, 3537, 63, 208, 3634} \[ \frac{\sqrt{b c-a d} \left (2 a^3 b^3 \left (4 c d (A-C)+B \left (4 c^2+3 d^2\right )\right )-3 a^2 b^4 \left (8 A c^2-6 A d^2-16 B c d-8 c^2 C+21 C d^2\right )+a^4 b^2 d (d (A-46 C)+4 B c)+3 a^5 b B d^2-15 a^6 C d^2-a b^5 \left (56 c d (A-C)+B \left (24 c^2-35 d^2\right )\right )-b^6 \left (4 c (5 B d+2 c C)-A \left (8 c^2-15 d^2\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{4 b^{7/2} f \left (a^2+b^2\right )^3}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{2 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^2}+\frac{(c+d \tan (e+f x))^{3/2} \left (a^2 b^2 (3 A d+4 B c-13 C d)+a^3 b B d-5 a^4 C d-a b^3 (8 A c-9 B d-8 c C)-b^4 (5 A d+4 B c)\right )}{4 b^2 f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))}-\frac{d \sqrt{c+d \tan (e+f x)} \left (a^2 b^2 (d (A-31 C)+4 B c)+3 a^3 b B d-15 a^4 C d-a b^3 (8 A c-11 B d-8 c C)-b^4 (7 A d+4 B c+8 C d)\right )}{4 b^3 f \left (a^2+b^2\right )^2}-\frac{(c-i d)^{5/2} (A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (b+i a)^3}+\frac{(c+i d)^{5/2} (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (-b+i a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((A - I*B - C)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((I*a + b)^3*f)) + ((A + I*B
- C)*(c + I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((I*a - b)^3*f) + (Sqrt[b*c - a*d]*(3*a
^5*b*B*d^2 - 15*a^6*C*d^2 + a^4*b^2*d*(4*B*c + (A - 46*C)*d) - 3*a^2*b^4*(8*A*c^2 - 8*c^2*C - 16*B*c*d - 6*A*d
^2 + 21*C*d^2) - a*b^5*(56*c*(A - C)*d + B*(24*c^2 - 35*d^2)) - b^6*(4*c*(2*c*C + 5*B*d) - A*(8*c^2 - 15*d^2))
+ 2*a^3*b^3*(4*c*(A - C)*d + B*(4*c^2 + 3*d^2)))*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])
/(4*b^(7/2)*(a^2 + b^2)^3*f) - (d*(3*a^3*b*B*d - 15*a^4*C*d - a*b^3*(8*A*c - 8*c*C - 11*B*d) + a^2*b^2*(4*B*c
+ (A - 31*C)*d) - b^4*(4*B*c + 7*A*d + 8*C*d))*Sqrt[c + d*Tan[e + f*x]])/(4*b^3*(a^2 + b^2)^2*f) + ((a^3*b*B*d
- 5*a^4*C*d - b^4*(4*B*c + 5*A*d) - a*b^3*(8*A*c - 8*c*C - 9*B*d) + a^2*b^2*(4*B*c + 3*A*d - 13*C*d))*(c + d*
Tan[e + f*x])^(3/2))/(4*b^2*(a^2 + b^2)^2*f*(a + b*Tan[e + f*x])) - ((A*b^2 - a*(b*B - a*C))*(c + d*Tan[e + f*
x])^(5/2))/(2*b*(a^2 + b^2)*f*(a + b*Tan[e + f*x])^2)

Rule 3645

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

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[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{(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^3} \, dx &=-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}+\frac{\int \frac{(c+d \tan (e+f x))^{3/2} \left (\frac{1}{2} \left (2 (b B-a C) \left (2 b c-\frac{5 a d}{2}\right )+2 A b \left (2 a c+\frac{5 b d}{2}\right )\right )-2 b ((A-C) (b c-a d)-B (a c+b d)) \tan (e+f x)+\frac{1}{2} \left (A b^2-a b B+5 a^2 C+4 b^2 C\right ) d \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^2} \, dx}{2 b \left (a^2+b^2\right )}\\ &=\frac{\left (a^3 b B d-5 a^4 C d-b^4 (4 B c+5 A d)-a b^3 (8 A c-8 c C-9 B d)+a^2 b^2 (4 B c+3 A d-13 C d)\right ) (c+d \tan (e+f x))^{3/2}}{4 b^2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}+\frac{\int \frac{\sqrt{c+d \tan (e+f x)} \left (\frac{1}{4} \left (b (2 a c+3 b d) \left (5 a^2 C d+b^2 (4 B c+5 A d)+a b (4 A c-4 c C-5 B d)\right )+(2 b c-3 a d) \left (a^2 b B d-5 a^3 C d-A b^2 (4 b c-3 a d)+4 b^3 (c C+B d)+4 a b^2 (B c-2 C d)\right )\right )+2 b^2 \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)-\frac{1}{4} d \left (3 a^3 b B d-15 a^4 C d-a b^3 (8 A c-8 c C-11 B d)+a^2 b^2 (4 B c+(A-31 C) d)-b^4 (4 B c+7 A d+8 C d)\right ) \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx}{2 b^2 \left (a^2+b^2\right )^2}\\ &=-\frac{d \left (3 a^3 b B d-15 a^4 C d-a b^3 (8 A c-8 c C-11 B d)+a^2 b^2 (4 B c+(A-31 C) d)-b^4 (4 B c+7 A d+8 C d)\right ) \sqrt{c+d \tan (e+f x)}}{4 b^3 \left (a^2+b^2\right )^2 f}+\frac{\left (a^3 b B d-5 a^4 C d-b^4 (4 B c+5 A d)-a b^3 (8 A c-8 c C-9 B d)+a^2 b^2 (4 B c+3 A d-13 C d)\right ) (c+d \tan (e+f x))^{3/2}}{4 b^2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}+\frac{\int \frac{\frac{1}{8} \left (-15 a^5 C d^3+3 a^4 b d^2 (5 c C+B d)+a^3 b^2 d^2 (B c+(A-31 C) d)-b^5 c \left (8 A c^2-8 c^2 C-20 B c d-15 A d^2\right )+a^2 b^3 \left (8 A c^3-8 c^3 C-20 B c^2 d-17 A c d^2+47 c C d^2+11 B d^3\right )+a b^4 \left (16 B c^3+40 A c^2 d-40 c^2 C d-31 B c d^2-7 A d^3-8 C d^3\right )\right )+b^3 \left (2 a b \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3-A \left (c^3-3 c d^2\right )\right )+a^2 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )-b^2 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right ) \tan (e+f x)-\frac{1}{8} d \left (15 a^5 C d^2-3 a^4 b d (5 c C+B d)-a^3 b^2 d (B c+(A-31 C) d)+b^5 \left (4 B c^2+9 A c d-24 c C d-8 B d^2\right )-a^2 b^3 \left (4 B c^2+7 A c d+23 c C d+3 B d^2\right )+a b^4 \left (8 A c^2-8 c^2 C-17 B c d-9 A d^2+24 C d^2\right )\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{b^3 \left (a^2+b^2\right )^2}\\ &=-\frac{d \left (3 a^3 b B d-15 a^4 C d-a b^3 (8 A c-8 c C-11 B d)+a^2 b^2 (4 B c+(A-31 C) d)-b^4 (4 B c+7 A d+8 C d)\right ) \sqrt{c+d \tan (e+f x)}}{4 b^3 \left (a^2+b^2\right )^2 f}+\frac{\left (a^3 b B d-5 a^4 C d-b^4 (4 B c+5 A d)-a b^3 (8 A c-8 c C-9 B d)+a^2 b^2 (4 B c+3 A d-13 C d)\right ) (c+d \tan (e+f x))^{3/2}}{4 b^2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}+\frac{\int \frac{-b^3 \left (3 a b^2 \left (A c^3-c^3 C-3 B c^2 d-3 A c d^2+3 c C d^2+B d^3\right )+a^3 \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3-A \left (c^3-3 c d^2\right )\right )-3 a^2 b \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )+b^3 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right )-b^3 \left (3 a^2 b \left (A c^3-c^3 C-3 B c^2 d-3 A c d^2+3 c C d^2+B d^3\right )+b^3 \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3-A \left (c^3-3 c d^2\right )\right )-a^3 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )+3 a b^2 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{b^3 \left (a^2+b^2\right )^3}-\frac{\left ((b c-a d) \left (3 a^5 b B d^2-15 a^6 C d^2+a^4 b^2 d (4 B c+(A-46 C) d)-3 a^2 b^4 \left (8 A c^2-8 c^2 C-16 B c d-6 A d^2+21 C d^2\right )-a b^5 \left (56 c (A-C) d+B \left (24 c^2-35 d^2\right )\right )-b^6 \left (4 c (2 c C+5 B d)-A \left (8 c^2-15 d^2\right )\right )+2 a^3 b^3 \left (4 c (A-C) d+B \left (4 c^2+3 d^2\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}{8 b^3 \left (a^2+b^2\right )^3}\\ &=-\frac{d \left (3 a^3 b B d-15 a^4 C d-a b^3 (8 A c-8 c C-11 B d)+a^2 b^2 (4 B c+(A-31 C) d)-b^4 (4 B c+7 A d+8 C d)\right ) \sqrt{c+d \tan (e+f x)}}{4 b^3 \left (a^2+b^2\right )^2 f}+\frac{\left (a^3 b B d-5 a^4 C d-b^4 (4 B c+5 A d)-a b^3 (8 A c-8 c C-9 B d)+a^2 b^2 (4 B c+3 A d-13 C d)\right ) (c+d \tan (e+f x))^{3/2}}{4 b^2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}+\frac{\left ((A-i B-C) (c-i d)^3\right ) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a-i b)^3}+\frac{\left ((A+i B-C) (c+i d)^3\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a+i b)^3}-\frac{\left ((b c-a d) \left (3 a^5 b B d^2-15 a^6 C d^2+a^4 b^2 d (4 B c+(A-46 C) d)-3 a^2 b^4 \left (8 A c^2-8 c^2 C-16 B c d-6 A d^2+21 C d^2\right )-a b^5 \left (56 c (A-C) d+B \left (24 c^2-35 d^2\right )\right )-b^6 \left (4 c (2 c C+5 B d)-A \left (8 c^2-15 d^2\right )\right )+2 a^3 b^3 \left (4 c (A-C) d+B \left (4 c^2+3 d^2\right )\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{8 b^3 \left (a^2+b^2\right )^3 f}\\ &=-\frac{d \left (3 a^3 b B d-15 a^4 C d-a b^3 (8 A c-8 c C-11 B d)+a^2 b^2 (4 B c+(A-31 C) d)-b^4 (4 B c+7 A d+8 C d)\right ) \sqrt{c+d \tan (e+f x)}}{4 b^3 \left (a^2+b^2\right )^2 f}+\frac{\left (a^3 b B d-5 a^4 C d-b^4 (4 B c+5 A d)-a b^3 (8 A c-8 c C-9 B d)+a^2 b^2 (4 B c+3 A d-13 C d)\right ) (c+d \tan (e+f x))^{3/2}}{4 b^2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}+\frac{\left ((A-i B-C) (c-i d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (i a+b)^3 f}-\frac{\left ((A+i B-C) (c+i d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (i a-b)^3 f}-\frac{\left ((b c-a d) \left (3 a^5 b B d^2-15 a^6 C d^2+a^4 b^2 d (4 B c+(A-46 C) d)-3 a^2 b^4 \left (8 A c^2-8 c^2 C-16 B c d-6 A d^2+21 C d^2\right )-a b^5 \left (56 c (A-C) d+B \left (24 c^2-35 d^2\right )\right )-b^6 \left (4 c (2 c C+5 B d)-A \left (8 c^2-15 d^2\right )\right )+2 a^3 b^3 \left (4 c (A-C) d+B \left (4 c^2+3 d^2\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 )}{4 b^3 \left (a^2+b^2\right )^3 d f}\\ &=\frac{\sqrt{b c-a d} \left (3 a^5 b B d^2-15 a^6 C d^2+a^4 b^2 d (4 B c+(A-46 C) d)-3 a^2 b^4 \left (8 A c^2-8 c^2 C-16 B c d-6 A d^2+21 C d^2\right )-a b^5 \left (56 c (A-C) d+B \left (24 c^2-35 d^2\right )\right )-b^6 \left (4 c (2 c C+5 B d)-A \left (8 c^2-15 d^2\right )\right )+2 a^3 b^3 \left (4 c (A-C) d+B \left (4 c^2+3 d^2\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{4 b^{7/2} \left (a^2+b^2\right )^3 f}-\frac{d \left (3 a^3 b B d-15 a^4 C d-a b^3 (8 A c-8 c C-11 B d)+a^2 b^2 (4 B c+(A-31 C) d)-b^4 (4 B c+7 A d+8 C d)\right ) \sqrt{c+d \tan (e+f x)}}{4 b^3 \left (a^2+b^2\right )^2 f}+\frac{\left (a^3 b B d-5 a^4 C d-b^4 (4 B c+5 A d)-a b^3 (8 A c-8 c C-9 B d)+a^2 b^2 (4 B c+3 A d-13 C d)\right ) (c+d \tan (e+f x))^{3/2}}{4 b^2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac{\left ((A-i B-C) (c-i d)^3\right ) \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)^3 d f}-\frac{\left ((A+i B-C) (c+i d)^3\right ) \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)^3 d f}\\ &=-\frac{(A-i B-C) (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(i a+b)^3 f}+\frac{(A+i B-C) (c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(i a-b)^3 f}+\frac{\sqrt{b c-a d} \left (3 a^5 b B d^2-15 a^6 C d^2+a^4 b^2 d (4 B c+(A-46 C) d)-3 a^2 b^4 \left (8 A c^2-8 c^2 C-16 B c d-6 A d^2+21 C d^2\right )-a b^5 \left (56 c (A-C) d+B \left (24 c^2-35 d^2\right )\right )-b^6 \left (4 c (2 c C+5 B d)-A \left (8 c^2-15 d^2\right )\right )+2 a^3 b^3 \left (4 c (A-C) d+B \left (4 c^2+3 d^2\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{4 b^{7/2} \left (a^2+b^2\right )^3 f}-\frac{d \left (3 a^3 b B d-15 a^4 C d-a b^3 (8 A c-8 c C-11 B d)+a^2 b^2 (4 B c+(A-31 C) d)-b^4 (4 B c+7 A d+8 C d)\right ) \sqrt{c+d \tan (e+f x)}}{4 b^3 \left (a^2+b^2\right )^2 f}+\frac{\left (a^3 b B d-5 a^4 C d-b^4 (4 B c+5 A d)-a b^3 (8 A c-8 c C-9 B d)+a^2 b^2 (4 B c+3 A d-13 C d)\right ) (c+d \tan (e+f x))^{3/2}}{4 b^2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}\\ \end{align*}

Mathematica [B]  time = 6.89837, size = 18214, normalized size = 28.33 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

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Maple [B]  time = 0.282, size = 20663, normalized size = 32.1 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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