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

[Out]

-(I*A+B-I*C)*(c-I*d)^(5/2)*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a-I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))/
f/(a-I*b)^(1/2)-(B-I*(A-C))*(c+I*d)^(5/2)*arctanh((c+I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*tan(
f*x+e))^(1/2))/f/(a+I*b)^(1/2)-1/8*(5*a^3*C*d^3-3*a^2*b*d^2*(2*B*d+5*C*c)+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(d^(1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d*tan(
f*x+e))^(1/2))/b^(7/2)/f/d^(1/2)+1/8*(8*b^2*d*(B*c+(A-C)*d)+(-a*d+b*c)*(6*B*b*d-5*C*a*d+5*C*b*c))*(a+b*tan(f*x
+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/b^3/f+1/12*(6*B*b*d-5*C*a*d+5*C*b*c)*(a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))
^(3/2)/b^2/f+1/3*C*(a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(5/2)/b/f

________________________________________________________________________________________

Rubi [A]  time = 6.23, antiderivative size = 505, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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 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]

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*}

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Mathematica [A]  time = 8.97, size = 780, normalized size = 1.54 \[ \frac {\frac {\frac {-\frac {3 \sqrt {b} \sqrt {c-\frac {a d}{b}} \left (5 a^3 C d^3-3 a^2 b d^2 (2 B d+5 c C)+a b^2 d \left (8 d^2 (A-C)+20 B c d+15 c^2 C\right )-\left (b^3 \left (40 c d^2 (A-C)+30 B c^2 d-16 B d^3+5 c^3 C\right )\right )\right ) \sqrt {\frac {b c+b d \tan (e+f x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c-\frac {a d}{b}}}\right )}{4 \sqrt {d} \sqrt {c+d \tan (e+f x)}}+\frac {6 b^3 \left (\sqrt {-b^2} \left (A c^3-3 A c d^2-3 B c^2 d+B d^3-c^3 C+3 c C d^2\right )+b d (A-C) \left (3 c^2-d^2\right )+b B \left (c^3-3 c d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {-b^2} d}{b}-c} \sqrt {a+b \tan (e+f x)}}{\sqrt {\sqrt {-b^2}-a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {\sqrt {-b^2}-a} \sqrt {\frac {\sqrt {-b^2} d}{b}-c}}-\frac {6 b^3 \left (-\sqrt {-b^2} \left (A c^3-3 A c d^2-3 B c^2 d+B d^3-c^3 C+3 c C d^2\right )+b d (A-C) \left (3 c^2-d^2\right )+b B \left (c^3-3 c d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {-b^2} d}{b}+c} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+\sqrt {-b^2}} \sqrt {\frac {\sqrt {-b^2} d}{b}+c}}}{b^2 f}+\frac {3 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left ((b c-a d) (-5 a C d+6 b B d+5 b c C)+8 b^2 d (d (A-C)+B c)\right )}{4 b f}}{2 b}+\frac {(-5 a C d+6 b B d+5 b c C) \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{4 b f}}{3 b}+\frac {C \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 b f} \]

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))*ArcTanh[(Sq
rt[-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[-c + (Sqrt[-b^2]*d)/b]) - (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))*ArcTanh[(Sqrt[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[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*T
an[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*T
an[e + f*x]]))/(b^2*f))/(2*b))/(3*b)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}} \left (A +B \tan \left (f x +e \right )+C \left (\tan ^{2}\left (f x +e \right )\right )\right )}{\sqrt {a +b \tan \left (f x +e \right )}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\sqrt {a + b \tan {\left (e + f x \right )}}}\, dx \]

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]

Integral((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)

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