3.2.16 \(\int \frac {(a+b \tan (e+f x))^3 (A+B \tan (e+f x)+C \tan ^2(e+f x))}{(c+d \tan (e+f x))^{3/2}} \, dx\) [116]

Optimal. Leaf size=511 \[ -\frac {(a-i b)^3 (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}-\frac {(i a-b)^3 (A+i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(c+i d)^{3/2} f}-\frac {2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {2 b \left (6 a^2 d^2 \left (12 c^2 C-5 B c d+(5 A+7 C) d^2\right )-15 a b d \left (8 c^3 C-6 B c^2 d+c (3 A+5 C) d^2-3 B d^3\right )+b^2 \left (48 c^4 C-40 B c^3 d+6 c^2 (5 A+3 C) d^2-25 B c d^3+15 (A-C) d^4\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 d^4 \left (c^2+d^2\right ) f}-\frac {2 b^2 \left (4 (b c-a d) \left (6 c^2 C-5 B c d+(5 A+C) d^2\right )-5 d^2 ((A-C) (b c-a d)+B (a c+b d))\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{15 d^3 \left (c^2+d^2\right ) f}+\frac {2 b \left (6 c^2 C-5 B c d+(5 A+C) d^2\right ) (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d^2 \left (c^2+d^2\right ) f} \]

[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)^(3/2)/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)^(3/2)/f+2/15*b*(6*a^2*d^2*(12*c^2*C-5*B*c*d+(5*A+7*C)*d^2)-15
*a*b*d*(8*c^3*C-6*B*c^2*d+c*(3*A+5*C)*d^2-3*B*d^3)+b^2*(48*c^4*C-40*B*c^3*d+6*c^2*(5*A+3*C)*d^2-25*B*c*d^3+15*
(A-C)*d^4))*(c+d*tan(f*x+e))^(1/2)/d^4/(c^2+d^2)/f-2/15*b^2*(4*(-a*d+b*c)*(6*c^2*C-5*B*c*d+(5*A+C)*d^2)-5*d^2*
((A-C)*(-a*d+b*c)+B*(a*c+b*d)))*(c+d*tan(f*x+e))^(1/2)*tan(f*x+e)/d^3/(c^2+d^2)/f+2/5*b*(6*c^2*C-5*B*c*d+(5*A+
C)*d^2)*(c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2/d^2/(c^2+d^2)/f-2*(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))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 1.66, antiderivative size = 511, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {3726, 3728, 3718, 3711, 3620, 3618, 65, 214} \begin {gather*} \frac {2 b \sqrt {c+d \tan (e+f x)} \left (6 a^2 d^2 \left (d^2 (5 A+7 C)-5 B c d+12 c^2 C\right )-15 a b d \left (c d^2 (3 A+5 C)-6 B c^2 d-3 B d^3+8 c^3 C\right )+b^2 \left (6 c^2 d^2 (5 A+3 C)+15 d^4 (A-C)-40 B c^3 d-25 B c d^3+48 c^4 C\right )\right )}{15 d^4 f \left (c^2+d^2\right )}-\frac {2 b^2 \tan (e+f x) \sqrt {c+d \tan (e+f x)} \left (4 (b c-a d) \left (d^2 (5 A+C)-5 B c d+6 c^2 C\right )-5 d^2 ((A-C) (b c-a d)+B (a c+b d))\right )}{15 d^3 f \left (c^2+d^2\right )}+\frac {2 b \left (d^2 (5 A+C)-5 B c d+6 c^2 C\right ) (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d^2 f \left (c^2+d^2\right )}-\frac {2 \left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^3}{d f \left (c^2+d^2\right ) \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)^{3/2}}-\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)^{3/2}} \end {gather*}

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])^(3/2),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)^(3/2)*f)) - ((I*a -
b)^3*(A + I*B - C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((c + I*d)^(3/2)*f) - (2*(c^2*C - B*c*d +
A*d^2)*(a + b*Tan[e + f*x])^3)/(d*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]) + (2*b*(6*a^2*d^2*(12*c^2*C - 5*B*c
*d + (5*A + 7*C)*d^2) - 15*a*b*d*(8*c^3*C - 6*B*c^2*d + c*(3*A + 5*C)*d^2 - 3*B*d^3) + b^2*(48*c^4*C - 40*B*c^
3*d + 6*c^2*(5*A + 3*C)*d^2 - 25*B*c*d^3 + 15*(A - C)*d^4))*Sqrt[c + d*Tan[e + f*x]])/(15*d^4*(c^2 + d^2)*f) -
(2*b^2*(4*(b*c - a*d)*(6*c^2*C - 5*B*c*d + (5*A + C)*d^2) - 5*d^2*((A - C)*(b*c - a*d) + B*(a*c + b*d)))*Tan[
e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(15*d^3*(c^2 + d^2)*f) + (2*b*(6*c^2*C - 5*B*c*d + (5*A + C)*d^2)*(a + b*Ta
n[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])/(5*d^2*(c^2 + d^2)*f)

Rule 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^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 3620

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 3711

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 3718

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 3726

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*Ta
n[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 3728

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])))

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))^{3/2}} \, dx &=-\frac {2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {2 \int \frac {(a+b \tan (e+f x))^2 \left (\frac {1}{2} \left (A d (a c+6 b d)+2 \left (3 b c-\frac {a d}{2}\right ) (c C-B d)\right )+\frac {1}{2} d ((A-C) (b c-a d)+B (a c+b d)) \tan (e+f x)+\frac {1}{2} b \left (6 c^2 C-5 B c d+(5 A+C) d^2\right ) \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx}{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}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {2 b \left (6 c^2 C-5 B c d+(5 A+C) d^2\right ) (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d^2 \left (c^2+d^2\right ) f}+\frac {4 \int \frac {(a+b \tan (e+f x)) \left (\frac {1}{4} \left (-b (4 b c+a d) \left (6 c^2 C-5 B c d+(5 A+C) d^2\right )+5 a d (A d (a c+6 b d)+(6 b c-a d) (c C-B d))\right )+\frac {5}{4} 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)-\frac {1}{4} b \left (4 (b c-a d) \left (6 c^2 C-5 B c d+(5 A+C) d^2\right )-5 d^2 ((A-C) (b c-a d)+B (a c+b d))\right ) \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx}{5 d^2 \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}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}-\frac {2 b^2 \left (4 (b c-a d) \left (6 c^2 C-5 B c d+(5 A+C) d^2\right )-5 d^2 ((A-C) (b c-a d)+B (a c+b d))\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{15 d^3 \left (c^2+d^2\right ) f}+\frac {2 b \left (6 c^2 C-5 B c d+(5 A+C) d^2\right ) (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d^2 \left (c^2+d^2\right ) f}-\frac {8 \int \frac {\frac {1}{8} \left (-15 a^3 d^3 (A c-c C+B d)-3 a^2 b d^2 \left (24 c^2 C-25 B c d+(25 A-C) d^2\right )+30 a b^2 c d \left (4 c^2 C-3 B c d+(3 A+C) d^2\right )-2 b^3 c \left (24 c^3 C-20 B c^2 d+3 c (5 A+3 C) d^2-5 B d^3\right )\right )-\frac {15}{8} d^3 \left (3 a^2 b (A c-c C+B d)-b^3 (A c-c C+B d)+a^3 (B c-(A-C) d)-3 a b^2 (B c-(A-C) d)\right ) \tan (e+f x)-\frac {1}{8} b \left (6 a^2 d^2 \left (12 c^2 C-5 B c d+(5 A+7 C) d^2\right )-15 a b d \left (8 c^3 C-6 B c^2 d+c (3 A+5 C) d^2-3 B d^3\right )+b^2 \left (48 c^4 C-40 B c^3 d+6 c^2 (5 A+3 C) d^2-25 B c d^3+15 (A-C) d^4\right )\right ) \tan ^2(e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{15 d^3 \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}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {2 b \left (6 a^2 d^2 \left (12 c^2 C-5 B c d+(5 A+7 C) d^2\right )-15 a b d \left (8 c^3 C-6 B c^2 d+c (3 A+5 C) d^2-3 B d^3\right )+b^2 \left (48 c^4 C-40 B c^3 d+6 c^2 (5 A+3 C) d^2-25 B c d^3+15 (A-C) d^4\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 d^4 \left (c^2+d^2\right ) f}-\frac {2 b^2 \left (4 (b c-a d) \left (6 c^2 C-5 B c d+(5 A+C) d^2\right )-5 d^2 ((A-C) (b c-a d)+B (a c+b d))\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{15 d^3 \left (c^2+d^2\right ) f}+\frac {2 b \left (6 c^2 C-5 B c d+(5 A+C) d^2\right ) (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d^2 \left (c^2+d^2\right ) f}-\frac {8 \int \frac {-\frac {15}{8} d^3 \left (a^3 (A c-c C+B d)-3 a b^2 (A c-c C+B d)-3 a^2 b (B c-(A-C) d)+b^3 (B c-(A-C) d)\right )-\frac {15}{8} d^3 \left (3 a^2 b (A c-c C+B d)-b^3 (A c-c C+B d)+a^3 (B c-(A-C) d)-3 a b^2 (B c-(A-C) d)\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{15 d^3 \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}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {2 b \left (6 a^2 d^2 \left (12 c^2 C-5 B c d+(5 A+7 C) d^2\right )-15 a b d \left (8 c^3 C-6 B c^2 d+c (3 A+5 C) d^2-3 B d^3\right )+b^2 \left (48 c^4 C-40 B c^3 d+6 c^2 (5 A+3 C) d^2-25 B c d^3+15 (A-C) d^4\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 d^4 \left (c^2+d^2\right ) f}-\frac {2 b^2 \left (4 (b c-a d) \left (6 c^2 C-5 B c d+(5 A+C) d^2\right )-5 d^2 ((A-C) (b c-a d)+B (a c+b d))\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{15 d^3 \left (c^2+d^2\right ) f}+\frac {2 b \left (6 c^2 C-5 B c d+(5 A+C) d^2\right ) (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d^2 \left (c^2+d^2\right ) 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)}+\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)}\\ &=-\frac {2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {2 b \left (6 a^2 d^2 \left (12 c^2 C-5 B c d+(5 A+7 C) d^2\right )-15 a b d \left (8 c^3 C-6 B c^2 d+c (3 A+5 C) d^2-3 B d^3\right )+b^2 \left (48 c^4 C-40 B c^3 d+6 c^2 (5 A+3 C) d^2-25 B c d^3+15 (A-C) d^4\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 d^4 \left (c^2+d^2\right ) f}-\frac {2 b^2 \left (4 (b c-a d) \left (6 c^2 C-5 B c d+(5 A+C) d^2\right )-5 d^2 ((A-C) (b c-a d)+B (a c+b d))\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{15 d^3 \left (c^2+d^2\right ) f}+\frac {2 b \left (6 c^2 C-5 B c d+(5 A+C) d^2\right ) (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d^2 \left (c^2+d^2\right ) f}+\frac {\left (i (a-i b)^3 (A-i B-C)\right ) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (c-i d) f}-\frac {\left (i (a+i b)^3 (A+i B-C)\right ) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (c+i d) f}\\ &=-\frac {2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {2 b \left (6 a^2 d^2 \left (12 c^2 C-5 B c d+(5 A+7 C) d^2\right )-15 a b d \left (8 c^3 C-6 B c^2 d+c (3 A+5 C) d^2-3 B d^3\right )+b^2 \left (48 c^4 C-40 B c^3 d+6 c^2 (5 A+3 C) d^2-25 B c d^3+15 (A-C) d^4\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 d^4 \left (c^2+d^2\right ) f}-\frac {2 b^2 \left (4 (b c-a d) \left (6 c^2 C-5 B c d+(5 A+C) d^2\right )-5 d^2 ((A-C) (b c-a d)+B (a c+b d))\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{15 d^3 \left (c^2+d^2\right ) f}+\frac {2 b \left (6 c^2 C-5 B c d+(5 A+C) d^2\right ) (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d^2 \left (c^2+d^2\right ) f}-\frac {\left ((a-i b)^3 (A-i B-C)\right ) \text {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) d f}-\frac {\left ((a+i b)^3 (A+i B-C)\right ) \text {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) 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)^{3/2} 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)^{3/2} f}-\frac {2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {2 b \left (6 a^2 d^2 \left (12 c^2 C-5 B c d+(5 A+7 C) d^2\right )-15 a b d \left (8 c^3 C-6 B c^2 d+c (3 A+5 C) d^2-3 B d^3\right )+b^2 \left (48 c^4 C-40 B c^3 d+6 c^2 (5 A+3 C) d^2-25 B c d^3+15 (A-C) d^4\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 d^4 \left (c^2+d^2\right ) f}-\frac {2 b^2 \left (4 (b c-a d) \left (6 c^2 C-5 B c d+(5 A+C) d^2\right )-5 d^2 ((A-C) (b c-a d)+B (a c+b d))\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{15 d^3 \left (c^2+d^2\right ) f}+\frac {2 b \left (6 c^2 C-5 B c d+(5 A+C) d^2\right ) (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d^2 \left (c^2+d^2\right ) f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 6.50, size = 920, normalized size = 1.80 \begin {gather*} \frac {2 C (a+b \tan (e+f x))^3}{5 d f \sqrt {c+d \tan (e+f x)}}+\frac {2 \left (\frac {(-6 b c C+5 b B d+6 a C d) (a+b \tan (e+f x))^2}{3 d f \sqrt {c+d \tan (e+f x)}}+\frac {2 \left (\frac {\left (15 b (A b+a B-b C) d^2+4 (b c-a d) (6 b c C-5 b B d-6 a C d)\right ) (a+b \tan (e+f x))}{2 d f \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {2 \left (-48 b^3 c^3 C+40 b^3 B c^2 d+144 a b^2 c^2 C d-30 A b^3 c d^2-110 a b^2 B c d^2-144 a^2 b c C d^2+30 b^3 c C d^2+60 a A b^2 d^3+85 a^2 b B d^3-15 b^3 B d^3+48 a^3 C d^3-60 a b^2 C d^3\right )}{d \sqrt {c+d \tan (e+f x)}}+\frac {2 \left (\frac {1}{2} \left (45 a^2 A b d^3-15 A b^3 d^3+15 a^3 B d^3-45 a b^2 B d^3-45 a^2 b C d^3+15 b^3 C d^3\right ) \left (-\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}\right )+\frac {\left (-\frac {1}{2} c d \left (45 a^2 A b d^3-15 A b^3 d^3+15 a^3 B d^3-45 a b^2 B d^3-45 a^2 b C d^3+15 b^3 C d^3\right )+d^2 \left (\frac {1}{2} \left (-48 b^3 c^3 C+40 b^3 B c^2 d+144 a b^2 c^2 C d-30 A b^3 c d^2-110 a b^2 B c d^2-144 a^2 b c C d^2+30 b^3 c C d^2+15 a^3 A d^3+15 a A b^2 d^3+40 a^2 b B d^3+33 a^3 C d^3-15 a b^2 C d^3\right )+\frac {1}{2} \left (48 b^3 c^3 C-40 b^3 B c^2 d-144 a b^2 c^2 C d+30 A b^3 c d^2+110 a b^2 B c d^2+144 a^2 b c C d^2-30 b^3 c C d^2-60 a A b^2 d^3-85 a^2 b B d^3+15 b^3 B d^3-48 a^3 C d^3+60 a b^2 C d^3\right )\right )\right ) \left (-\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {c+d \tan (e+f x)}{c-i d}\right )}{(i c+d) \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 )}{(i c-d) \sqrt {c+d \tan (e+f x)}}\right )}{d}\right )}{d}}{4 d f}\right )}{3 d}\right )}{5 d} \end {gather*}

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])^(3/2),x]

[Out]

(2*C*(a + b*Tan[e + f*x])^3)/(5*d*f*Sqrt[c + d*Tan[e + f*x]]) + (2*(((-6*b*c*C + 5*b*B*d + 6*a*C*d)*(a + b*Tan
[e + f*x])^2)/(3*d*f*Sqrt[c + d*Tan[e + f*x]]) + (2*(((15*b*(A*b + a*B - b*C)*d^2 + 4*(b*c - a*d)*(6*b*c*C - 5
*b*B*d - 6*a*C*d))*(a + b*Tan[e + f*x]))/(2*d*f*Sqrt[c + d*Tan[e + f*x]]) + ((-2*(-48*b^3*c^3*C + 40*b^3*B*c^2
*d + 144*a*b^2*c^2*C*d - 30*A*b^3*c*d^2 - 110*a*b^2*B*c*d^2 - 144*a^2*b*c*C*d^2 + 30*b^3*c*C*d^2 + 60*a*A*b^2*
d^3 + 85*a^2*b*B*d^3 - 15*b^3*B*d^3 + 48*a^3*C*d^3 - 60*a*b^2*C*d^3))/(d*Sqrt[c + d*Tan[e + f*x]]) + (2*(((45*
a^2*A*b*d^3 - 15*A*b^3*d^3 + 15*a^3*B*d^3 - 45*a*b^2*B*d^3 - 45*a^2*b*C*d^3 + 15*b^3*C*d^3)*(((-I)*ArcTanh[Sqr
t[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/Sqrt[c - I*d] + (I*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/Sqrt
[c + I*d]))/2 + ((-1/2*(c*d*(45*a^2*A*b*d^3 - 15*A*b^3*d^3 + 15*a^3*B*d^3 - 45*a*b^2*B*d^3 - 45*a^2*b*C*d^3 +
15*b^3*C*d^3)) + d^2*((-48*b^3*c^3*C + 40*b^3*B*c^2*d + 144*a*b^2*c^2*C*d - 30*A*b^3*c*d^2 - 110*a*b^2*B*c*d^2
- 144*a^2*b*c*C*d^2 + 30*b^3*c*C*d^2 + 15*a^3*A*d^3 + 15*a*A*b^2*d^3 + 40*a^2*b*B*d^3 + 33*a^3*C*d^3 - 15*a*b
^2*C*d^3)/2 + (48*b^3*c^3*C - 40*b^3*B*c^2*d - 144*a*b^2*c^2*C*d + 30*A*b^3*c*d^2 + 110*a*b^2*B*c*d^2 + 144*a^
2*b*c*C*d^2 - 30*b^3*c*C*d^2 - 60*a*A*b^2*d^3 - 85*a^2*b*B*d^3 + 15*b^3*B*d^3 - 48*a^3*C*d^3 + 60*a*b^2*C*d^3)
/2))*(-(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]])))/d))/d
)/(4*d*f)))/(3*d)))/(5*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(13577\) vs. \(2(476)=952\).
time = 0.66, size = 13578, normalized size = 26.57

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(13578\)
default	\(\text {Expression too large to display}\)	\(13578\)

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))^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {3}{2}}}\, dx \end {gather*}

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))**(3/2),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))**(3/2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

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))^(3/2),x)

[Out]

\text{Hanged}

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