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

Optimal. Leaf size=505 \[ -\frac{\left (-3 a^2 b d^2 (2 B d+5 c C)+5 a^3 C d^3+a b^2 d \left (8 d^2 (A-C)+20 B c d+15 c^2 C\right )+b^3 \left (-\left (40 c d^2 (A-C)+30 B c^2 d-16 B d^3+5 c^3 C\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{8 b^{7/2} \sqrt{d} f}+\frac{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left ((b c-a d) (-5 a C d+6 b B d+5 b c C)+8 b^2 d (d (A-C)+B c)\right )}{8 b^3 f}-\frac{(c-i d)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a-i b}}-\frac{(c+i d)^{5/2} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a+i b}}+\frac{(-5 a C d+6 b B d+5 b c C) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 b f} \]

[Out]

-(((I*A + B - I*C)*(c - I*d)^(5/2)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*
Tan[e + f*x]])])/(Sqrt[a - I*b]*f)) - ((B - I*(A - C))*(c + I*d)^(5/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e
+ f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a + I*b]*f) - ((5*a^3*C*d^3 - 3*a^2*b*d^2*(5*c*C +
2*B*d) + a*b^2*d*(15*c^2*C + 20*B*c*d + 8*(A - C)*d^2) - b^3*(5*c^3*C + 30*B*c^2*d + 40*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^(7/2)*Sqrt[d]*f) + (
(8*b^2*d*(B*c + (A - C)*d) + (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*Ta
n[e + f*x]])/(8*b^3*f) + ((5*b*c*C + 6*b*B*d - 5*a*C*d)*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(
12*b^2*f) + (C*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(5/2))/(3*b*f)

________________________________________________________________________________________

Rubi [A]  time = 6.23032, 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} \[ -\frac{\left (-3 a^2 b d^2 (2 B d+5 c C)+5 a^3 C d^3+a b^2 d \left (8 d^2 (A-C)+20 B c d+15 c^2 C\right )+b^3 \left (-\left (40 c d^2 (A-C)+30 B c^2 d-16 B d^3+5 c^3 C\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{8 b^{7/2} \sqrt{d} f}+\frac{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left ((b c-a d) (-5 a C d+6 b B d+5 b c C)+8 b^2 d (d (A-C)+B c)\right )}{8 b^3 f}-\frac{(c-i d)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a-i b}}-\frac{(c+i d)^{5/2} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a+i b}}+\frac{(-5 a C d+6 b B d+5 b c C) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((I*A + B - I*C)*(c - I*d)^(5/2)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*
Tan[e + f*x]])])/(Sqrt[a - I*b]*f)) - ((B - I*(A - C))*(c + I*d)^(5/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e
+ f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a + I*b]*f) - ((5*a^3*C*d^3 - 3*a^2*b*d^2*(5*c*C +
2*B*d) + a*b^2*d*(15*c^2*C + 20*B*c*d + 8*(A - C)*d^2) - b^3*(5*c^3*C + 30*B*c^2*d + 40*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^(7/2)*Sqrt[d]*f) + (
(8*b^2*d*(B*c + (A - C)*d) + (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*Ta
n[e + f*x]])/(8*b^3*f) + ((5*b*c*C + 6*b*B*d - 5*a*C*d)*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(
12*b^2*f) + (C*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(5/2))/(3*b*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{(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt{a+b \tan (e+f x)}} \, dx &=\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 b f}+\frac{\int \frac{(c+d \tan (e+f x))^{3/2} \left (\frac{1}{2} (6 A b c-C (b c+5 a d))+3 b (B c+(A-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{a+b \tan (e+f x)}} \, dx}{3 b}\\ &=\frac{(5 b c C+6 b B d-5 a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 b f}+\frac{\int \frac{\sqrt{c+d \tan (e+f x)} \left (\frac{1}{4} (-(b c+3 a d) (5 b c C+6 b B d-5 a C d)+4 b c (6 A b c-C (b c+5 a d)))+6 b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \tan (e+f x)+\frac{3}{4} \left (8 b^2 d (B c+(A-C) d)+(b c-a d) (5 b c C+6 b B d-5 a C d)\right ) \tan ^2(e+f x)\right )}{\sqrt{a+b \tan (e+f x)}} \, dx}{6 b^2}\\ &=\frac{\left (8 b^2 d (B c+(A-C) d)+(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 b^3 f}+\frac{(5 b c C+6 b B d-5 a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 b f}+\frac{\int \frac{-\frac{3}{8} \left (5 a^3 C d^3-3 a^2 b d^2 (5 c C+2 B d)+b^3 c \left (11 c^2 C+18 B c d-8 C d^2\right )+a b^2 d \left (15 c^2 C+20 B c d-8 C d^2\right )-8 A b^2 \left (2 b c^3-b c d^2-a d^3\right )\right )+6 b^3 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right ) \tan (e+f x)+\frac{3}{8} \left ((b c-a d) \left (8 b^2 d (B c+(A-C) d)+(b c-a d) (5 b c C+6 b B d-5 a C d)\right )+16 b^3 d \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right ) \tan ^2(e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{6 b^3}\\ &=\frac{\left (8 b^2 d (B c+(A-C) d)+(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 b^3 f}+\frac{(5 b c C+6 b B d-5 a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 b f}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{3}{8} \left (5 a^3 C d^3-3 a^2 b d^2 (5 c C+2 B d)+b^3 c \left (11 c^2 C+18 B c d-8 C d^2\right )+a b^2 d \left (15 c^2 C+20 B c d-8 C d^2\right )-8 A b^2 \left (2 b c^3-b c d^2-a d^3\right )\right )+6 b^3 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right ) x+\frac{3}{8} \left ((b c-a d) \left (8 b^2 d (B c+(A-C) d)+(b c-a d) (5 b c C+6 b B d-5 a C d)\right )+16 b^3 d \left (2 c (A-C) d+B \left (c^2-d^2\right )\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^3 f}\\ &=\frac{\left (8 b^2 d (B c+(A-C) d)+(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 b^3 f}+\frac{(5 b c C+6 b B d-5 a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 b f}+\frac{\operatorname{Subst}\left (\int \left (-\frac{3 \left (5 a^3 C d^3-3 a^2 b d^2 (5 c C+2 B d)+a b^2 d \left (15 c^2 C+20 B c d+8 (A-C) d^2\right )-b^3 \left (5 c^3 C+30 B c^2 d+40 c (A-C) d^2-16 B d^3\right )\right )}{8 \sqrt{a+b x} \sqrt{c+d x}}+\frac{6 \left (-b^3 \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3-A \left (c^3-3 c d^2\right )\right )+b^3 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 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^3 f}\\ &=\frac{\left (8 b^2 d (B c+(A-C) d)+(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 b^3 f}+\frac{(5 b c C+6 b B d-5 a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 b f}+\frac{\operatorname{Subst}\left (\int \frac{-b^3 \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3-A \left (c^3-3 c d^2\right )\right )+b^3 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 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^3 f}-\frac{\left (5 a^3 C d^3-3 a^2 b d^2 (5 c C+2 B d)+a b^2 d \left (15 c^2 C+20 B c d+8 (A-C) d^2\right )-b^3 \left (5 c^3 C+30 B c^2 d+40 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^3 f}\\ &=\frac{\left (8 b^2 d (B c+(A-C) d)+(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 b^3 f}+\frac{(5 b c C+6 b B d-5 a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 b f}+\frac{\operatorname{Subst}\left (\int \left (\frac{-i b^3 \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3-A \left (c^3-3 c d^2\right )\right )-b^3 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )}{2 (i-x) \sqrt{a+b x} \sqrt{c+d x}}+\frac{-i b^3 \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3-A \left (c^3-3 c d^2\right )\right )+b^3 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )}{2 (i+x) \sqrt{a+b x} \sqrt{c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{b^3 f}-\frac{\left (5 a^3 C d^3-3 a^2 b d^2 (5 c C+2 B d)+a b^2 d \left (15 c^2 C+20 B c d+8 (A-C) d^2\right )-b^3 \left (5 c^3 C+30 B c^2 d+40 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^4 f}\\ &=\frac{\left (8 b^2 d (B c+(A-C) d)+(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 b^3 f}+\frac{(5 b c C+6 b B d-5 a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 b f}+\frac{\left ((i A+B-i C) (c-i 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 f}-\frac{\left (5 a^3 C d^3-3 a^2 b d^2 (5 c C+2 B d)+a b^2 d \left (15 c^2 C+20 B c d+8 (A-C) d^2\right )-b^3 \left (5 c^3 C+30 B c^2 d+40 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^4 f}-\frac{\left (i b^3 \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3-A \left (c^3-3 c d^2\right )\right )+b^3 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\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 b^3 f}\\ &=-\frac{\left (5 a^3 C d^3-3 a^2 b d^2 (5 c C+2 B d)+a b^2 d \left (15 c^2 C+20 B c d+8 (A-C) d^2\right )-b^3 \left (5 c^3 C+30 B c^2 d+40 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^{7/2} \sqrt{d} f}+\frac{\left (8 b^2 d (B c+(A-C) d)+(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 b^3 f}+\frac{(5 b c C+6 b B d-5 a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 b f}+\frac{\left ((i A+B-i C) (c-i 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 )}{f}-\frac{\left (i b^3 \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3-A \left (c^3-3 c d^2\right )\right )+b^3 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\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 )}{b^3 f}\\ &=-\frac{(i A+B-i C) (c-i d)^{5/2} \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 )}{\sqrt{a-i b} f}+\frac{(i A-B-i C) (c+i d)^{5/2} \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 )}{\sqrt{a+i b} f}-\frac{\left (5 a^3 C d^3-3 a^2 b d^2 (5 c C+2 B d)+a b^2 d \left (15 c^2 C+20 B c d+8 (A-C) d^2\right )-b^3 \left (5 c^3 C+30 B c^2 d+40 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^{7/2} \sqrt{d} f}+\frac{\left (8 b^2 d (B c+(A-C) d)+(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 b^3 f}+\frac{(5 b c C+6 b B d-5 a C d) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 b f}\\ \end{align*}

Mathematica [A]  time = 8.55599, size = 780, normalized size = 1.54 \[ \frac{\frac{\frac{-\frac{3 \sqrt{b} \sqrt{c-\frac{a d}{b}} \left (-3 a^2 b d^2 (2 B d+5 c C)+5 a^3 C d^3+a b^2 d \left (8 d^2 (A-C)+20 B c d+15 c^2 C\right )+b^3 \left (-\left (40 c d^2 (A-C)+30 B c^2 d-16 B d^3+5 c^3 C\right )\right )\right ) \sqrt{\frac{b c+b d \tan (e+f x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c-\frac{a d}{b}}}\right )}{4 \sqrt{d} \sqrt{c+d \tan (e+f x)}}+\frac{6 b^3 \left (\sqrt{-b^2} \left (A c^3-3 A c d^2-3 B c^2 d+B d^3-c^3 C+3 c C d^2\right )+b d (A-C) \left (3 c^2-d^2\right )+b B \left (c^3-3 c d^2\right )\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{b d}{\sqrt{-b^2}}+c} \sqrt{a+b \tan (e+f x)}}{\sqrt{\sqrt{-b^2}-a} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{\sqrt{-b^2}-a} \sqrt{\frac{b d}{\sqrt{-b^2}}+c}}-\frac{6 b^3 \left (-\sqrt{-b^2} \left (A c^3-3 A c d^2-3 B c^2 d+B d^3-c^3 C+3 c C d^2\right )+b d (A-C) \left (3 c^2-d^2\right )+b B \left (c^3-3 c d^2\right )\right ) \tan ^{-1}\left (\frac{\sqrt{-\frac{\sqrt{-b^2} d+b c}{b}} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+\sqrt{-b^2}} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a+\sqrt{-b^2}} \sqrt{-\frac{\sqrt{-b^2} d+b c}{b}}}}{b^2 f}+\frac{3 \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left ((b c-a d) (-5 a C d+6 b B d+5 b c C)+8 b^2 d (d (A-C)+B c)\right )}{4 b f}}{2 b}+\frac{(-5 a C d+6 b B d+5 b c C) \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{4 b f}}{3 b}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(5/2))/(3*b*f) + (((5*b*c*C + 6*b*B*d - 5*a*C*d)*Sqrt[a + b*T
an[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(4*b*f) + ((3*(8*b^2*d*(B*c + (A - C)*d) + (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*b*f) + ((6*b^3*(b*(A - C)*d*(3*c^2 - d^
2) + b*B*(c^3 - 3*c*d^2) + Sqrt[-b^2]*(A*c^3 - c^3*C - 3*B*c^2*d - 3*A*c*d^2 + 3*c*C*d^2 + B*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*b^3*(b*(A - C)*d*(3*c^2 - d^2) + b*B*(c^3 - 3*c*d^2) - Sqrt[-b^
2]*(A*c^3 - c^3*C - 3*B*c^2*d - 3*A*c*d^2 + 3*c*C*d^2 + B*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 - 3*a^2*b*d^2*(5*c*C + 2*B*d) + a*b^2*d*(15*c^2*C +
20*B*c*d + 8*(A - C)*d^2) - b^3*(5*c^3*C + 30*B*c^2*d + 40*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^2*f))/(2*b))/(3*b)

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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 ( c+d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}{\frac{1}{\sqrt{a+b\tan \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: TypeError