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

Optimal. Leaf size=508 \[ \frac{\left (-a^2 b d^2 (2 B d+3 c C)+a^3 C d^3+a b^2 d \left (8 d^2 (A-C)+12 B c d+3 c^2 C\right )+b^3 \left (-\left (-24 c d^2 (A-C)-6 B c^2 d+16 B d^3+c^3 C\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{8 b^{5/2} d^{3/2} f}+\frac{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left (8 b d^2 (a B+A b-b C)-(b c-a d) (-a C d-6 b B d+b c C)\right )}{8 b^2 d f}-\frac{\sqrt{a-i b} (c-i d)^{3/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}-\frac{\sqrt{a+i b} (c+i d)^{3/2} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}-\frac{(-a C d-6 b B d+b c C) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b d f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f} \]

[Out]

-((Sqrt[a - I*b]*(I*A + B - I*C)*(c - I*d)^(3/2)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*
b]*Sqrt[c + d*Tan[e + f*x]])])/f) - (Sqrt[a + I*b]*(B - I*(A - C))*(c + I*d)^(3/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt
[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/f + ((a^3*C*d^3 - a^2*b*d^2*(3*c*C + 2*B*d) +
a*b^2*d*(3*c^2*C + 12*B*c*d + 8*(A - C)*d^2) - b^3*(c^3*C - 6*B*c^2*d - 24*c*(A - C)*d^2 + 16*B*d^3))*ArcTanh
[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/(8*b^(5/2)*d^(3/2)*f) + ((8*b*(A*b +
a*B - b*C)*d^2 - (b*c - a*d)*(b*c*C - 6*b*B*d - a*C*d))*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(8*
b^2*d*f) - ((b*c*C - 6*b*B*d - a*C*d)*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(12*b*d*f) + (C*Sqr
t[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(5/2))/(3*d*f)

________________________________________________________________________________________

Rubi [A]  time = 7.48825, antiderivative size = 508, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 49, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {3647, 3655, 6725, 63, 217, 206, 93, 208} \[ \frac{\left (-a^2 b d^2 (2 B d+3 c C)+a^3 C d^3+a b^2 d \left (8 d^2 (A-C)+12 B c d+3 c^2 C\right )+b^3 \left (-\left (-24 c d^2 (A-C)-6 B c^2 d+16 B d^3+c^3 C\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{8 b^{5/2} d^{3/2} f}+\frac{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left (8 b d^2 (a B+A b-b C)-(b c-a d) (-a C d-6 b B d+b c C)\right )}{8 b^2 d f}-\frac{\sqrt{a-i b} (c-i d)^{3/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}-\frac{\sqrt{a+i b} (c+i d)^{3/2} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}-\frac{(-a C d-6 b B d+b c C) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b d f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[a - I*b]*(I*A + B - I*C)*(c - I*d)^(3/2)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*
b]*Sqrt[c + d*Tan[e + f*x]])])/f) - (Sqrt[a + I*b]*(B - I*(A - C))*(c + I*d)^(3/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt
[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/f + ((a^3*C*d^3 - a^2*b*d^2*(3*c*C + 2*B*d) +
a*b^2*d*(3*c^2*C + 12*B*c*d + 8*(A - C)*d^2) - b^3*(c^3*C - 6*B*c^2*d - 24*c*(A - C)*d^2 + 16*B*d^3))*ArcTanh
[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/(8*b^(5/2)*d^(3/2)*f) + ((8*b*(A*b +
a*B - b*C)*d^2 - (b*c - a*d)*(b*c*C - 6*b*B*d - a*C*d))*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(8*
b^2*d*f) - ((b*c*C - 6*b*B*d - a*C*d)*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(12*b*d*f) + (C*Sqr
t[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(5/2))/(3*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 3655

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] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[((a + b*ff*x)^m*(c + d*ff*x)^n*(A + B*ff*x + C*ff^2*x^2))/(1 + ff^2*x^2), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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 \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{\int \frac{(c+d \tan (e+f x))^{3/2} \left (\frac{1}{2} (-b c C+a (6 A-5 C) d)+3 (A b+a B-b C) d \tan (e+f x)-\frac{1}{2} (b c C-6 b B d-a C d) \tan ^2(e+f x)\right )}{\sqrt{a+b \tan (e+f x)}} \, dx}{3 d}\\ &=-\frac{(b c C-6 b B d-a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b d f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{\int \frac{\sqrt{c+d \tan (e+f x)} \left (-\frac{3}{4} \left (a^2 C d^2-2 a b d (4 A c-3 c C-3 B d)+b^2 c (c C+2 B d)\right )+6 b d (A b c+a B c-b c C+a A d-b B d-a C d) \tan (e+f x)+\frac{3}{4} \left (8 b (A b+a B-b C) d^2-(b c-a d) (b c C-6 b B d-a C d)\right ) \tan ^2(e+f x)\right )}{\sqrt{a+b \tan (e+f x)}} \, dx}{6 b d}\\ &=\frac{\left (8 b (A b+a B-b C) d^2-(b c-a d) (b c C-6 b B d-a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b^2 d f}-\frac{(b c C-6 b B d-a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b d f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{\int \frac{\frac{3}{8} \left (a^3 C d^3-a^2 b d^2 (3 c C+2 B d)-b^3 c \left (c^2 C+10 B c d+8 (A-C) d^2\right )-a b^2 d \left (13 c^2 C+20 B c d-8 C d^2-8 A \left (2 c^2-d^2\right )\right )\right )+6 b^2 d \left (2 a A c d-2 a c C d+A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )-b \left (c^2 C+2 B c d-C d^2\right )\right ) \tan (e+f x)+\frac{3}{8} \left (a^3 C d^3-a^2 b d^2 (3 c C+2 B d)+a b^2 d \left (3 c^2 C+12 B c d+8 (A-C) d^2\right )-b^3 \left (c^3 C-6 B c^2 d-24 c (A-C) d^2+16 B d^3\right )\right ) \tan ^2(e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{6 b^2 d}\\ &=\frac{\left (8 b (A b+a B-b C) d^2-(b c-a d) (b c C-6 b B d-a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b^2 d f}-\frac{(b c C-6 b B d-a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b d f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{\operatorname{Subst}\left (\int \frac{\frac{3}{8} \left (a^3 C d^3-a^2 b d^2 (3 c C+2 B d)-b^3 c \left (c^2 C+10 B c d+8 (A-C) d^2\right )-a b^2 d \left (13 c^2 C+20 B c d-8 C d^2-8 A \left (2 c^2-d^2\right )\right )\right )+6 b^2 d \left (2 a A c d-2 a c C d+A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )-b \left (c^2 C+2 B c d-C d^2\right )\right ) x+\frac{3}{8} \left (a^3 C d^3-a^2 b d^2 (3 c C+2 B d)+a b^2 d \left (3 c^2 C+12 B c d+8 (A-C) d^2\right )-b^3 \left (c^3 C-6 B c^2 d-24 c (A-C) d^2+16 B d^3\right )\right ) x^2}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{6 b^2 d f}\\ &=\frac{\left (8 b (A b+a B-b C) d^2-(b c-a d) (b c C-6 b B d-a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b^2 d f}-\frac{(b c C-6 b B d-a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b d f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{\operatorname{Subst}\left (\int \left (\frac{3 \left (a^3 C d^3-a^2 b d^2 (3 c C+2 B d)+a b^2 d \left (3 c^2 C+12 B c d+8 (A-C) d^2\right )-b^3 \left (c^3 C-6 B c^2 d-24 c (A-C) d^2+16 B d^3\right )\right )}{8 \sqrt{a+b x} \sqrt{c+d x}}+\frac{6 \left (-b^2 d \left (a \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )+b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right )+b^2 d \left (2 a A c d-2 a c C d+A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )-b \left (c^2 C+2 B c d-C d^2\right )\right ) x\right )}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{6 b^2 d f}\\ &=\frac{\left (8 b (A b+a B-b C) d^2-(b c-a d) (b c C-6 b B d-a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b^2 d f}-\frac{(b c C-6 b B d-a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b d f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{\operatorname{Subst}\left (\int \frac{-b^2 d \left (a \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )+b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right )+b^2 d \left (2 a A c d-2 a c C d+A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )-b \left (c^2 C+2 B c d-C d^2\right )\right ) x}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{b^2 d f}+\frac{\left (a^3 C d^3-a^2 b d^2 (3 c C+2 B d)+a b^2 d \left (3 c^2 C+12 B c d+8 (A-C) d^2\right )-b^3 \left (c^3 C-6 B c^2 d-24 c (A-C) d^2+16 B d^3\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{16 b^2 d f}\\ &=\frac{\left (8 b (A b+a B-b C) d^2-(b c-a d) (b c C-6 b B d-a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b^2 d f}-\frac{(b c C-6 b B d-a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b d f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{\operatorname{Subst}\left (\int \left (\frac{-b^2 d \left (2 a A c d-2 a c C d+A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )-b \left (c^2 C+2 B c d-C d^2\right )\right )-i b^2 d \left (a \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )+b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right )}{2 (i-x) \sqrt{a+b x} \sqrt{c+d x}}+\frac{b^2 d \left (2 a A c d-2 a c C d+A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )-b \left (c^2 C+2 B c d-C d^2\right )\right )-i b^2 d \left (a \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )+b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right )}{2 (i+x) \sqrt{a+b x} \sqrt{c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{b^2 d f}+\frac{\left (a^3 C d^3-a^2 b d^2 (3 c C+2 B d)+a b^2 d \left (3 c^2 C+12 B c d+8 (A-C) d^2\right )-b^3 \left (c^3 C-6 B c^2 d-24 c (A-C) d^2+16 B d^3\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b \tan (e+f x)}\right )}{8 b^3 d f}\\ &=\frac{\left (8 b (A b+a B-b C) d^2-(b c-a d) (b c C-6 b B d-a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b^2 d f}-\frac{(b c C-6 b B d-a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b d f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{\left ((i a+b) (A-i B-C) (c-i d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(i+x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac{\left ((i a-b) (A+i B-C) (c+i d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac{\left (a^3 C d^3-a^2 b d^2 (3 c C+2 B d)+a b^2 d \left (3 c^2 C+12 B c d+8 (A-C) d^2\right )-b^3 \left (c^3 C-6 B c^2 d-24 c (A-C) d^2+16 B d^3\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{8 b^3 d f}\\ &=\frac{\left (a^3 C d^3-a^2 b d^2 (3 c C+2 B d)+a b^2 d \left (3 c^2 C+12 B c d+8 (A-C) d^2\right )-b^3 \left (c^3 C-6 B c^2 d-24 c (A-C) d^2+16 B d^3\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{8 b^{5/2} d^{3/2} f}+\frac{\left (8 b (A b+a B-b C) d^2-(b c-a d) (b c C-6 b B d-a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b^2 d f}-\frac{(b c C-6 b B d-a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b d f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{\left ((i a+b) (A-i B-C) (c-i d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-a+i b-(-c+i d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{\left ((i a-b) (A+i B-C) (c+i d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+i b-(c+i d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{f}\\ &=-\frac{\sqrt{a-i b} (i A+B-i C) (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}-\frac{\sqrt{a+i b} (B-i (A-C)) (c+i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{\left (a^3 C d^3-a^2 b d^2 (3 c C+2 B d)+a b^2 d \left (3 c^2 C+12 B c d+8 (A-C) d^2\right )-b^3 \left (c^3 C-6 B c^2 d-24 c (A-C) d^2+16 B d^3\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{8 b^{5/2} d^{3/2} f}+\frac{\left (8 b (A b+a B-b C) d^2-(b c-a d) (b c C-6 b B d-a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b^2 d f}-\frac{(b c C-6 b B d-a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b d f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\\ \end{align*}

Mathematica [A]  time = 8.79454, size = 867, normalized size = 1.71 \[ \frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{\frac{(-b c C+a d C+6 b B d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{4 b f}+\frac{\frac{3 \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left (8 b (A b-C b+a B) d^2-(b c-a d) (b c C-a d C-6 b B d)\right )}{4 b f}+\frac{\frac{6 d \left (b \left (2 a A c d-2 a c C d+A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )-b \left (C c^2+2 B d c-C d^2\right )\right )-\sqrt{-b^2} \left (a \left (C c^2+2 B d c-C d^2-A \left (c^2-d^2\right )\right )+b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right )\right ) \tan ^{-1}\left (\frac{\sqrt{c+\frac{b d}{\sqrt{-b^2}}} \sqrt{a+b \tan (e+f x)}}{\sqrt{\sqrt{-b^2}-a} \sqrt{c+d \tan (e+f x)}}\right ) b^2}{\sqrt{\sqrt{-b^2}-a} \sqrt{c+\frac{b d}{\sqrt{-b^2}}}}-\frac{6 d \left (b \left (2 a A c d-2 a c C d+A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )-b \left (C c^2+2 B d c-C d^2\right )\right )+\sqrt{-b^2} \left (a \left (C c^2+2 B d c-C d^2-A \left (c^2-d^2\right )\right )+b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right )\right ) \tan ^{-1}\left (\frac{\sqrt{-\frac{b c+\sqrt{-b^2} d}{b}} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+\sqrt{-b^2}} \sqrt{c+d \tan (e+f x)}}\right ) b^2}{\sqrt{a+\sqrt{-b^2}} \sqrt{-\frac{b c+\sqrt{-b^2} d}{b}}}+\frac{3 \sqrt{c-\frac{a d}{b}} \left (-\left (C c^3-6 B d c^2-24 (A-C) d^2 c+16 B d^3\right ) b^3+a d \left (3 C c^2+12 B d c+8 (A-C) d^2\right ) b^2-a^2 d^2 (3 c C+2 B d) b+a^3 C d^3\right ) \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c-\frac{a d}{b}}}\right ) \sqrt{\frac{b c+b d \tan (e+f x)}{b c-a d}} \sqrt{b}}{4 \sqrt{d} \sqrt{c+d \tan (e+f x)}}}{b^2 f}}{2 b}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(5/2))/(3*d*f) + (((-(b*c*C) + 6*b*B*d + a*C*d)*Sqrt[a + b*Ta
n[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(4*b*f) + ((3*(8*b*(A*b + a*B - b*C)*d^2 - (b*c - a*d)*(b*c*C - 6*b*B*
d - a*C*d))*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(4*b*f) + ((6*b^2*d*(b*(2*a*A*c*d - 2*a*c*C*d +
A*b*(c^2 - d^2) + a*B*(c^2 - d^2) - b*(c^2*C + 2*B*c*d - C*d^2)) - Sqrt[-b^2]*(a*(c^2*C + 2*B*c*d - C*d^2 - A
*(c^2 - d^2)) + b*(2*c*(A - C)*d + B*(c^2 - d^2))))*ArcTan[(Sqrt[c + (b*d)/Sqrt[-b^2]]*Sqrt[a + b*Tan[e + f*x]
])/(Sqrt[-a + Sqrt[-b^2]]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[-a + Sqrt[-b^2]]*Sqrt[c + (b*d)/Sqrt[-b^2]]) - (6*
b^2*d*(b*(2*a*A*c*d - 2*a*c*C*d + A*b*(c^2 - d^2) + a*B*(c^2 - d^2) - b*(c^2*C + 2*B*c*d - C*d^2)) + Sqrt[-b^2
]*(a*(c^2*C + 2*B*c*d - C*d^2 - A*(c^2 - d^2)) + b*(2*c*(A - C)*d + B*(c^2 - d^2))))*ArcTan[(Sqrt[-((b*c + Sqr
t[-b^2]*d)/b)]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + Sqrt[-b^2]]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a + Sqrt[-b^2
]]*Sqrt[-((b*c + Sqrt[-b^2]*d)/b)]) + (3*Sqrt[b]*Sqrt[c - (a*d)/b]*(a^3*C*d^3 - a^2*b*d^2*(3*c*C + 2*B*d) + a*
b^2*d*(3*c^2*C + 12*B*c*d + 8*(A - C)*d^2) - b^3*(c^3*C - 6*B*c^2*d - 24*c*(A - C)*d^2 + 16*B*d^3))*ArcSinh[(S
qrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c - (a*d)/b])]*Sqrt[(b*c + b*d*Tan[e + f*x])/(b*c - a*d)])/(4*S
qrt[d]*Sqrt[c + d*Tan[e + f*x]]))/(b^2*f))/(2*b))/(3*d)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a+b\tan \left ( fx+e \right ) } \left ( c+d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ( A+B\tan \left ( fx+e \right ) +C \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x)

[Out]

int((a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} \sqrt{b \tan \left (f x + e\right ) + a}{\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))^(1/2)*(c+d*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

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

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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))^(1/2)*(c+d*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [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))**(1/2)*(c+d*tan(f*x+e))**(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)**2),x)

[Out]

Timed out

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Giac [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))^(1/2)*(c+d*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="giac")

[Out]

Timed out