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

Optimal. Leaf size=585 \[ \frac {2 b \sqrt {c+d \tan (e+f x)} \left (6 a^2 d^3 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+3 a b d \left (-c^2 d^2 (A-17 C)+d^4 (5 A+3 C)-2 B c^3 d-8 B c d^3+8 c^4 C\right )-b^2 \left (2 c^3 d^2 (A+15 C)+8 c d^4 (A+C)-8 B c^4 d-17 B c^2 d^3-3 B d^5+16 c^5 C\right )\right )}{3 d^4 f \left (c^2+d^2\right )^2}+\frac {2 b^2 \tan (e+f x) \sqrt {c+d \tan (e+f x)} \left (3 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (c^2 d^2 (A+15 C)+d^4 (7 A+C)-4 B c^3 d-10 B c d^3+8 c^4 C\right )\right )}{3 d^3 f \left (c^2+d^2\right )^2}-\frac {2 \left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^3}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac {2 (a+b \tan (e+f x))^2 \left (a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (2 A d^4-B c^3 d-3 B c d^3+2 c^4 C+4 c^2 C d^2\right )\right )}{d^2 f \left (c^2+d^2\right )^2 \sqrt {c+d \tan (e+f x)}}-\frac {(-b+i a)^3 (A+i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (c+i d)^{5/2}}-\frac {(a-i b)^3 (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (c-i d)^{5/2}} \]

[Out]

-(a-I*b)^3*(I*A+B-I*C)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(5/2)/f-(I*a-b)^3*(A+I*B-C)*arcta
nh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(c+I*d)^(5/2)/f+2/3*b*(3*a*b*d*(8*c^4*C-2*B*c^3*d-c^2*(A-17*C)*d^2-8*
B*c*d^3+(5*A+3*C)*d^4)-b^2*(16*c^5*C-8*B*c^4*d+2*c^3*(A+15*C)*d^2-17*B*c^2*d^3+8*c*(A+C)*d^4-3*B*d^5)+6*a^2*d^
3*(2*c*(A-C)*d-B*(c^2-d^2)))*(c+d*tan(f*x+e))^(1/2)/d^4/(c^2+d^2)^2/f+2/3*b^2*(b*(8*c^4*C-4*B*c^3*d+c^2*(A+15*
C)*d^2-10*B*c*d^3+(7*A+C)*d^4)+3*a*d^2*(2*c*(A-C)*d-B*(c^2-d^2)))*(c+d*tan(f*x+e))^(1/2)*tan(f*x+e)/d^3/(c^2+d
^2)^2/f-2*(b*(2*A*d^4-B*c^3*d-3*B*c*d^3+2*C*c^4+4*C*c^2*d^2)+a*d^2*(2*c*(A-C)*d-B*(c^2-d^2)))*(a+b*tan(f*x+e))
^2/d^2/(c^2+d^2)^2/f/(c+d*tan(f*x+e))^(1/2)-2/3*(A*d^2-B*c*d+C*c^2)*(a+b*tan(f*x+e))^3/d/(c^2+d^2)/f/(c+d*tan(
f*x+e))^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 2.97, antiderivative size = 585, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.149, Rules used = {3645, 3637, 3630, 3539, 3537, 63, 208} \[ \frac {2 b \sqrt {c+d \tan (e+f x)} \left (6 a^2 d^3 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+3 a b d \left (-c^2 d^2 (A-17 C)+d^4 (5 A+3 C)-2 B c^3 d-8 B c d^3+8 c^4 C\right )-b^2 \left (2 c^3 d^2 (A+15 C)+8 c d^4 (A+C)-17 B c^2 d^3-8 B c^4 d-3 B d^5+16 c^5 C\right )\right )}{3 d^4 f \left (c^2+d^2\right )^2}+\frac {2 b^2 \tan (e+f x) \sqrt {c+d \tan (e+f x)} \left (3 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (c^2 d^2 (A+15 C)+d^4 (7 A+C)-4 B c^3 d-10 B c d^3+8 c^4 C\right )\right )}{3 d^3 f \left (c^2+d^2\right )^2}-\frac {2 (a+b \tan (e+f x))^2 \left (a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (2 A d^4-B c^3 d-3 B c d^3+4 c^2 C d^2+2 c^4 C\right )\right )}{d^2 f \left (c^2+d^2\right )^2 \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^3}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac {(-b+i a)^3 (A+i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (c+i d)^{5/2}}-\frac {(a-i b)^3 (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (c-i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a - I*b)^3*(I*A + B - I*C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(5/2)*f)) - ((I*a -
b)^3*(A + I*B - C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((c + I*d)^(5/2)*f) - (2*(c^2*C - B*c*d +
A*d^2)*(a + b*Tan[e + f*x])^3)/(3*d*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) - (2*(b*(2*c^4*C - B*c^3*d + 4*
c^2*C*d^2 - 3*B*c*d^3 + 2*A*d^4) + a*d^2*(2*c*(A - C)*d - B*(c^2 - d^2)))*(a + b*Tan[e + f*x])^2)/(d^2*(c^2 +
d^2)^2*f*Sqrt[c + d*Tan[e + f*x]]) + (2*b*(3*a*b*d*(8*c^4*C - 2*B*c^3*d - c^2*(A - 17*C)*d^2 - 8*B*c*d^3 + (5*
A + 3*C)*d^4) - b^2*(16*c^5*C - 8*B*c^4*d + 2*c^3*(A + 15*C)*d^2 - 17*B*c^2*d^3 + 8*c*(A + C)*d^4 - 3*B*d^5) +
6*a^2*d^3*(2*c*(A - C)*d - B*(c^2 - d^2)))*Sqrt[c + d*Tan[e + f*x]])/(3*d^4*(c^2 + d^2)^2*f) + (2*b^2*(b*(8*c
^4*C - 4*B*c^3*d + c^2*(A + 15*C)*d^2 - 10*B*c*d^3 + (7*A + C)*d^4) + 3*a*d^2*(2*c*(A - C)*d - B*(c^2 - d^2)))
*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(3*d^3*(c^2 + d^2)^2*f)

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 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 3637

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(b*C*Tan[e + f*x]*(c + d*Tan[e + f*x])
^(n + 1))/(d*f*(n + 2)), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] &&  !LtQ[n, -1]

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]

Rubi steps

\begin {align*} \int \frac {(a+b \tan (e+f x))^3 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac {2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \int \frac {(a+b \tan (e+f x))^2 \left (\frac {3}{2} \left (A d (a c+2 b d)+\frac {2}{3} \left (3 b c-\frac {3 a d}{2}\right ) (c C-B d)\right )+\frac {3}{2} d ((A-C) (b c-a d)+B (a c+b d)) \tan (e+f x)+\frac {3}{2} b \left (2 c^2 C-B c d+(A+C) d^2\right ) \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx}{3 d \left (c^2+d^2\right )}\\ &=-\frac {2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (b \left (2 c^4 C-B c^3 d+4 c^2 C d^2-3 B c d^3+2 A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^2}{d^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {4 \int \frac {(a+b \tan (e+f x)) \left (-\frac {3}{4} \left ((4 b c-a d) \left (a d^2 (B c-(A-C) d)-b (2 c C-B d) \left (c^2+d^2\right )\right )-d (a c+4 b d) \left (a d (A c-c C+B d)+2 b \left (c^2 C-B c d+A d^2\right )\right )\right )-\frac {3}{4} d^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 {3}{4} b \left (b \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx}{3 d^2 \left (c^2+d^2\right )^2}\\ &=-\frac {2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (b \left (2 c^4 C-B c^3 d+4 c^2 C d^2-3 B c d^3+2 A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^2}{d^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {2 b^2 \left (b \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )+3 a d^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)}}{3 d^3 \left (c^2+d^2\right )^2 f}-\frac {8 \int \frac {-\frac {3}{8} \left (6 a b^2 d \left (4 c^4 C-B c^3 d-2 c^2 (A-5 C) d^2-7 B c d^3+4 A d^4\right )-2 b^3 c \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )-3 a^3 d^3 \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )+15 a^2 b d^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )+\frac {9}{8} d^3 \left (3 a^2 b \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )-b^3 \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )+a^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )-3 a b^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \tan (e+f x)-\frac {3}{8} b \left (3 a b d \left (8 c^4 C-2 B c^3 d-c^2 (A-17 C) d^2-8 B c d^3+(5 A+3 C) d^4\right )-b^2 \left (16 c^5 C-8 B c^4 d+2 c^3 (A+15 C) d^2-17 B c^2 d^3+8 c (A+C) d^4-3 B d^5\right )+6 a^2 d^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \tan ^2(e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{9 d^3 \left (c^2+d^2\right )^2}\\ &=-\frac {2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (b \left (2 c^4 C-B c^3 d+4 c^2 C d^2-3 B c d^3+2 A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^2}{d^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {2 b \left (3 a b d \left (8 c^4 C-2 B c^3 d-c^2 (A-17 C) d^2-8 B c d^3+(5 A+3 C) d^4\right )-b^2 \left (16 c^5 C-8 B c^4 d+2 c^3 (A+15 C) d^2-17 B c^2 d^3+8 c (A+C) d^4-3 B d^5\right )+6 a^2 d^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 d^4 \left (c^2+d^2\right )^2 f}+\frac {2 b^2 \left (b \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )+3 a d^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)}}{3 d^3 \left (c^2+d^2\right )^2 f}-\frac {8 \int \frac {\frac {9}{8} d^3 \left (a^3 \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )-3 a b^2 \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )-3 a^2 b \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )+b^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )+\frac {9}{8} d^3 \left (3 a^2 b \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )-b^3 \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )+a^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )-3 a 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}{9 d^3 \left (c^2+d^2\right )^2}\\ &=-\frac {2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (b \left (2 c^4 C-B c^3 d+4 c^2 C d^2-3 B c d^3+2 A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^2}{d^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {2 b \left (3 a b d \left (8 c^4 C-2 B c^3 d-c^2 (A-17 C) d^2-8 B c d^3+(5 A+3 C) d^4\right )-b^2 \left (16 c^5 C-8 B c^4 d+2 c^3 (A+15 C) d^2-17 B c^2 d^3+8 c (A+C) d^4-3 B d^5\right )+6 a^2 d^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 d^4 \left (c^2+d^2\right )^2 f}+\frac {2 b^2 \left (b \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )+3 a d^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)}}{3 d^3 \left (c^2+d^2\right )^2 f}+\frac {\left ((a-i b)^3 (A-i B-C)\right ) \int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (c-i d)^2}+\frac {\left ((a+i b)^3 (A+i B-C)\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (c+i d)^2}\\ &=-\frac {2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (b \left (2 c^4 C-B c^3 d+4 c^2 C d^2-3 B c d^3+2 A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^2}{d^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {2 b \left (3 a b d \left (8 c^4 C-2 B c^3 d-c^2 (A-17 C) d^2-8 B c d^3+(5 A+3 C) d^4\right )-b^2 \left (16 c^5 C-8 B c^4 d+2 c^3 (A+15 C) d^2-17 B c^2 d^3+8 c (A+C) d^4-3 B d^5\right )+6 a^2 d^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 d^4 \left (c^2+d^2\right )^2 f}+\frac {2 b^2 \left (b \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )+3 a d^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)}}{3 d^3 \left (c^2+d^2\right )^2 f}+\frac {\left (i (a-i b)^3 (A-i B-C)\right ) \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (c-i d)^2 f}-\frac {\left (i (a+i b)^3 (A+i B-C)\right ) \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (c+i d)^2 f}\\ &=-\frac {2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (b \left (2 c^4 C-B c^3 d+4 c^2 C d^2-3 B c d^3+2 A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^2}{d^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {2 b \left (3 a b d \left (8 c^4 C-2 B c^3 d-c^2 (A-17 C) d^2-8 B c d^3+(5 A+3 C) d^4\right )-b^2 \left (16 c^5 C-8 B c^4 d+2 c^3 (A+15 C) d^2-17 B c^2 d^3+8 c (A+C) d^4-3 B d^5\right )+6 a^2 d^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 d^4 \left (c^2+d^2\right )^2 f}+\frac {2 b^2 \left (b \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )+3 a d^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)}}{3 d^3 \left (c^2+d^2\right )^2 f}-\frac {\left ((a-i b)^3 (A-i B-C)\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 )}{(c-i d)^2 d f}-\frac {\left ((a+i b)^3 (A+i B-C)\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 )}{(c+i d)^2 d f}\\ &=-\frac {(a-i b)^3 (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}-\frac {(i a-b)^3 (A+i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(c+i d)^{5/2} f}-\frac {2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (b \left (2 c^4 C-B c^3 d+4 c^2 C d^2-3 B c d^3+2 A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^2}{d^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {2 b \left (3 a b d \left (8 c^4 C-2 B c^3 d-c^2 (A-17 C) d^2-8 B c d^3+(5 A+3 C) d^4\right )-b^2 \left (16 c^5 C-8 B c^4 d+2 c^3 (A+15 C) d^2-17 B c^2 d^3+8 c (A+C) d^4-3 B d^5\right )+6 a^2 d^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 d^4 \left (c^2+d^2\right )^2 f}+\frac {2 b^2 \left (b \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )+3 a d^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)}}{3 d^3 \left (c^2+d^2\right )^2 f}\\ \end {align*}

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Mathematica [C]  time = 6.91, size = 670, normalized size = 1.15 \[ \frac {2 C (a+b \tan (e+f x))^3}{3 d f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (-\frac {3 (-2 a C d-b B d+2 b c C) (a+b \tan (e+f x))^2}{d f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (-\frac {3 (a+b \tan (e+f x)) \left (b d^2 (a B+A b-b C)+4 (b c-a d) (-2 a C d-b B d+2 b c C)\right )}{2 d f (c+d \tan (e+f x))^{3/2}}-\frac {3 \left (-\frac {2 \left (16 a^3 C d^3+9 a^2 b B d^3-48 a^2 b c C d^2-18 a b^2 B c d^2+48 a b^2 c^2 C d-2 A b^3 c d^2+8 b^3 B c^2 d+b^3 B d^3-16 b^3 c^3 C+2 b^3 c C d^2\right )}{3 d (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (\frac {\left (\frac {3}{2} c d^4 \left (a^3 B+3 a^2 b (A-C)-3 a b^2 B-b^3 (A-C)\right )+\frac {3}{2} d^5 \left (-\left (a^3 (A-C)\right )+3 a^2 b B+3 a b^2 (A-C)-b^3 B\right )\right ) \left (\frac {\, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {c+d \tan (e+f x)}{c+i d}\right )}{3 (-d+i c) (c+d \tan (e+f x))^{3/2}}-\frac {\, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {c+d \tan (e+f x)}{c-i d}\right )}{3 (d+i c) (c+d \tan (e+f x))^{3/2}}\right )}{d}-\frac {3}{2} d^3 \left (a^3 B+3 a^2 b (A-C)-3 a b^2 B-b^3 (A-C)\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {c+d \tan (e+f x)}{c+i d}\right )}{(-d+i c) \sqrt {c+d \tan (e+f x)}}-\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {c+d \tan (e+f x)}{c-i d}\right )}{(d+i c) \sqrt {c+d \tan (e+f x)}}\right )\right )}{3 d}\right )}{4 d f}\right )}{d}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*C*(a + b*Tan[e + f*x])^3)/(3*d*f*(c + d*Tan[e + f*x])^(3/2)) + (2*((-3*(2*b*c*C - b*B*d - 2*a*C*d)*(a + b*T
an[e + f*x])^2)/(d*f*(c + d*Tan[e + f*x])^(3/2)) + (2*((-3*(b*(A*b + a*B - b*C)*d^2 + 4*(b*c - a*d)*(2*b*c*C -
b*B*d - 2*a*C*d))*(a + b*Tan[e + f*x]))/(2*d*f*(c + d*Tan[e + f*x])^(3/2)) - (3*((-2*(-16*b^3*c^3*C + 8*b^3*B
*c^2*d + 48*a*b^2*c^2*C*d - 2*A*b^3*c*d^2 - 18*a*b^2*B*c*d^2 - 48*a^2*b*c*C*d^2 + 2*b^3*c*C*d^2 + 9*a^2*b*B*d^
3 + b^3*B*d^3 + 16*a^3*C*d^3))/(3*d*(c + d*Tan[e + f*x])^(3/2)) + (2*((((3*c*(a^3*B - 3*a*b^2*B + 3*a^2*b*(A -
C) - b^3*(A - C))*d^4)/2 + (3*(3*a^2*b*B - b^3*B - a^3*(A - C) + 3*a*b^2*(A - C))*d^5)/2)*(-1/3*Hypergeometri
c2F1[-3/2, 1, -1/2, (c + d*Tan[e + f*x])/(c - I*d)]/((I*c + d)*(c + d*Tan[e + f*x])^(3/2)) + Hypergeometric2F1
[-3/2, 1, -1/2, (c + d*Tan[e + f*x])/(c + I*d)]/(3*(I*c - d)*(c + d*Tan[e + f*x])^(3/2))))/d - (3*(a^3*B - 3*a
*b^2*B + 3*a^2*b*(A - C) - b^3*(A - C))*d^3*(-(Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I*d)]
/((I*c + d)*Sqrt[c + d*Tan[e + f*x]])) + Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c + I*d)]/((I*c
- d)*Sqrt[c + d*Tan[e + f*x]])))/2))/(3*d)))/(4*d*f)))/d))/(3*d)

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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))^3*(A+B*tan(f*x+e)+C*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))^3*(A+B*tan(f*x+e)+C*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 = 0.62, size = 85156, normalized size = 145.57 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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))^3*(A + B*tan(e + f*x) + C*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 \[ \int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{3} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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