3.52 \(\int \frac{\sqrt{c+d x} (A+B x+C x^2)}{(a+b x)^3 \sqrt{e+f x}} \, dx\)

Optimal. Leaf size=484 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x} \sqrt{b e-a f}}{\sqrt{e+f x} \sqrt{b c-a d}}\right ) \left (3 a^2 b^2 C \left (c^2 f^2+10 c d e f+5 d^2 e^2\right )-4 a^3 b C d f (3 c f+5 d e)+8 a^4 C d^2 f^2-a b^3 \left (2 c d \left (2 A f^2-B e f+12 C e^2\right )+d^2 e (3 B e-4 A f)+c^2 f (8 C e-B f)\right )-b^4 \left (c^2 \left (-\left (3 A f^2-4 B e f+8 C e^2\right )\right )-2 c d e (2 B e-A f)+A d^2 e^2\right )\right )}{4 b^3 (b c-a d)^{3/2} (b e-a f)^{5/2}}+\frac{\sqrt{c+d x} \sqrt{e+f x} \left (-a^2 b C (5 c f+7 d e)+4 a^3 C d f+a b^2 (-4 A d f+B c f+3 B d e+8 c C e)-b^3 (-3 A c f-A d e+4 B c e)\right )}{4 b^2 (a+b x) (b c-a d) (b e-a f)^2}-\frac{(c+d x)^{3/2} \sqrt{e+f x} \left (A b^2-a (b B-a C)\right )}{2 b (a+b x)^2 (b c-a d) (b e-a f)}+\frac{2 C \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d} \sqrt{e+f x}}\right )}{b^3 \sqrt{f}} \]

[Out]

((4*a^3*C*d*f - a^2*b*C*(7*d*e + 5*c*f) - b^3*(4*B*c*e - A*d*e - 3*A*c*f) + a*b^2*(8*c*C*e + 3*B*d*e + B*c*f -
4*A*d*f))*Sqrt[c + d*x]*Sqrt[e + f*x])/(4*b^2*(b*c - a*d)*(b*e - a*f)^2*(a + b*x)) - ((A*b^2 - a*(b*B - a*C))
*(c + d*x)^(3/2)*Sqrt[e + f*x])/(2*b*(b*c - a*d)*(b*e - a*f)*(a + b*x)^2) + (2*C*Sqrt[d]*ArcTanh[(Sqrt[f]*Sqrt
[c + d*x])/(Sqrt[d]*Sqrt[e + f*x])])/(b^3*Sqrt[f]) - ((8*a^4*C*d^2*f^2 - 4*a^3*b*C*d*f*(5*d*e + 3*c*f) + 3*a^2
*b^2*C*(5*d^2*e^2 + 10*c*d*e*f + c^2*f^2) - a*b^3*(d^2*e*(3*B*e - 4*A*f) + c^2*f*(8*C*e - B*f) + 2*c*d*(12*C*e
^2 - B*e*f + 2*A*f^2)) - b^4*(A*d^2*e^2 - 2*c*d*e*(2*B*e - A*f) - c^2*(8*C*e^2 - 4*B*e*f + 3*A*f^2)))*ArcTanh[
(Sqrt[b*e - a*f]*Sqrt[c + d*x])/(Sqrt[b*c - a*d]*Sqrt[e + f*x])])/(4*b^3*(b*c - a*d)^(3/2)*(b*e - a*f)^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 1.56326, antiderivative size = 484, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1613, 149, 157, 63, 217, 206, 93, 208} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x} \sqrt{b e-a f}}{\sqrt{e+f x} \sqrt{b c-a d}}\right ) \left (3 a^2 b^2 C \left (c^2 f^2+10 c d e f+5 d^2 e^2\right )-4 a^3 b C d f (3 c f+5 d e)+8 a^4 C d^2 f^2-a b^3 \left (2 c d \left (2 A f^2-B e f+12 C e^2\right )+d^2 e (3 B e-4 A f)+c^2 f (8 C e-B f)\right )-b^4 \left (c^2 \left (-\left (3 A f^2-4 B e f+8 C e^2\right )\right )-2 c d e (2 B e-A f)+A d^2 e^2\right )\right )}{4 b^3 (b c-a d)^{3/2} (b e-a f)^{5/2}}+\frac{\sqrt{c+d x} \sqrt{e+f x} \left (-a^2 b C (5 c f+7 d e)+4 a^3 C d f+a b^2 (-4 A d f+B c f+3 B d e+8 c C e)-b^3 (-3 A c f-A d e+4 B c e)\right )}{4 b^2 (a+b x) (b c-a d) (b e-a f)^2}-\frac{(c+d x)^{3/2} \sqrt{e+f x} \left (A b^2-a (b B-a C)\right )}{2 b (a+b x)^2 (b c-a d) (b e-a f)}+\frac{2 C \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d} \sqrt{e+f x}}\right )}{b^3 \sqrt{f}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[c + d*x]*(A + B*x + C*x^2))/((a + b*x)^3*Sqrt[e + f*x]),x]

[Out]

((4*a^3*C*d*f - a^2*b*C*(7*d*e + 5*c*f) - b^3*(4*B*c*e - A*d*e - 3*A*c*f) + a*b^2*(8*c*C*e + 3*B*d*e + B*c*f -
4*A*d*f))*Sqrt[c + d*x]*Sqrt[e + f*x])/(4*b^2*(b*c - a*d)*(b*e - a*f)^2*(a + b*x)) - ((A*b^2 - a*(b*B - a*C))
*(c + d*x)^(3/2)*Sqrt[e + f*x])/(2*b*(b*c - a*d)*(b*e - a*f)*(a + b*x)^2) + (2*C*Sqrt[d]*ArcTanh[(Sqrt[f]*Sqrt
[c + d*x])/(Sqrt[d]*Sqrt[e + f*x])])/(b^3*Sqrt[f]) - ((8*a^4*C*d^2*f^2 - 4*a^3*b*C*d*f*(5*d*e + 3*c*f) + 3*a^2
*b^2*C*(5*d^2*e^2 + 10*c*d*e*f + c^2*f^2) - a*b^3*(d^2*e*(3*B*e - 4*A*f) + c^2*f*(8*C*e - B*f) + 2*c*d*(12*C*e
^2 - B*e*f + 2*A*f^2)) - b^4*(A*d^2*e^2 - 2*c*d*e*(2*B*e - A*f) - c^2*(8*C*e^2 - 4*B*e*f + 3*A*f^2)))*ArcTanh[
(Sqrt[b*e - a*f]*Sqrt[c + d*x])/(Sqrt[b*c - a*d]*Sqrt[e + f*x])])/(4*b^3*(b*c - a*d)^(3/2)*(b*e - a*f)^(5/2))

Rule 1613

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> With[{
Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[(b*R*(a + b*x)^(m + 1)
*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e
- a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*ExpandToSum[(m + 1)*(b*c - a*d)*(b*e - a*f)*Qx + a*d*f
*R*(m + 1) - b*R*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*R*(m + n + p + 3)*x, x], x], x]] /; FreeQ[{a, b,
c, d, e, f, n, p}, x] && PolyQ[Px, x] && ILtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 149

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\sqrt{c+d x} \left (A+B x+C x^2\right )}{(a+b x)^3 \sqrt{e+f x}} \, dx &=-\frac{\left (A b^2-a (b B-a C)\right ) (c+d x)^{3/2} \sqrt{e+f x}}{2 b (b c-a d) (b e-a f) (a+b x)^2}-\frac{\int \frac{\sqrt{c+d x} \left (-\frac{a^2 C (3 d e+c f)+b^2 (4 B c e-A d e-3 A c f)-a b (4 c C e+3 B d e+B c f-4 A d f)}{2 b}-\frac{2 C (b c-a d) (b e-a f) x}{b}\right )}{(a+b x)^2 \sqrt{e+f x}} \, dx}{2 (b c-a d) (b e-a f)}\\ &=\frac{\left (4 a^3 C d f-a^2 b C (7 d e+5 c f)-b^3 (4 B c e-A d e-3 A c f)+a b^2 (8 c C e+3 B d e+B c f-4 A d f)\right ) \sqrt{c+d x} \sqrt{e+f x}}{4 b^2 (b c-a d) (b e-a f)^2 (a+b x)}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d x)^{3/2} \sqrt{e+f x}}{2 b (b c-a d) (b e-a f) (a+b x)^2}-\frac{\int \frac{\frac{4 a^3 C d f (d e+c f)-a^2 b C \left (7 d^2 e^2+14 c d e f+3 c^2 f^2\right )+a b^2 \left (d^2 e (3 B e-4 A f)+c^2 f (8 C e-B f)+2 c d \left (8 C e^2-B e f+2 A f^2\right )\right )+b^3 \left (A d^2 e^2-2 c d e (2 B e-A f)-c^2 \left (8 C e^2-4 B e f+3 A f^2\right )\right )}{4 b}-\frac{2 C d (b c-a d) (b e-a f)^2 x}{b}}{(a+b x) \sqrt{c+d x} \sqrt{e+f x}} \, dx}{2 b (b c-a d) (b e-a f)^2}\\ &=\frac{\left (4 a^3 C d f-a^2 b C (7 d e+5 c f)-b^3 (4 B c e-A d e-3 A c f)+a b^2 (8 c C e+3 B d e+B c f-4 A d f)\right ) \sqrt{c+d x} \sqrt{e+f x}}{4 b^2 (b c-a d) (b e-a f)^2 (a+b x)}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d x)^{3/2} \sqrt{e+f x}}{2 b (b c-a d) (b e-a f) (a+b x)^2}+\frac{(C d) \int \frac{1}{\sqrt{c+d x} \sqrt{e+f x}} \, dx}{b^3}+\frac{\left (8 a^4 C d^2 f^2-4 a^3 b C d f (5 d e+3 c f)+3 a^2 b^2 C \left (5 d^2 e^2+10 c d e f+c^2 f^2\right )-a b^3 \left (d^2 e (3 B e-4 A f)+c^2 f (8 C e-B f)+2 c d \left (12 C e^2-B e f+2 A f^2\right )\right )-b^4 \left (A d^2 e^2-2 c d e (2 B e-A f)-c^2 \left (8 C e^2-4 B e f+3 A f^2\right )\right )\right ) \int \frac{1}{(a+b x) \sqrt{c+d x} \sqrt{e+f x}} \, dx}{8 b^3 (b c-a d) (b e-a f)^2}\\ &=\frac{\left (4 a^3 C d f-a^2 b C (7 d e+5 c f)-b^3 (4 B c e-A d e-3 A c f)+a b^2 (8 c C e+3 B d e+B c f-4 A d f)\right ) \sqrt{c+d x} \sqrt{e+f x}}{4 b^2 (b c-a d) (b e-a f)^2 (a+b x)}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d x)^{3/2} \sqrt{e+f x}}{2 b (b c-a d) (b e-a f) (a+b x)^2}+\frac{(2 C) \operatorname{Subst}\left (\int \frac{1}{\sqrt{e-\frac{c f}{d}+\frac{f x^2}{d}}} \, dx,x,\sqrt{c+d x}\right )}{b^3}+\frac{\left (8 a^4 C d^2 f^2-4 a^3 b C d f (5 d e+3 c f)+3 a^2 b^2 C \left (5 d^2 e^2+10 c d e f+c^2 f^2\right )-a b^3 \left (d^2 e (3 B e-4 A f)+c^2 f (8 C e-B f)+2 c d \left (12 C e^2-B e f+2 A f^2\right )\right )-b^4 \left (A d^2 e^2-2 c d e (2 B e-A f)-c^2 \left (8 C e^2-4 B e f+3 A f^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-b c+a d-(-b e+a f) x^2} \, dx,x,\frac{\sqrt{c+d x}}{\sqrt{e+f x}}\right )}{4 b^3 (b c-a d) (b e-a f)^2}\\ &=\frac{\left (4 a^3 C d f-a^2 b C (7 d e+5 c f)-b^3 (4 B c e-A d e-3 A c f)+a b^2 (8 c C e+3 B d e+B c f-4 A d f)\right ) \sqrt{c+d x} \sqrt{e+f x}}{4 b^2 (b c-a d) (b e-a f)^2 (a+b x)}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d x)^{3/2} \sqrt{e+f x}}{2 b (b c-a d) (b e-a f) (a+b x)^2}-\frac{\left (8 a^4 C d^2 f^2-4 a^3 b C d f (5 d e+3 c f)+3 a^2 b^2 C \left (5 d^2 e^2+10 c d e f+c^2 f^2\right )-a b^3 \left (d^2 e (3 B e-4 A f)+c^2 f (8 C e-B f)+2 c d \left (12 C e^2-B e f+2 A f^2\right )\right )-b^4 \left (A d^2 e^2-2 c d e (2 B e-A f)-c^2 \left (8 C e^2-4 B e f+3 A f^2\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b e-a f} \sqrt{c+d x}}{\sqrt{b c-a d} \sqrt{e+f x}}\right )}{4 b^3 (b c-a d)^{3/2} (b e-a f)^{5/2}}+\frac{(2 C) \operatorname{Subst}\left (\int \frac{1}{1-\frac{f x^2}{d}} \, dx,x,\frac{\sqrt{c+d x}}{\sqrt{e+f x}}\right )}{b^3}\\ &=\frac{\left (4 a^3 C d f-a^2 b C (7 d e+5 c f)-b^3 (4 B c e-A d e-3 A c f)+a b^2 (8 c C e+3 B d e+B c f-4 A d f)\right ) \sqrt{c+d x} \sqrt{e+f x}}{4 b^2 (b c-a d) (b e-a f)^2 (a+b x)}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d x)^{3/2} \sqrt{e+f x}}{2 b (b c-a d) (b e-a f) (a+b x)^2}+\frac{2 C \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d} \sqrt{e+f x}}\right )}{b^3 \sqrt{f}}-\frac{\left (8 a^4 C d^2 f^2-4 a^3 b C d f (5 d e+3 c f)+3 a^2 b^2 C \left (5 d^2 e^2+10 c d e f+c^2 f^2\right )-a b^3 \left (d^2 e (3 B e-4 A f)+c^2 f (8 C e-B f)+2 c d \left (12 C e^2-B e f+2 A f^2\right )\right )-b^4 \left (A d^2 e^2-2 c d e (2 B e-A f)-c^2 \left (8 C e^2-4 B e f+3 A f^2\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b e-a f} \sqrt{c+d x}}{\sqrt{b c-a d} \sqrt{e+f x}}\right )}{4 b^3 (b c-a d)^{3/2} (b e-a f)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 6.30474, size = 535, normalized size = 1.11 \[ -\frac{(c+d x)^{3/2} \sqrt{e+f x} \left (A b^2-a (b B-a C)\right )}{2 b (a+b x)^2 (b c-a d) (b e-a f)}+\frac{\left (A b^2-a (b B-a C)\right ) (-4 a d f+3 b c f+b d e) \left (\frac{\sqrt{c+d x} \sqrt{e+f x}}{(a+b x) (b e-a f)}-\frac{(d e-c f) \tan ^{-1}\left (\frac{\sqrt{c+d x} \sqrt{b e-a f}}{\sqrt{e+f x} \sqrt{a d-b c}}\right )}{\sqrt{a d-b c} (b e-a f)^{3/2}}\right )}{4 b^2 (b c-a d) (b e-a f)}-\frac{\sqrt{c+d x} \sqrt{e+f x} (b B-2 a C)}{b^2 (a+b x) (b e-a f)}+\frac{(b B-2 a C) (d e-c f) \tan ^{-1}\left (\frac{\sqrt{c+d x} \sqrt{b e-a f}}{\sqrt{e+f x} \sqrt{a d-b c}}\right )}{b^2 \sqrt{a d-b c} (b e-a f)^{3/2}}-\frac{2 C \sqrt{a d-b c} \tan ^{-1}\left (\frac{\sqrt{c+d x} \sqrt{b e-a f}}{\sqrt{e+f x} \sqrt{a d-b c}}\right )}{b^3 \sqrt{b e-a f}}+\frac{2 C \sqrt{d e-c f} \sqrt{\frac{d (e+f x)}{d e-c f}} \sinh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d e-c f}}\right )}{b^3 \sqrt{f} \sqrt{e+f x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[c + d*x]*(A + B*x + C*x^2))/((a + b*x)^3*Sqrt[e + f*x]),x]

[Out]

-(((b*B - 2*a*C)*Sqrt[c + d*x]*Sqrt[e + f*x])/(b^2*(b*e - a*f)*(a + b*x))) - ((A*b^2 - a*(b*B - a*C))*(c + d*x
)^(3/2)*Sqrt[e + f*x])/(2*b*(b*c - a*d)*(b*e - a*f)*(a + b*x)^2) + (2*C*Sqrt[d*e - c*f]*Sqrt[(d*(e + f*x))/(d*
e - c*f)]*ArcSinh[(Sqrt[f]*Sqrt[c + d*x])/Sqrt[d*e - c*f]])/(b^3*Sqrt[f]*Sqrt[e + f*x]) - (2*C*Sqrt[-(b*c) + a
*d]*ArcTan[(Sqrt[b*e - a*f]*Sqrt[c + d*x])/(Sqrt[-(b*c) + a*d]*Sqrt[e + f*x])])/(b^3*Sqrt[b*e - a*f]) + ((b*B
- 2*a*C)*(d*e - c*f)*ArcTan[(Sqrt[b*e - a*f]*Sqrt[c + d*x])/(Sqrt[-(b*c) + a*d]*Sqrt[e + f*x])])/(b^2*Sqrt[-(b
*c) + a*d]*(b*e - a*f)^(3/2)) + ((A*b^2 - a*(b*B - a*C))*(b*d*e + 3*b*c*f - 4*a*d*f)*((Sqrt[c + d*x]*Sqrt[e +
f*x])/((b*e - a*f)*(a + b*x)) - ((d*e - c*f)*ArcTan[(Sqrt[b*e - a*f]*Sqrt[c + d*x])/(Sqrt[-(b*c) + a*d]*Sqrt[e
+ f*x])])/(Sqrt[-(b*c) + a*d]*(b*e - a*f)^(3/2))))/(4*b^2*(b*c - a*d)*(b*e - a*f))

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Maple [B]  time = 0.088, size = 9100, normalized size = 18.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+B*x+A)*(d*x+c)^(1/2)/(b*x+a)^3/(f*x+e)^(1/2),x)

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(d*x+c)^(1/2)/(b*x+a)^3/(f*x+e)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(d*x+c)^(1/2)/(b*x+a)^3/(f*x+e)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)*(d*x+c)**(1/2)/(b*x+a)**3/(f*x+e)**(1/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(d*x+c)^(1/2)/(b*x+a)^3/(f*x+e)^(1/2),x, algorithm="giac")

[Out]

Timed out