\(\int \frac {(c+d \tan (e+f x))^{5/2} (A+B \tan (e+f x)+C \tan ^2(e+f x))}{a+b \tan (e+f x)} \, dx\) [107]

Optimal result

Integrand size = 47, antiderivative size = 336 \[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx=-\frac {(i A+B-i C) (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(a-i b) f}+\frac {(i A-B-i C) (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(a+i b) f}-\frac {2 \left (A b^2-a (b B-a C)\right ) (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{7/2} \left (a^2+b^2\right ) f}+\frac {2 \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 f}+\frac {2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 b f} \]

[Out]

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

Rubi [A] (verified)

Time = 3.13 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.149, Rules used = {3728, 3734, 3620, 3618, 65, 214, 3715} \[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx=-\frac {2 (b c-a d)^{5/2} \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{7/2} f \left (a^2+b^2\right )}-\frac {(c-i d)^{5/2} (i A+B-i C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (a-i b)}+\frac {(c+i d)^{5/2} (i A-B-i C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (a+i b)}+\frac {2 \sqrt {c+d \tan (e+f x)} \left ((b c-a d) (-a C d+b B d+b c C)+b^2 d (d (A-C)+B c)\right )}{b^3 f}+\frac {2 (-a C d+b B d+b c C) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 b f} \]

[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]),x]

[Out]

-(((I*A + B - I*C)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((a - I*b)*f)) + ((I*A - B
- I*C)*(c + I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)*f) - (2*(A*b^2 - a*(b*B -
a*C))*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/(b^(7/2)*(a^2 + b^2)*f) +
(2*(b^2*d*(B*c + (A - C)*d) + (b*c - a*d)*(b*c*C + b*B*d - a*C*d))*Sqrt[c + d*Tan[e + f*x]])/(b^3*f) + (2*(b*
c*C + b*B*d - a*C*d)*(c + d*Tan[e + f*x])^(3/2))/(3*b^2*f) + (2*C*(c + d*Tan[e + f*x])^(5/2))/(5*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 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 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Rule 3620

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

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 3734

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[(c + d*Tan[e + f*x])^n*((1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*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[n, 0] &&  !LeQ[n, -
1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 C (c+d \tan (e+f x))^{5/2}}{5 b f}+\frac {2 \int \frac {(c+d \tan (e+f x))^{3/2} \left (\frac {5}{2} (A b c-a C d)+\frac {5}{2} b (B c+(A-C) d) \tan (e+f x)+\frac {5}{2} (b c C+b B d-a C d) \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx}{5 b} \\ & = \frac {2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 b f}+\frac {4 \int \frac {\sqrt {c+d \tan (e+f x)} \left (\frac {15}{4} \left (A b^2 c^2+a d (a C d-b (2 c C+B d))\right )+\frac {15}{4} b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \tan (e+f x)+\frac {15}{4} \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx}{15 b^2} \\ & = \frac {2 \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 f}+\frac {2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 b f}+\frac {8 \int \frac {\frac {15}{8} \left (A b^2 \left (b c^3-a d^3\right )-a d \left (a^2 C d^2-a b d (3 c C+B d)+b^2 \left (3 c^2 C+3 B c d-C d^2\right )\right )\right )+\frac {15}{8} 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 {15}{8} \left ((b c-a d) \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right )+b^3 d \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{15 b^3} \\ & = \frac {2 \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 f}+\frac {2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 b f}+\frac {8 \int \frac {-\frac {15}{8} b^3 \left (a \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 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right )+\frac {15}{8} b^3 \left (a A d \left (3 c^2-d^2\right )-A b \left (c^3-3 c d^2\right )+b \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3\right )-a \left (C d \left (3 c^2-d^2\right )-B \left (c^3-3 c d^2\right )\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{15 b^3 \left (a^2+b^2\right )}+\frac {\left (\left (A b^2-a (b B-a C)\right ) (b c-a d)^3\right ) \int \frac {1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{b^3 \left (a^2+b^2\right )} \\ & = \frac {2 \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 f}+\frac {2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 b f}+\frac {\left ((A-i B-C) (c-i d)^3\right ) \int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (a-i b)}+\frac {\left ((A+i B-C) (c+i d)^3\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (a+i b)}+\frac {\left (\left (A b^2-a (b B-a C)\right ) (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{b^3 \left (a^2+b^2\right ) f} \\ & = \frac {2 \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 f}+\frac {2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 b f}+\frac {\left ((i A+B-i C) (c-i d)^3\right ) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (a-i b) f}-\frac {\left (i (A+i B-C) (c+i d)^3\right ) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (a+i b) f}+\frac {\left (2 \left (A b^2-a (b B-a C)\right ) (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{b^3 \left (a^2+b^2\right ) d f} \\ & = -\frac {2 \left (A b^2-a (b B-a C)\right ) (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{7/2} \left (a^2+b^2\right ) f}+\frac {2 \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 f}+\frac {2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 b f}-\frac {\left ((A-i B-C) (c-i d)^3\right ) \text {Subst}\left (\int \frac {1}{-1-\frac {i c}{d}+\frac {i x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{(a-i b) d f}-\frac {\left ((A+i B-C) (c+i d)^3\right ) \text {Subst}\left (\int \frac {1}{-1+\frac {i c}{d}-\frac {i x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{(a+i b) d f} \\ & = -\frac {(i A+B-i C) (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(a-i b) f}-\frac {(A+i B-C) (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(i a-b) f}-\frac {2 \left (A b^2-a (b B-a C)\right ) (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{7/2} \left (a^2+b^2\right ) f}+\frac {2 \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 f}+\frac {2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 b f} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.76 (sec) , antiderivative size = 322, normalized size of antiderivative = 0.96 \[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx=\frac {\frac {15 \left (b^{7/2} (-i a+b) (A-i B-C) (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+b^{7/2} (i a+b) (A+i B-C) (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )-2 \left (A b^2+a (-b B+a C)\right ) (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )\right )}{b^{5/2} \left (a^2+b^2\right )}+\frac {30 \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^2}+\frac {10 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{b}+6 C (c+d \tan (e+f x))^{5/2}}{15 b f} \]

[In]

Integrate[((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]),x]

[Out]

((15*(b^(7/2)*((-I)*a + b)*(A - I*B - C)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + b^(
7/2)*(I*a + b)*(A + I*B - C)*(c + I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] - 2*(A*b^2 + a*(-
(b*B) + a*C))*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]]))/(b^(5/2)*(a^2 +
b^2)) + (30*(b^2*d*(B*c + (A - C)*d) + (b*c - a*d)*(b*c*C + b*B*d - a*C*d))*Sqrt[c + d*Tan[e + f*x]])/b^2 + (1
0*(b*c*C + b*B*d - a*C*d)*(c + d*Tan[e + f*x])^(3/2))/b + 6*C*(c + d*Tan[e + f*x])^(5/2))/(15*b*f)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(8697\) vs. \(2(294)=588\).

Time = 0.18 (sec) , antiderivative size = 8698, normalized size of antiderivative = 25.89

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(8698\)
default	\(\text {Expression too large to display}\)	\(8698\)

[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)),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F(-1)]

Timed out. \[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx=\text {Timed out} \]

[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)),x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx=\text {Timed out} \]

[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)),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx=\text {Exception raised: ValueError} \]

[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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

Giac [F(-1)]

Timed out. \[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx=\text {Timed out} \]

[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)),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx=\text {Hanged} \]

[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)),x)

[Out]

\text{Hanged}