\(\int \frac {\sqrt {c+d \tan (e+f x)} (A+B \tan (e+f x)+C \tan ^2(e+f x))}{(a+b \tan (e+f x))^{5/2}} \, dx\) [133]

Optimal result

Integrand size = 49, antiderivative size = 370 \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^{5/2}} \, dx=-\frac {(i A+B-i C) \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{5/2} f}-\frac {(B-i (A-C)) \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{5/2} f}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (2 a^3 b B d+a^4 C d+b^4 (3 B c+A d)+2 a b^3 (3 A c-3 c C-2 B d)-a^2 b^2 (3 B c+5 A d-7 C d)\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right )^2 (b c-a d) f \sqrt {a+b \tan (e+f x)}} \]

[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)/
(a-I*b)^(5/2)/f-(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)/(a+I*b)^(5/2)/f-2/3*(2*a^3*b*B*d+a^4*C*d+b^4*(A*d+3*B*c)+2*a*b^3*(3*A*c-2*B*d-3*C*c)-a^2*b^2*(5
*A*d+3*B*c-7*C*d))*(c+d*tan(f*x+e))^(1/2)/b/(a^2+b^2)^2/(-a*d+b*c)/f/(a+b*tan(f*x+e))^(1/2)-2/3*(A*b^2-a*(B*b-
C*a))*(c+d*tan(f*x+e))^(1/2)/b/(a^2+b^2)/f/(a+b*tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 2.26 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3726, 3730, 3697, 3696, 95, 214} \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^{5/2}} \, dx=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (a^4 C d+2 a^3 b B d-a^2 b^2 (5 A d+3 B c-7 C d)+2 a b^3 (3 A c-2 B d-3 c C)+b^4 (A d+3 B c)\right )}{3 b f \left (a^2+b^2\right )^2 (b c-a d) \sqrt {a+b \tan (e+f x)}}-\frac {\sqrt {c-i d} (i A+B-i C) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{5/2}}-\frac {\sqrt {c+i d} (B-i (A-C)) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{5/2}} \]

[In]

Int[(Sqrt[c + d*Tan[e + f*x]]*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2))/(a + b*Tan[e + f*x])^(5/2),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]])])/((a - I*b)^(5/2)*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]])])/((a + I*b)^(5/2)*f) - (2*(A*b^2 - a*(b*B - a*C))*Sqrt[c + d
*Tan[e + f*x]])/(3*b*(a^2 + b^2)*f*(a + b*Tan[e + f*x])^(3/2)) - (2*(2*a^3*b*B*d + a^4*C*d + b^4*(3*B*c + A*d)
+ 2*a*b^3*(3*A*c - 3*c*C - 2*B*d) - a^2*b^2*(3*B*c + 5*A*d - 7*C*d))*Sqrt[c + d*Tan[e + f*x]])/(3*b*(a^2 + b^
2)^2*(b*c - a*d)*f*Sqrt[a + b*Tan[e + f*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 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 3696

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

Rule 3697

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

Rule 3726

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] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Ta
n[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3730

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] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Ta
n[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

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

Mathematica [A] (verified)

Time = 7.10 (sec) , antiderivative size = 600, normalized size of antiderivative = 1.62 \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^{5/2}} \, dx=-\frac {C \sqrt {c+d \tan (e+f x)}}{b f (a+b \tan (e+f x))^{3/2}}-\frac {-\frac {2 \left (\frac {1}{2} b^2 (-2 A b c+3 b c C-a C d)-a \left (-b^2 (B c+(A-C) d)-\frac {1}{2} a (b c C-2 b B d-a C d)\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (-\frac {3 b (b c-a d) \left (\frac {(a+i b)^2 (i A+B-i C) \sqrt {-c+i d} \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+i b}}+\frac {(a-i b)^2 (B-i (A-C)) \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b}}\right )}{2 \left (a^2+b^2\right ) f}-\frac {2 \left (\frac {1}{2} b^2 (b c-a d) \left (a^2 C d+b^2 (3 B c+A d)+a b (3 A c-3 c C-B d)\right )-a \left (\frac {1}{2} a \left (2 A b^2-2 a b B-a^2 C-3 b^2 C\right ) d (b c-a d)-\frac {3}{2} b^2 (b c-a d) (A b c-a B c-b c C-a A d-b B d+a C d)\right )\right ) \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)}}\right )}{3 \left (a^2+b^2\right ) (b c-a d)}}{b} \]

[In]

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

[Out]

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

Maple [F(-1)]

Timed out.

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

[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))^(5/2),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))^(5/2),x)

Fricas [F(-1)]

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

[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))^(5/2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

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

[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))**(5/2),x)

[Out]

Integral(sqrt(c + d*tan(e + f*x))*(A + B*tan(e + f*x) + C*tan(e + f*x)**2)/(a + b*tan(e + f*x))**(5/2), x)

Maxima [F(-2)]

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

[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))^(5/2),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(((2*b*d+2*a*c)^2>0)', see `ass
ume?` for mo

Giac [F(-1)]

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

[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))^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

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

[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))^(5/2),x)

[Out]

\text{Hanged}