\(\int \frac {\sqrt {a+b \tan (e+f x)} (A+B \tan (e+f x)+C \tan ^2(e+f x))}{(c+d \tan (e+f x))^{5/2}} \, dx\) [161]

Optimal result

Integrand size = 49, antiderivative size = 373 \[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {\sqrt {a-i b} (i A+B-i C) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c-i d)^{5/2} f}-\frac {\sqrt {a+i b} (B-i (A-C)) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c+i d)^{5/2} f}-\frac {2 \left (c^2 C-B c d+A d^2\right ) \sqrt {a+b \tan (e+f x)}}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (b \left (c^4 C+2 B c^3 d-c^2 (5 A-7 C) d^2-4 B c d^3+A d^4\right )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 d (b c-a d) \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \]

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

Rubi [A] (verified)

Time = 2.24 (sec) , antiderivative size = 373, 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} \[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {\sqrt {a-i b} (i A+B-i C) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (c-i d)^{5/2}}-\frac {\sqrt {a+i b} (B-i (A-C)) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (c+i d)^{5/2}}-\frac {2 \left (A d^2-B c d+c^2 C\right ) \sqrt {a+b \tan (e+f x)}}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {2 \sqrt {a+b \tan (e+f x)} \left (3 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (-c^2 d^2 (5 A-7 C)+A d^4+2 B c^3 d-4 B c d^3+c^4 C\right )\right )}{3 d f \left (c^2+d^2\right )^2 (b c-a d) \sqrt {c+d \tan (e+f x)}} \]

[In]

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

[Out]

-((Sqrt[a - I*b]*(I*A + B - I*C)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Ta
n[e + f*x]])])/((c - I*d)^(5/2)*f)) - (Sqrt[a + I*b]*(B - I*(A - C))*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e +
f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/((c + I*d)^(5/2)*f) - (2*(c^2*C - B*c*d + A*d^2)*Sqrt[a + b
*Tan[e + f*x]])/(3*d*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) + (2*(b*(c^4*C + 2*B*c^3*d - c^2*(5*A - 7*C)*d^
2 - 4*B*c*d^3 + A*d^4) + 3*a*d^2*(2*c*(A - C)*d - B*(c^2 - d^2)))*Sqrt[a + b*Tan[e + f*x]])/(3*d*(b*c - a*d)*(
c^2 + d^2)^2*f*Sqrt[c + d*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 (c^2 C-B c d+A d^2\right ) \sqrt {a+b \tan (e+f x)}}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \int \frac {\frac {1}{2} (A d (3 a c+b d)+(b c-3 a d) (c C-B d))+\frac {3}{2} d ((A-C) (b c-a d)+B (a c+b d)) \tan (e+f x)+\frac {1}{2} b \left (c^2 C+2 B c d-(2 A-3 C) d^2\right ) \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx}{3 d \left (c^2+d^2\right )} \\ & = -\frac {2 \left (c^2 C-B c d+A d^2\right ) \sqrt {a+b \tan (e+f x)}}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (b \left (c^4 C+2 B c^3 d-c^2 (5 A-7 C) d^2-4 B c d^3+A d^4\right )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 d (b c-a d) \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {4 \int \frac {-\frac {3}{4} d (b c-a d) \left (a \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )-b \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )-\frac {3}{4} d (b c-a d) \left (2 a A c d-2 a c C d-A b \left (c^2-d^2\right )-a B \left (c^2-d^2\right )+b \left (c^2 C-2 B c d-C d^2\right )\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{3 d (b c-a d) \left (c^2+d^2\right )^2} \\ & = -\frac {2 \left (c^2 C-B c d+A d^2\right ) \sqrt {a+b \tan (e+f x)}}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (b \left (c^4 C+2 B c^3 d-c^2 (5 A-7 C) d^2-4 B c d^3+A d^4\right )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 d (b c-a d) \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {((a-i b) (A-i B-C)) \int \frac {1+i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (c-i d)^2}+\frac {((a+i b) (A+i B-C)) \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (c+i d)^2} \\ & = -\frac {2 \left (c^2 C-B c d+A d^2\right ) \sqrt {a+b \tan (e+f x)}}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (b \left (c^4 C+2 B c^3 d-c^2 (5 A-7 C) d^2-4 B c d^3+A d^4\right )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 d (b c-a d) \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {((a-i b) (A-i B-C)) \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 (c-i d)^2 f}+\frac {((a+i b) (A+i B-C)) \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 (c+i d)^2 f} \\ & = -\frac {2 \left (c^2 C-B c d+A d^2\right ) \sqrt {a+b \tan (e+f x)}}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (b \left (c^4 C+2 B c^3 d-c^2 (5 A-7 C) d^2-4 B c d^3+A d^4\right )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 d (b c-a d) \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {((a-i b) (A-i B-C)) \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 )}{(c-i d)^2 f}+\frac {((a+i b) (A+i B-C)) \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 )}{(c+i d)^2 f} \\ & = -\frac {\sqrt {a-i b} (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 )}{(c-i d)^{5/2} f}-\frac {\sqrt {a+i b} (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 )}{(c+i d)^{5/2} f}-\frac {2 \left (c^2 C-B c d+A d^2\right ) \sqrt {a+b \tan (e+f x)}}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (b \left (c^4 C+2 B c^3 d-c^2 (5 A-7 C) d^2-4 B c d^3+A d^4\right )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 d (b c-a d) \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.11 (sec) , antiderivative size = 609, normalized size of antiderivative = 1.63 \[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {C \sqrt {a+b \tan (e+f x)}}{d f (c+d \tan (e+f x))^{3/2}}-\frac {-\frac {2 \left (\frac {1}{2} d^2 (-b c C-a (2 A-3 C) d)-c \left (-\left ((A b+a B-b C) d^2\right )-\frac {1}{2} c (-b c C-2 b B d+a C d)\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 (-b c+a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (-\frac {3 d (b c-a d)^2 \left (\frac {\sqrt {-a+i b} (i A+B-i C) (c+i d)^2 \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-c+i d}}+\frac {\sqrt {a+i b} (B-i (A-C)) (c-i d)^2 \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c+i d}}\right )}{2 (-b c+a d) \left (c^2+d^2\right ) f}-\frac {2 \left (-\frac {1}{2} d^2 (b c-a d) \left (3 a d (A c-c C+B d)+b \left (c^2 C-B c d+A d^2\right )\right )-c \left (-\frac {3}{2} d^2 (b c-a d) (A b c+a B c-b c C-a A d+b B d+a C d)+\frac {1}{2} b c (b c-a d) \left (c^2 C+2 B c d-(2 A-3 C) d^2\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{(-b c+a d) \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}\right )}{3 (-b c+a d) \left (c^2+d^2\right )}}{d} \]

[In]

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

[Out]

-((C*Sqrt[a + b*Tan[e + f*x]])/(d*f*(c + d*Tan[e + f*x])^(3/2))) - ((-2*((d^2*(-(b*c*C) - a*(2*A - 3*C)*d))/2
- c*(-((A*b + a*B - b*C)*d^2) - (c*(-(b*c*C) - 2*b*B*d + a*C*d))/2))*Sqrt[a + b*Tan[e + f*x]])/(3*(-(b*c) + a*
d)*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) - (2*((-3*d*(b*c - a*d)^2*((Sqrt[-a + I*b]*(I*A + B - I*C)*(c + I
*d)^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[-c +
I*d] + (Sqrt[a + I*b]*(B - I*(A - C))*(c - I*d)^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[c + I*d]))/(2*(-(b*c) + a*d)*(c^2 + d^2)*f) - (2*(-1/2*(d^2*(b*c - a*d)*(
3*a*d*(A*c - c*C + B*d) + b*(c^2*C - B*c*d + A*d^2))) - c*((-3*d^2*(b*c - a*d)*(A*b*c + a*B*c - b*c*C - a*A*d
+ b*B*d + a*C*d))/2 + (b*c*(b*c - a*d)*(c^2*C + 2*B*c*d - (2*A - 3*C)*d^2))/2))*Sqrt[a + b*Tan[e + f*x]])/((-(
b*c) + a*d)*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]])))/(3*(-(b*c) + a*d)*(c^2 + d^2)))/d

Maple [F(-1)]

Timed out.

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

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*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. \[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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