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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 2.46494, antiderivative size = 511, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.17, Rules used = {3645, 3647, 3637, 3630, 3539, 3537, 63, 208} \[ \frac{2 b \sqrt{c+d \tan (e+f x)} \left (6 a^2 d^2 \left (d^2 (5 A+7 C)-5 B c d+12 c^2 C\right )-15 a b d \left (c d^2 (3 A+5 C)-6 B c^2 d-3 B d^3+8 c^3 C\right )+b^2 \left (6 c^2 d^2 (5 A+3 C)+15 d^4 (A-C)-40 B c^3 d-25 B c d^3+48 c^4 C\right )\right )}{15 d^4 f \left (c^2+d^2\right )}-\frac{2 b^2 \tan (e+f x) \sqrt{c+d \tan (e+f x)} \left (4 (b c-a d) \left (d^2 (5 A+C)-5 B c d+6 c^2 C\right )-5 d^2 ((A-C) (b c-a d)+B (a c+b d))\right )}{15 d^3 f \left (c^2+d^2\right )}+\frac{2 b \left (d^2 (5 A+C)-5 B c d+6 c^2 C\right ) (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d^2 f \left (c^2+d^2\right )}-\frac{2 \left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^3}{d f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{(-b+i a)^3 (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{3/2}}-\frac{(a-i b)^3 (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}} \]

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 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 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 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 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]

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 ) \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) 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) 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 ) \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) 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) 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*}

Mathematica [C]  time = 6.7738, size = 920, normalized size = 1.8 \[ \frac{2 C (a+b \tan (e+f x))^3}{5 d f \sqrt{c+d \tan (e+f x)}}+\frac{2 \left (\frac{(-6 b c C+6 a d C+5 b B d) (a+b \tan (e+f x))^2}{3 d f \sqrt{c+d \tan (e+f x)}}+\frac{2 \left (\frac{\left (15 b (A b-C b+a B) d^2+4 (b c-a d) (6 b c C-6 a d C-5 b B d)\right ) (a+b \tan (e+f x))}{2 d f \sqrt{c+d \tan (e+f x)}}+\frac{\frac{2 \left (\frac{1}{2} \left (-15 A b^3 d^3+45 a^2 A b d^3+15 a^3 B d^3-45 a b^2 B d^3+15 b^3 C d^3-45 a^2 b C d^3\right ) \left (\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d}}\right )+\frac{\left (d^2 \left (\frac{1}{2} \left (-30 A c d^2 b^3+30 c C d^2 b^3-48 c^3 C b^3+40 B c^2 d b^3+15 a A d^3 b^2-15 a C d^3 b^2-110 a B c d^2 b^2+144 a c^2 C d b^2+40 a^2 B d^3 b-144 a^2 c C d^2 b+15 a^3 A d^3+33 a^3 C d^3\right )+\frac{1}{2} \left (15 B d^3 b^3+30 A c d^2 b^3-30 c C d^2 b^3+48 c^3 C b^3-40 B c^2 d b^3-60 a A d^3 b^2+60 a C d^3 b^2+110 a B c d^2 b^2-144 a c^2 C d b^2-85 a^2 B d^3 b+144 a^2 c C d^2 b-48 a^3 C d^3\right )\right )-\frac{1}{2} c d \left (-15 A b^3 d^3+45 a^2 A b d^3+15 a^3 B d^3-45 a b^2 B d^3+15 b^3 C d^3-45 a^2 b C d^3\right )\right ) \left (\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c+i d}\right )}{(i c-d) \sqrt{c+d \tan (e+f x)}}-\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c-i d}\right )}{(i c+d) \sqrt{c+d \tan (e+f x)}}\right )}{d}\right )}{d}-\frac{2 \left (-15 B d^3 b^3-30 A c d^2 b^3+30 c C d^2 b^3-48 c^3 C b^3+40 B c^2 d b^3+60 a A d^3 b^2-60 a C d^3 b^2-110 a B c d^2 b^2+144 a c^2 C d b^2+85 a^2 B d^3 b-144 a^2 c C d^2 b+48 a^3 C d^3\right )}{d \sqrt{c+d \tan (e+f x)}}}{4 d f}\right )}{3 d}\right )}{5 d} \]

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])^(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 + ((-(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))/2 + 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]  time = 0.244, size = 49725, normalized size = 97.3 \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))^3*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(3/2),x)

[Out]

result too large to display

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Maxima [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))^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)]  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))^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., size = 0, normalized size = 0. \begin{align*} \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{align*}

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]  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 (b \tan \left (f x + e\right ) + a\right )}^{3}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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]

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