\(\int \frac {\sqrt {c+d \tan (e+f x)} (A+B \tan (e+f x)+C \tan ^2(e+f x))}{\sqrt {a+b \tan (e+f x)}} \, dx\) [131]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F(-1)]
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F(-1)]
   Mupad [F(-1)]

Optimal result

Integrand size = 49, antiderivative size = 287 \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {a+b \tan (e+f x)}} \, dx=-\frac {(i A+B-i C) \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a-i b} f}-\frac {(B-i (A-C)) \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} f}+\frac {(b c C+2 b B d-a C d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{3/2} \sqrt {d} f}+\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f} \]

[Out]

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

Rubi [A] (verified)

Time = 3.15 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {3728, 3736, 6857, 65, 223, 212, 95, 214} \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {a+b \tan (e+f x)}} \, dx=-\frac {\sqrt {c-i d} (i A+B-i C) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a-i b}}-\frac {\sqrt {c+i d} (B-i (A-C)) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a+i b}}+\frac {(-a C d+2 b B d+b c C) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{3/2} \sqrt {d} f}+\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f} \]

[In]

Int[(Sqrt[c + d*Tan[e + f*x]]*(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)*Sqrt[c - I*d]*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Ta
n[e + f*x]])])/(Sqrt[a - I*b]*f)) - ((B - I*(A - C))*Sqrt[c + I*d]*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*c*C + 2*b*B*d - a*C*d)*ArcTanh[(Sqrt[
d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/(b^(3/2)*Sqrt[d]*f) + (C*Sqrt[a + b*Tan[e +
f*x]]*Sqrt[c + d*Tan[e + f*x]])/(b*f)

Rule 65

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 95

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 212

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

Rule 214

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

Rule 223

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 3728

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 3736

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 6857

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*} \text {integral}& = \frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\int \frac {\frac {1}{2} (2 A b c-C (b c+a d))+b (B c+(A-C) d) \tan (e+f x)+\frac {1}{2} (b c C+2 b B d-a C d) \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{b} \\ & = \frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\text {Subst}\left (\int \frac {\frac {1}{2} (2 A b c-C (b c+a d))+b (B c+(A-C) d) x+\frac {1}{2} (b c C+2 b B d-a C d) x^2}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{b f} \\ & = \frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\text {Subst}\left (\int \left (\frac {b c C+2 b B d-a C d}{2 \sqrt {a+b x} \sqrt {c+d x}}+\frac {b (A c-c C-B d)+b (B c+(A-C) d) x}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{b f} \\ & = \frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\text {Subst}\left (\int \frac {b (A c-c C-B d)+b (B c+(A-C) d) x}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{b f}+\frac {(b c C+2 b B d-a C d) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 b f} \\ & = \frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\text {Subst}\left (\int \left (\frac {i b (A c-c C-B d)-b (B c+(A-C) d)}{2 (i-x) \sqrt {a+b x} \sqrt {c+d x}}+\frac {i b (A c-c C-B d)+b (B c+(A-C) d)}{2 (i+x) \sqrt {a+b x} \sqrt {c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{b f}+\frac {(b c C+2 b B d-a C d) \text {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 )}{b^2 f} \\ & = \frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {((A-i B-C) (i c+d)) \text {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 {(b c C+2 b B d-a C d) \text {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 )}{b^2 f}+\frac {(i b (A c-c C-B d)-b (B c+(A-C) d)) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 b f} \\ & = \frac {(b c C+2 b B d-a C d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{3/2} \sqrt {d} f}+\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {((A-i B-C) (i c+d)) \text {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 {(i b (A c-c C-B d)-b (B c+(A-C) d)) \text {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 f} \\ & = -\frac {(i A+B-i C) \sqrt {c-i d} \text {arctanh}\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 {(B-i (A-C)) \sqrt {c+i d} \text {arctanh}\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 {(b c C+2 b B d-a C d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{3/2} \sqrt {d} f}+\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.70 (sec) , antiderivative size = 441, normalized size of antiderivative = 1.54 \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {a+b \tan (e+f x)}} \, dx=\frac {\frac {b \left (b B c+b (A-C) d+\sqrt {-b^2} (A c-c C-B d)\right ) \text {arctanh}\left (\frac {\sqrt {-c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+\sqrt {-b^2}} \sqrt {-c+\frac {\sqrt {-b^2} d}{b}}}+\frac {b \left (\sqrt {-b^2} (A c-c C-B d)-b (B c+(A-C) d)\right ) \text {arctanh}\left (\frac {\sqrt {c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+\frac {\sqrt {-b^2} d}{b}}}+b C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}+\frac {\sqrt {b} \sqrt {c-\frac {a d}{b}} (b c C+2 b B d-a C d) \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c-\frac {a d}{b}}}\right ) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}{\sqrt {d} \sqrt {c+d \tan (e+f x)}}}{b^2 f} \]

[In]

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

[Out]

((b*(b*B*c + b*(A - C)*d + Sqrt[-b^2]*(A*c - c*C - B*d))*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
]) + (b*(Sqrt[-b^2]*(A*c - c*C - B*d) - b*(B*c + (A - C)*d))*ArcTanh[(Sqrt[c + (Sqrt[-b^2]*d)/b]*Sqrt[a + b*Ta
n[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
]) + b*C*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]] + (Sqrt[b]*Sqrt[c - (a*d)/b]*(b*c*C + 2*b*B*d - a*C
*d)*ArcSinh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c - (a*d)/b])]*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c
 - a*d)])/(Sqrt[d]*Sqrt[c + d*Tan[e + f*x]]))/(b^2*f)

Maple [F(-1)]

Timed out.

\[\int \frac {\sqrt {c +d \tan \left (f x +e \right )}\, \left (A +B \tan \left (f x +e \right )+C \tan \left (f x +e \right )^{2}\right )}{\sqrt {a +b \tan \left (f x +e \right )}}d x\]

[In]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39018 vs. \(2 (225) = 450\).

Time = 105.64 (sec) , antiderivative size = 78051, normalized size of antiderivative = 271.95 \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {a+b \tan (e+f x)}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {a+b \tan (e+f x)}} \, dx=\int \frac {\sqrt {c + d \tan {\left (e + f x \right )}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\sqrt {a + b \tan {\left (e + f x \right )}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {a+b \tan (e+f x)}} \, dx=\int { \frac {{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} \sqrt {d \tan \left (f x + e\right ) + c}}{\sqrt {b \tan \left (f x + e\right ) + a}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {a+b \tan (e+f x)}} \, dx=\text {Timed out} \]

[In]

integrate((c+d*tan(f*x+e))^(1/2)*(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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {a+b \tan (e+f x)}} \, dx=\text {Hanged} \]

[In]

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

[Out]

\text{Hanged}