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

Optimal. Leaf size=505 \[ \frac{\left (-15 a^2 b d^2 (c C-2 B d)+5 a^3 C d^3+5 a b^2 d \left (8 d^2 (A-C)-4 B c d+3 c^2 C\right )+b^3 \left (-\left (8 c d^2 (A-C)-6 B c^2 d+16 B d^3+5 c^3 C\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{8 \sqrt{b} d^{7/2} f}+\frac{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left (8 b d^2 (a B+A b-b C)+(b c-a d) (-5 a C d-6 b B d+5 b c C)\right )}{8 d^3 f}-\frac{(a-i b)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{c-i d}}-\frac{(a+i b)^{5/2} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{c+i d}}-\frac{(-5 a C d-6 b B d+5 b c C) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 d^2 f}+\frac{C (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 d f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 5.95301, antiderivative size = 505, 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} \[ \frac{\left (-15 a^2 b d^2 (c C-2 B d)+5 a^3 C d^3+5 a b^2 d \left (8 d^2 (A-C)-4 B c d+3 c^2 C\right )+b^3 \left (-\left (8 c d^2 (A-C)-6 B c^2 d+16 B d^3+5 c^3 C\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{8 \sqrt{b} d^{7/2} f}+\frac{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left (8 b d^2 (a B+A b-b C)+(b c-a d) (-5 a C d-6 b B d+5 b c C)\right )}{8 d^3 f}-\frac{(a-i b)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{c-i d}}-\frac{(a+i b)^{5/2} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{c+i d}}-\frac{(-5 a C d-6 b B d+5 b c C) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 d^2 f}+\frac{C (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a - I*b)^(5/2)*(I*A + B - I*C)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*
Tan[e + f*x]])])/(Sqrt[c - I*d]*f)) - ((a + I*b)^(5/2)*(B - I*(A - C))*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e
+ f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[c + I*d]*f) + ((5*a^3*C*d^3 - 15*a^2*b*d^2*(c*C - 2
*B*d) + 5*a*b^2*d*(3*c^2*C - 4*B*c*d + 8*(A - C)*d^2) - b^3*(5*c^3*C - 6*B*c^2*d + 8*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*Sqrt[b]*d^(7/2)*f) + ((8*
b*(A*b + a*B - b*C)*d^2 + (b*c - a*d)*(5*b*c*C - 6*b*B*d - 5*a*C*d))*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e
+ f*x]])/(8*d^3*f) - ((5*b*c*C - 6*b*B*d - 5*a*C*d)*(a + b*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]])/(12*
d^2*f) + (C*(a + b*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]])/(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 \frac{(a+b \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx &=\frac{C (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{\int \frac{(a+b \tan (e+f x))^{3/2} \left (\frac{1}{2} (-5 b c C+a (6 A-C) d)+3 (A b+a B-b C) d \tan (e+f x)-\frac{1}{2} (5 b c C-6 b B d-5 a C d) \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d}\\ &=-\frac{(5 b c C-6 b B d-5 a C d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 d^2 f}+\frac{C (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{\int \frac{\sqrt{a+b \tan (e+f x)} \left (\frac{1}{4} (-4 a d (5 b c C-a (6 A-C) d)+(3 b c+a d) (5 b c C-6 b B d-5 a C d))+6 \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2 \tan (e+f x)+\frac{3}{4} \left (8 b (A b+a B-b C) d^2+(b c-a d) (5 b c C-6 b B d-5 a C d)\right ) \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{6 d^2}\\ &=\frac{\left (8 b (A b+a B-b C) d^2+(b c-a d) (5 b c C-6 b B d-5 a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d^3 f}-\frac{(5 b c C-6 b B d-5 a C d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 d^2 f}+\frac{C (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{\int \frac{\frac{3}{8} \left (a^3 (16 A-11 C) d^3-3 a^2 b d^2 (5 c C+6 B d)+a b^2 d \left (15 c^2 C-20 B c d-8 (A-C) d^2\right )-b^3 c \left (5 c^2 C-6 B c d+8 (A-C) d^2\right )\right )+6 \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{3}{8} \left (16 b \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^3-(b c-a d) \left (8 b (A b+a B-b C) d^2+(b c-a d) (5 b c C-6 b B d-5 a C d)\right )\right ) \tan ^2(e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{6 d^3}\\ &=\frac{\left (8 b (A b+a B-b C) d^2+(b c-a d) (5 b c C-6 b B d-5 a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d^3 f}-\frac{(5 b c C-6 b B d-5 a C d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 d^2 f}+\frac{C (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{\operatorname{Subst}\left (\int \frac{\frac{3}{8} \left (a^3 (16 A-11 C) d^3-3 a^2 b d^2 (5 c C+6 B d)+a b^2 d \left (15 c^2 C-20 B c d-8 (A-C) d^2\right )-b^3 c \left (5 c^2 C-6 B c d+8 (A-C) d^2\right )\right )+6 \left (a^3 B-3 a b^2 B+3 a^2 b (A-C)-b^3 (A-C)\right ) d^3 x+\frac{3}{8} \left (16 b \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^3-(b c-a d) \left (8 b (A b+a B-b C) d^2+(b c-a d) (5 b c C-6 b B d-5 a C d)\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 d^3 f}\\ &=\frac{\left (8 b (A b+a B-b C) d^2+(b c-a d) (5 b c C-6 b B d-5 a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d^3 f}-\frac{(5 b c C-6 b B d-5 a C d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 d^2 f}+\frac{C (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{\operatorname{Subst}\left (\int \left (\frac{3 \left (5 a^3 C d^3-15 a^2 b d^2 (c C-2 B d)+5 a b^2 d \left (3 c^2 C-4 B c d+8 (A-C) d^2\right )-b^3 \left (5 c^3 C-6 B c^2 d+8 c (A-C) d^2+16 B d^3\right )\right )}{8 \sqrt{a+b x} \sqrt{c+d x}}+\frac{6 \left (-\left (3 a^2 b B-b^3 B-a^3 (A-C)+3 a b^2 (A-C)\right ) d^3+\left (a^3 B-3 a b^2 B+3 a^2 b (A-C)-b^3 (A-C)\right ) d^3 x\right )}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{6 d^3 f}\\ &=\frac{\left (8 b (A b+a B-b C) d^2+(b c-a d) (5 b c C-6 b B d-5 a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d^3 f}-\frac{(5 b c C-6 b B d-5 a C d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 d^2 f}+\frac{C (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{\operatorname{Subst}\left (\int \frac{-\left (3 a^2 b B-b^3 B-a^3 (A-C)+3 a b^2 (A-C)\right ) d^3+\left (a^3 B-3 a b^2 B+3 a^2 b (A-C)-b^3 (A-C)\right ) d^3 x}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{d^3 f}+\frac{\left (5 a^3 C d^3-15 a^2 b d^2 (c C-2 B d)+5 a b^2 d \left (3 c^2 C-4 B c d+8 (A-C) d^2\right )-b^3 \left (5 c^3 C-6 B c^2 d+8 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 d^3 f}\\ &=\frac{\left (8 b (A b+a B-b C) d^2+(b c-a d) (5 b c C-6 b B d-5 a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d^3 f}-\frac{(5 b c C-6 b B d-5 a C d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 d^2 f}+\frac{C (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{\operatorname{Subst}\left (\int \left (\frac{-i \left (3 a^2 b B-b^3 B-a^3 (A-C)+3 a b^2 (A-C)\right ) d^3-\left (a^3 B-3 a b^2 B+3 a^2 b (A-C)-b^3 (A-C)\right ) d^3}{2 (i-x) \sqrt{a+b x} \sqrt{c+d x}}+\frac{-i \left (3 a^2 b B-b^3 B-a^3 (A-C)+3 a b^2 (A-C)\right ) d^3+\left (a^3 B-3 a b^2 B+3 a^2 b (A-C)-b^3 (A-C)\right ) d^3}{2 (i+x) \sqrt{a+b x} \sqrt{c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{d^3 f}+\frac{\left (5 a^3 C d^3-15 a^2 b d^2 (c C-2 B d)+5 a b^2 d \left (3 c^2 C-4 B c d+8 (A-C) d^2\right )-b^3 \left (5 c^3 C-6 B c^2 d+8 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 d^3 f}\\ &=\frac{\left (8 b (A b+a B-b C) d^2+(b c-a d) (5 b c C-6 b B d-5 a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d^3 f}-\frac{(5 b c C-6 b B d-5 a C d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 d^2 f}+\frac{C (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{\left ((a-i b)^3 (i A+B-i C)\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 \left (3 a^2 b B-b^3 B-a^3 (A-C)+3 a b^2 (A-C)\right ) d^3+\left (a^3 B-3 a b^2 B+3 a^2 b (A-C)-b^3 (A-C)\right ) d^3\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 d^3 f}+\frac{\left (5 a^3 C d^3-15 a^2 b d^2 (c C-2 B d)+5 a b^2 d \left (3 c^2 C-4 B c d+8 (A-C) d^2\right )-b^3 \left (5 c^3 C-6 B c^2 d+8 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 d^3 f}\\ &=\frac{\left (5 a^3 C d^3-15 a^2 b d^2 (c C-2 B d)+5 a b^2 d \left (3 c^2 C-4 B c d+8 (A-C) d^2\right )-b^3 \left (5 c^3 C-6 B c^2 d+8 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 \sqrt{b} d^{7/2} f}+\frac{\left (8 b (A b+a B-b C) d^2+(b c-a d) (5 b c C-6 b B d-5 a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d^3 f}-\frac{(5 b c C-6 b B d-5 a C d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 d^2 f}+\frac{C (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{\left ((a-i b)^3 (i A+B-i C)\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 \left (3 a^2 b B-b^3 B-a^3 (A-C)+3 a b^2 (A-C)\right ) d^3+\left (a^3 B-3 a b^2 B+3 a^2 b (A-C)-b^3 (A-C)\right ) d^3\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 )}{d^3 f}\\ &=-\frac{(a-i b)^{5/2} (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 )}{\sqrt{c-i d} f}+\frac{(a+i b)^{5/2} (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 )}{\sqrt{c+i d} f}+\frac{\left (5 a^3 C d^3-15 a^2 b d^2 (c C-2 B d)+5 a b^2 d \left (3 c^2 C-4 B c d+8 (A-C) d^2\right )-b^3 \left (5 c^3 C-6 B c^2 d+8 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 \sqrt{b} d^{7/2} f}+\frac{\left (8 b (A b+a B-b C) d^2+(b c-a d) (5 b c C-6 b B d-5 a C d)\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d^3 f}-\frac{(5 b c C-6 b B d-5 a C d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 d^2 f}+\frac{C (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 d f}\\ \end{align*}

Mathematica [A]  time = 8.34208, size = 785, normalized size = 1.55 \[ \frac{\frac{\frac{\frac{3 \sqrt{b} \sqrt{c-\frac{a d}{b}} \left (-15 a^2 b d^2 (c C-2 B d)+5 a^3 C d^3+5 a b^2 d \left (8 d^2 (A-C)-4 B c d+3 c^2 C\right )+b^3 \left (-\left (8 c d^2 (A-C)-6 B c^2 d+16 B d^3+5 c^3 C\right )\right )\right ) \sqrt{\frac{b c+b d \tan (e+f x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c-\frac{a d}{b}}}\right )}{4 \sqrt{d} \sqrt{c+d \tan (e+f x)}}-\frac{6 d^3 \left (\sqrt{-b^2} \left (a^3 (-(A-C))+3 a^2 b B+3 a b^2 (A-C)-b^3 B\right )-b \left (3 a^2 b (A-C)+a^3 B-3 a b^2 B-b^3 (A-C)\right )\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{b d}{\sqrt{-b^2}}+c} \sqrt{a+b \tan (e+f x)}}{\sqrt{\sqrt{-b^2}-a} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{\sqrt{-b^2}-a} \sqrt{\frac{b d}{\sqrt{-b^2}}+c}}-\frac{6 d^3 \left (\sqrt{-b^2} \left (a^3 (-(A-C))+3 a^2 b B+3 a b^2 (A-C)-b^3 B\right )+b \left (3 a^2 b (A-C)+a^3 B-3 a b^2 B-b^3 (A-C)\right )\right ) \tan ^{-1}\left (\frac{\sqrt{-\frac{\sqrt{-b^2} d+b c}{b}} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+\sqrt{-b^2}} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a+\sqrt{-b^2}} \sqrt{-\frac{\sqrt{-b^2} d+b c}{b}}}}{b d f}+\frac{3 \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left (8 b d^2 (a B+A b-b C)+(b c-a d) (-5 a C d-6 b B d+5 b c C)\right )}{4 d f}}{2 d}+\frac{(5 a C d+6 b B d-5 b c C) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{4 d f}}{3 d}+\frac{C (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*(a + b*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]])/(3*d*f) + (((-5*b*c*C + 6*b*B*d + 5*a*C*d)*(a + b*Tan[
e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]])/(4*d*f) + ((3*(8*b*(A*b + a*B - b*C)*d^2 + (b*c - a*d)*(5*b*c*C - 6*
b*B*d - 5*a*C*d))*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(4*d*f) + ((-6*(Sqrt[-b^2]*(3*a^2*b*B - b
^3*B - a^3*(A - C) + 3*a*b^2*(A - C)) - b*(a^3*B - 3*a*b^2*B + 3*a^2*b*(A - C) - b^3*(A - C)))*d^3*ArcTan[(Sqr
t[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*(Sqrt[-b^2]*(3*a^2*b*B - b^3*B - a^3*(A - C) + 3*a*b^2*(A - C))
+ b*(a^3*B - 3*a*b^2*B + 3*a^2*b*(A - C) - b^3*(A - C)))*d^3*ArcTan[(Sqrt[-((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]])])/(Sqrt[a + Sqrt[-b^2]]*Sqrt[-((b*c + Sqrt[-
b^2]*d)/b)]) + (3*Sqrt[b]*Sqrt[c - (a*d)/b]*(5*a^3*C*d^3 - 15*a^2*b*d^2*(c*C - 2*B*d) + 5*a*b^2*d*(3*c^2*C - 4
*B*c*d + 8*(A - C)*d^2) - b^3*(5*c^3*C - 6*B*c^2*d + 8*c*(A - C)*d^2 + 16*B*d^3))*ArcSinh[(Sqrt[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*Sqrt[d]*Sqrt[c + d*
Tan[e + f*x]]))/(b*d*f))/(2*d))/(3*d)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{(A+B\tan \left ( fx+e \right ) +C \left ( \tan \left ( fx+e \right ) \right ) ^{2}) \left ( a+b\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}{\frac{1}{\sqrt{c+d\tan \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1/2),x)

[Out]

int((a+b*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1/2),x)

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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))^(5/2)*(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))^(5/2)*(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(-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))**(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(c+d*tan(f*x+e))**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError