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

[Out]

-(A-I*B-C)*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(I*a+b)^3/f+(A+I*B-C)*(c+I*d)^(5/2)*arc
tanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(I*a-b)^3/f+1/4*(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^2-6*A*d^2-16*B*c*d-8*C*c^2+21*C*d^2)-a*b^5*(56*c*(A-C)*d+B*(24*c^2-35*d^2))-b^6*(4*c*(5
*B*d+2*C*c)-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(b^(1/2)*(c+d*tan(f*x+e))^(1/2)/
(-a*d+b*c)^(1/2))*(-a*d+b*c)^(1/2)/b^(7/2)/(a^2+b^2)^3/f-1/4*d*(3*a^3*b*B*d-15*a^4*C*d-a*b^3*(8*A*c-11*B*d-8*C
*c)+a^2*b^2*(4*B*c+(A-31*C)*d)-b^4*(7*A*d+4*B*c+8*C*d))*(c+d*tan(f*x+e))^(1/2)/b^3/(a^2+b^2)^2/f+1/4*(a^3*b*B*
d-5*a^4*C*d-b^4*(5*A*d+4*B*c)-a*b^3*(8*A*c-9*B*d-8*C*c)+a^2*b^2*(3*A*d+4*B*c-13*C*d))*(c+d*tan(f*x+e))^(3/2)/b
^2/(a^2+b^2)^2/f/(a+b*tan(f*x+e))-1/2*(A*b^2-a*(B*b-C*a))*(c+d*tan(f*x+e))^(5/2)/b/(a^2+b^2)/f/(a+b*tan(f*x+e)
)^2

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Rubi [A]  time = 6.07, antiderivative size = 643, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, 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 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 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]

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*}

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Mathematica [B]  time = 6.89, 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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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((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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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((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]

Timed out

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maple [B]  time = 0.84, size = 20663, normalized size = 32.14 \[ \text {output too large to display} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> 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(((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]

\text{Hanged}

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sympy [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((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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