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

Optimal. Leaf size=407 \[ \frac{2 \sqrt{c+d \tan (e+f x)} \left (-6 a^2 b d^2 (32 c C-49 B d)+72 a^3 C d^3+21 a b^2 d \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )+b^3 \left (-\left (70 c d^2 (A-C)-56 B c^2 d+105 B d^3+48 c^3 C\right )\right )\right )}{105 d^4 f}+\frac{2 b \tan (e+f x) \sqrt{c+d \tan (e+f x)} \left (35 b d^2 (a B+A b-b C)+4 (b c-a d) (-6 a C d-7 b B d+6 b c C)\right )}{105 d^3 f}-\frac{(-b+i a)^3 (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f \sqrt{c+i d}}+\frac{(b+i a)^3 (A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}}-\frac{2 (-6 a C d-7 b B d+6 b c C) (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{35 d^2 f}+\frac{2 C (a+b \tan (e+f x))^3 \sqrt{c+d \tan (e+f x)}}{7 d f} \]

[Out]

((I*a + b)^3*(A - I*B - C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) - ((I*a - b)^3*(
A + I*B - C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) + (2*(72*a^3*C*d^3 - 6*a^2*b*d
^2*(32*c*C - 49*B*d) + 21*a*b^2*d*(8*c^2*C - 10*B*c*d + 15*(A - C)*d^2) - b^3*(48*c^3*C - 56*B*c^2*d + 70*c*(A
- C)*d^2 + 105*B*d^3))*Sqrt[c + d*Tan[e + f*x]])/(105*d^4*f) + (2*b*(35*b*(A*b + a*B - b*C)*d^2 + 4*(b*c - a*
d)*(6*b*c*C - 7*b*B*d - 6*a*C*d))*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(105*d^3*f) - (2*(6*b*c*C - 7*b*B*d -
6*a*C*d)*(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])/(35*d^2*f) + (2*C*(a + b*Tan[e + f*x])^3*Sqrt[c + d
*Tan[e + f*x]])/(7*d*f)

________________________________________________________________________________________

Rubi [A]  time = 1.69893, antiderivative size = 407, normalized size of antiderivative = 1., 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 = {3647, 3637, 3630, 3539, 3537, 63, 208} \[ \frac{2 \sqrt{c+d \tan (e+f x)} \left (-6 a^2 b d^2 (32 c C-49 B d)+72 a^3 C d^3+21 a b^2 d \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )+b^3 \left (-\left (70 c d^2 (A-C)-56 B c^2 d+105 B d^3+48 c^3 C\right )\right )\right )}{105 d^4 f}+\frac{2 b \tan (e+f x) \sqrt{c+d \tan (e+f x)} \left (35 b d^2 (a B+A b-b C)+4 (b c-a d) (-6 a C d-7 b B d+6 b c C)\right )}{105 d^3 f}-\frac{(-b+i a)^3 (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f \sqrt{c+i d}}+\frac{(b+i a)^3 (A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}}-\frac{2 (-6 a C d-7 b B d+6 b c C) (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{35 d^2 f}+\frac{2 C (a+b \tan (e+f x))^3 \sqrt{c+d \tan (e+f x)}}{7 d f} \]

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))/Sqrt[c + d*Tan[e + f*x]],x]

[Out]

((I*a + b)^3*(A - I*B - C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) - ((I*a - b)^3*(
A + I*B - C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) + (2*(72*a^3*C*d^3 - 6*a^2*b*d
^2*(32*c*C - 49*B*d) + 21*a*b^2*d*(8*c^2*C - 10*B*c*d + 15*(A - C)*d^2) - b^3*(48*c^3*C - 56*B*c^2*d + 70*c*(A
- C)*d^2 + 105*B*d^3))*Sqrt[c + d*Tan[e + f*x]])/(105*d^4*f) + (2*b*(35*b*(A*b + a*B - b*C)*d^2 + 4*(b*c - a*
d)*(6*b*c*C - 7*b*B*d - 6*a*C*d))*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(105*d^3*f) - (2*(6*b*c*C - 7*b*B*d -
6*a*C*d)*(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])/(35*d^2*f) + (2*C*(a + b*Tan[e + f*x])^3*Sqrt[c + d
*Tan[e + f*x]])/(7*d*f)

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 )}{\sqrt{c+d \tan (e+f x)}} \, dx &=\frac{2 C (a+b \tan (e+f x))^3 \sqrt{c+d \tan (e+f x)}}{7 d f}+\frac{2 \int \frac{(a+b \tan (e+f x))^2 \left (\frac{1}{2} (-6 b c C+a (7 A-C) d)+\frac{7}{2} (A b+a B-b C) d \tan (e+f x)-\frac{1}{2} (6 b c C-7 b B d-6 a C d) \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{7 d}\\ &=-\frac{2 (6 b c C-7 b B d-6 a C d) (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{35 d^2 f}+\frac{2 C (a+b \tan (e+f x))^3 \sqrt{c+d \tan (e+f x)}}{7 d f}+\frac{4 \int \frac{(a+b \tan (e+f x)) \left (\frac{1}{4} (-5 a d (6 b c C-a (7 A-C) d)+(4 b c+a d) (6 b c C-7 b B d-6 a C d))+\frac{35}{4} \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2 \tan (e+f x)+\frac{1}{4} \left (35 b (A b+a B-b C) d^2+4 (b c-a d) (6 b c C-7 b B d-6 a C d)\right ) \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{35 d^2}\\ &=\frac{2 b \left (35 b (A b+a B-b C) d^2+4 (b c-a d) (6 b c C-7 b B d-6 a C d)\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{105 d^3 f}-\frac{2 (6 b c C-7 b B d-6 a C d) (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{35 d^2 f}+\frac{2 C (a+b \tan (e+f x))^3 \sqrt{c+d \tan (e+f x)}}{7 d f}-\frac{8 \int \frac{\frac{1}{8} \left (-3 a^3 (35 A-11 C) d^3-42 a b^2 c d (4 c C-5 B d)+3 a^2 b d^2 (64 c C+7 B d)+2 b^3 c \left (24 c^2 C-28 B c d+35 (A-C) d^2\right )\right )-\frac{105}{8} \left (a^3 B-3 a b^2 B+3 a^2 b (A-C)-b^3 (A-C)\right ) d^3 \tan (e+f x)-\frac{1}{8} \left (72 a^3 C d^3-6 a^2 b d^2 (32 c C-49 B d)+21 a b^2 d \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )-b^3 \left (48 c^3 C-56 B c^2 d+70 c (A-C) d^2+105 B d^3\right )\right ) \tan ^2(e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{105 d^3}\\ &=\frac{2 \left (72 a^3 C d^3-6 a^2 b d^2 (32 c C-49 B d)+21 a b^2 d \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )-b^3 \left (48 c^3 C-56 B c^2 d+70 c (A-C) d^2+105 B d^3\right )\right ) \sqrt{c+d \tan (e+f x)}}{105 d^4 f}+\frac{2 b \left (35 b (A b+a B-b C) d^2+4 (b c-a d) (6 b c C-7 b B d-6 a C d)\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{105 d^3 f}-\frac{2 (6 b c C-7 b B d-6 a C d) (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{35 d^2 f}+\frac{2 C (a+b \tan (e+f x))^3 \sqrt{c+d \tan (e+f x)}}{7 d f}-\frac{8 \int \frac{\frac{105}{8} \left (3 a^2 b B-b^3 B-a^3 (A-C)+3 a b^2 (A-C)\right ) d^3-\frac{105}{8} \left (a^3 B-3 a b^2 B+3 a^2 b (A-C)-b^3 (A-C)\right ) d^3 \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{105 d^3}\\ &=\frac{2 \left (72 a^3 C d^3-6 a^2 b d^2 (32 c C-49 B d)+21 a b^2 d \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )-b^3 \left (48 c^3 C-56 B c^2 d+70 c (A-C) d^2+105 B d^3\right )\right ) \sqrt{c+d \tan (e+f x)}}{105 d^4 f}+\frac{2 b \left (35 b (A b+a B-b C) d^2+4 (b c-a d) (6 b c C-7 b B d-6 a C d)\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{105 d^3 f}-\frac{2 (6 b c C-7 b B d-6 a C d) (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{35 d^2 f}+\frac{2 C (a+b \tan (e+f x))^3 \sqrt{c+d \tan (e+f x)}}{7 d f}+\frac{1}{2} \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+\frac{1}{2} \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\\ &=\frac{2 \left (72 a^3 C d^3-6 a^2 b d^2 (32 c C-49 B d)+21 a b^2 d \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )-b^3 \left (48 c^3 C-56 B c^2 d+70 c (A-C) d^2+105 B d^3\right )\right ) \sqrt{c+d \tan (e+f x)}}{105 d^4 f}+\frac{2 b \left (35 b (A b+a B-b C) d^2+4 (b c-a d) (6 b c C-7 b B d-6 a C d)\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{105 d^3 f}-\frac{2 (6 b c C-7 b B d-6 a C d) (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{35 d^2 f}+\frac{2 C (a+b \tan (e+f x))^3 \sqrt{c+d \tan (e+f x)}}{7 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 f}+\frac{\left ((a-i b)^3 (i A+B-i C)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 f}\\ &=\frac{2 \left (72 a^3 C d^3-6 a^2 b d^2 (32 c C-49 B d)+21 a b^2 d \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )-b^3 \left (48 c^3 C-56 B c^2 d+70 c (A-C) d^2+105 B d^3\right )\right ) \sqrt{c+d \tan (e+f x)}}{105 d^4 f}+\frac{2 b \left (35 b (A b+a B-b C) d^2+4 (b c-a d) (6 b c C-7 b B d-6 a C d)\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{105 d^3 f}-\frac{2 (6 b c C-7 b B d-6 a C d) (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{35 d^2 f}+\frac{2 C (a+b \tan (e+f x))^3 \sqrt{c+d \tan (e+f x)}}{7 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 )}{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 )}{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 )}{\sqrt{c-i d} 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 )}{\sqrt{c+i d} f}+\frac{2 \left (72 a^3 C d^3-6 a^2 b d^2 (32 c C-49 B d)+21 a b^2 d \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )-b^3 \left (48 c^3 C-56 B c^2 d+70 c (A-C) d^2+105 B d^3\right )\right ) \sqrt{c+d \tan (e+f x)}}{105 d^4 f}+\frac{2 b \left (35 b (A b+a B-b C) d^2+4 (b c-a d) (6 b c C-7 b B d-6 a C d)\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{105 d^3 f}-\frac{2 (6 b c C-7 b B d-6 a C d) (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{35 d^2 f}+\frac{2 C (a+b \tan (e+f x))^3 \sqrt{c+d \tan (e+f x)}}{7 d f}\\ \end{align*}

Mathematica [B]  time = 6.45581, size = 1200, normalized size = 2.95 \[ \frac{2 C \sqrt{c+d \tan (e+f x)} (a+b \tan (e+f x))^3}{7 d f}+\frac{2 \left (\frac{(-6 b c C+6 a d C+7 b B d) \sqrt{c+d \tan (e+f x)} (a+b \tan (e+f x))^2}{5 d f}+\frac{2 \left (\frac{b \left (35 b (A b-C b+a B) d^2+4 (b c-a d) (6 b c C-6 a d C-7 b B d)\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{6 d f}-\frac{2 \left (\frac{2 \sqrt{c+d \tan (e+f x)} \left (b \left (\frac{1}{4} c \left (35 b (A b-C b+a B) d^2+4 (b c-a d) (6 b c C-6 a d C-7 b B d)\right )-\frac{105}{8} \left (B a^2+2 b (A-C) a-b^2 B\right ) d^3\right )-\frac{3}{8} a d \left (35 b (A b-C b+a B) d^2+4 (b c-a d) (6 b c C-6 a d C-7 b B d)\right )\right )}{d f}+\frac{i \sqrt{c-i d} \left (\frac{1}{4} b c \left (35 b (A b-C b+a B) d^2+4 (b c-a d) (6 b c C-6 a d C-7 b B d)\right )+\frac{3}{8} a d \left (35 b (A b-C b+a B) d^2+4 (b c-a d) (6 b c C-6 a d C-7 b B d)\right )-\frac{3}{8} a d ((4 b c+a d) (6 b c C-6 a d C-7 b B d)-5 a d (6 b c C-a (7 A-C) d))-b \left (\frac{1}{4} c \left (35 b (A b-C b+a B) d^2+4 (b c-a d) (6 b c C-6 a d C-7 b B d)\right )-\frac{105}{8} \left (B a^2+2 b (A-C) a-b^2 B\right ) d^3\right )+\frac{3}{2} i d \left (\frac{35}{4} a \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2-\frac{1}{4} b \left (35 b (A b-C b+a B) d^2+4 (b c-a d) (6 b c C-6 a d C-7 b B d)\right )+\frac{1}{4} b ((4 b c+a d) (6 b c C-6 a d C-7 b B d)-5 a d (6 b c C-a (7 A-C) d))\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(i d-c) f}-\frac{i \sqrt{c+i d} \left (\frac{1}{4} b c \left (35 b (A b-C b+a B) d^2+4 (b c-a d) (6 b c C-6 a d C-7 b B d)\right )+\frac{3}{8} a d \left (35 b (A b-C b+a B) d^2+4 (b c-a d) (6 b c C-6 a d C-7 b B d)\right )-\frac{3}{8} a d ((4 b c+a d) (6 b c C-6 a d C-7 b B d)-5 a d (6 b c C-a (7 A-C) d))-b \left (\frac{1}{4} c \left (35 b (A b-C b+a B) d^2+4 (b c-a d) (6 b c C-6 a d C-7 b B d)\right )-\frac{105}{8} \left (B a^2+2 b (A-C) a-b^2 B\right ) d^3\right )-\frac{3}{2} i d \left (\frac{35}{4} a \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2-\frac{1}{4} b \left (35 b (A b-C b+a B) d^2+4 (b c-a d) (6 b c C-6 a d C-7 b B d)\right )+\frac{1}{4} b ((4 b c+a d) (6 b c C-6 a d C-7 b B d)-5 a d (6 b c C-a (7 A-C) d))\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(-c-i d) f}\right )}{3 d}\right )}{5 d}\right )}{7 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))/Sqrt[c + d*Tan[e + f*x]],x]

[Out]

(2*C*(a + b*Tan[e + f*x])^3*Sqrt[c + d*Tan[e + f*x]])/(7*d*f) + (2*(((-6*b*c*C + 7*b*B*d + 6*a*C*d)*(a + b*Tan
[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])/(5*d*f) + (2*((b*(35*b*(A*b + a*B - b*C)*d^2 + 4*(b*c - a*d)*(6*b*c*C -
7*b*B*d - 6*a*C*d))*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(6*d*f) - (2*((I*Sqrt[c - I*d]*((b*c*(35*b*(A*b +
a*B - b*C)*d^2 + 4*(b*c - a*d)*(6*b*c*C - 7*b*B*d - 6*a*C*d)))/4 + (3*a*d*(35*b*(A*b + a*B - b*C)*d^2 + 4*(b*c
- a*d)*(6*b*c*C - 7*b*B*d - 6*a*C*d)))/8 - (3*a*d*(-5*a*d*(6*b*c*C - a*(7*A - C)*d) + (4*b*c + a*d)*(6*b*c*C
- 7*b*B*d - 6*a*C*d)))/8 - b*((-105*(a^2*B - b^2*B + 2*a*b*(A - C))*d^3)/8 + (c*(35*b*(A*b + a*B - b*C)*d^2 +
4*(b*c - a*d)*(6*b*c*C - 7*b*B*d - 6*a*C*d)))/4) + ((3*I)/2)*d*((35*a*(a^2*B - b^2*B + 2*a*b*(A - C))*d^2)/4 -
(b*(35*b*(A*b + a*B - b*C)*d^2 + 4*(b*c - a*d)*(6*b*c*C - 7*b*B*d - 6*a*C*d)))/4 + (b*(-5*a*d*(6*b*c*C - a*(7
*A - C)*d) + (4*b*c + a*d)*(6*b*c*C - 7*b*B*d - 6*a*C*d)))/4))*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]]
)/((-c + I*d)*f) - (I*Sqrt[c + I*d]*((b*c*(35*b*(A*b + a*B - b*C)*d^2 + 4*(b*c - a*d)*(6*b*c*C - 7*b*B*d - 6*a
*C*d)))/4 + (3*a*d*(35*b*(A*b + a*B - b*C)*d^2 + 4*(b*c - a*d)*(6*b*c*C - 7*b*B*d - 6*a*C*d)))/8 - (3*a*d*(-5*
a*d*(6*b*c*C - a*(7*A - C)*d) + (4*b*c + a*d)*(6*b*c*C - 7*b*B*d - 6*a*C*d)))/8 - b*((-105*(a^2*B - b^2*B + 2*
a*b*(A - C))*d^3)/8 + (c*(35*b*(A*b + a*B - b*C)*d^2 + 4*(b*c - a*d)*(6*b*c*C - 7*b*B*d - 6*a*C*d)))/4) - ((3*
I)/2)*d*((35*a*(a^2*B - b^2*B + 2*a*b*(A - C))*d^2)/4 - (b*(35*b*(A*b + a*B - b*C)*d^2 + 4*(b*c - a*d)*(6*b*c*
C - 7*b*B*d - 6*a*C*d)))/4 + (b*(-5*a*d*(6*b*c*C - a*(7*A - C)*d) + (4*b*c + a*d)*(6*b*c*C - 7*b*B*d - 6*a*C*d
)))/4))*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((-c - I*d)*f) + (2*((-3*a*d*(35*b*(A*b + a*B - b*C)*
d^2 + 4*(b*c - a*d)*(6*b*c*C - 7*b*B*d - 6*a*C*d)))/8 + b*((-105*(a^2*B - b^2*B + 2*a*b*(A - C))*d^3)/8 + (c*(
35*b*(A*b + a*B - b*C)*d^2 + 4*(b*c - a*d)*(6*b*c*C - 7*b*B*d - 6*a*C*d)))/4))*Sqrt[c + d*Tan[e + f*x]])/(d*f)
))/(3*d)))/(5*d)))/(7*d)

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Maple [B]  time = 0.211, size = 25426, normalized size = 62.5 \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))^(1/2),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))^(1/2),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))^(1/2),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 )}{\sqrt{c + d \tan{\left (e + f x \right )}}}\, 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))**(1/2),x)

[Out]

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

________________________________________________________________________________________

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}}{\sqrt{d \tan \left (f x + e\right ) + c}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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