3.6.82 \(\int \frac {a+b x+c x^2}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx\)

Optimal. Leaf size=160 \[ \frac {2 \sqrt {f+g x} \left (c \left (6 d e f-4 d^2 g\right )-e (-2 a e g-b d g+3 b e f)\right )}{3 e^2 \sqrt {d+e x} (e f-d g)^2}-\frac {2 \sqrt {f+g x} \left (a+\frac {d (c d-b e)}{e^2}\right )}{3 (d+e x)^{3/2} (e f-d g)}+\frac {2 c \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{e^{5/2} \sqrt {g}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {949, 78, 63, 217, 206} \begin {gather*} \frac {2 \sqrt {f+g x} \left (c \left (6 d e f-4 d^2 g\right )-e (-2 a e g-b d g+3 b e f)\right )}{3 e^2 \sqrt {d+e x} (e f-d g)^2}-\frac {2 \sqrt {f+g x} \left (a+\frac {d (c d-b e)}{e^2}\right )}{3 (d+e x)^{3/2} (e f-d g)}+\frac {2 c \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{e^{5/2} \sqrt {g}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + (d*(c*d - b*e))/e^2)*Sqrt[f + g*x])/(3*(e*f - d*g)*(d + e*x)^(3/2)) + (2*(c*(6*d*e*f - 4*d^2*g) - e*(
3*b*e*f - b*d*g - 2*a*e*g))*Sqrt[f + g*x])/(3*e^2*(e*f - d*g)^2*Sqrt[d + e*x]) + (2*c*ArcTanh[(Sqrt[g]*Sqrt[d
+ e*x])/(Sqrt[e]*Sqrt[f + g*x])])/(e^(5/2)*Sqrt[g])

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 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 949

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{Qx = PolynomialQuotient[(a + b*x + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p,
d + e*x, x]}, Simp[(R*(d + e*x)^(m + 1)*(f + g*x)^(n + 1))/((m + 1)*(e*f - d*g)), x] + Dist[1/((m + 1)*(e*f -
d*g)), Int[(d + e*x)^(m + 1)*(f + g*x)^n*ExpandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /;
FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&& IGtQ[p, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b x+c x^2}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx &=-\frac {2 \left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{3 (e f-d g) (d+e x)^{3/2}}-\frac {2 \int \frac {\frac {c d (3 e f-d g)-e (3 b e f-b d g-2 a e g)}{2 e^2}-\frac {3}{2} c \left (f-\frac {d g}{e}\right ) x}{(d+e x)^{3/2} \sqrt {f+g x}} \, dx}{3 (e f-d g)}\\ &=-\frac {2 \left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{3 (e f-d g) (d+e x)^{3/2}}+\frac {2 \left (c \left (6 d e f-4 d^2 g\right )-e (3 b e f-b d g-2 a e g)\right ) \sqrt {f+g x}}{3 e^2 (e f-d g)^2 \sqrt {d+e x}}+\frac {c \int \frac {1}{\sqrt {d+e x} \sqrt {f+g x}} \, dx}{e^2}\\ &=-\frac {2 \left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{3 (e f-d g) (d+e x)^{3/2}}+\frac {2 \left (c \left (6 d e f-4 d^2 g\right )-e (3 b e f-b d g-2 a e g)\right ) \sqrt {f+g x}}{3 e^2 (e f-d g)^2 \sqrt {d+e x}}+\frac {(2 c) \operatorname {Subst}\left (\int \frac {1}{\sqrt {f-\frac {d g}{e}+\frac {g x^2}{e}}} \, dx,x,\sqrt {d+e x}\right )}{e^3}\\ &=-\frac {2 \left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{3 (e f-d g) (d+e x)^{3/2}}+\frac {2 \left (c \left (6 d e f-4 d^2 g\right )-e (3 b e f-b d g-2 a e g)\right ) \sqrt {f+g x}}{3 e^2 (e f-d g)^2 \sqrt {d+e x}}+\frac {(2 c) \operatorname {Subst}\left (\int \frac {1}{1-\frac {g x^2}{e}} \, dx,x,\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e^3}\\ &=-\frac {2 \left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{3 (e f-d g) (d+e x)^{3/2}}+\frac {2 \left (c \left (6 d e f-4 d^2 g\right )-e (3 b e f-b d g-2 a e g)\right ) \sqrt {f+g x}}{3 e^2 (e f-d g)^2 \sqrt {d+e x}}+\frac {2 c \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{e^{5/2} \sqrt {g}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.23, size = 173, normalized size = 1.08 \begin {gather*} \frac {2 \sqrt {f+g x} \left (2 g (d+e x) \left (g (a g-b f)+c f^2\right )-(e f-d g) \left (g (a g-b f)+c f^2\right )+(f+g x) (2 c f-b g) (e f-d g)-\frac {c (e f-d g)^3 \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};\frac {g (d+e x)}{d g-e f}\right )}{e^2 \sqrt {\frac {e (f+g x)}{e f-d g}}}\right )}{3 g^2 (d+e x)^{3/2} (e f-d g)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[f + g*x]*(-((e*f - d*g)*(c*f^2 + g*(-(b*f) + a*g))) + 2*g*(c*f^2 + g*(-(b*f) + a*g))*(d + e*x) + (2*c*
f - b*g)*(e*f - d*g)*(f + g*x) - (c*(e*f - d*g)^3*Hypergeometric2F1[-3/2, -3/2, -1/2, (g*(d + e*x))/(-(e*f) +
d*g)])/(e^2*Sqrt[(e*(f + g*x))/(e*f - d*g)])))/(3*g^2*(e*f - d*g)^2*(d + e*x)^(3/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.23, size = 161, normalized size = 1.01 \begin {gather*} \frac {2 c \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {g} \sqrt {d+e x}}\right )}{e^{5/2} \sqrt {g}}-\frac {2 \sqrt {f+g x} \left (\frac {a e^3 (f+g x)}{d+e x}-3 a e^2 g-\frac {b d e^2 (f+g x)}{d+e x}+3 b e^2 f+\frac {c d^2 e (f+g x)}{d+e x}+3 c d^2 g-6 c d e f\right )}{3 e^2 \sqrt {d+e x} (e f-d g)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x + c*x^2)/((d + e*x)^(5/2)*Sqrt[f + g*x]),x]

[Out]

(-2*Sqrt[f + g*x]*(-6*c*d*e*f + 3*b*e^2*f + 3*c*d^2*g - 3*a*e^2*g + (c*d^2*e*(f + g*x))/(d + e*x) - (b*d*e^2*(
f + g*x))/(d + e*x) + (a*e^3*(f + g*x))/(d + e*x)))/(3*e^2*(e*f - d*g)^2*Sqrt[d + e*x]) + (2*c*ArcTanh[(Sqrt[e
]*Sqrt[f + g*x])/(Sqrt[g]*Sqrt[d + e*x])])/(e^(5/2)*Sqrt[g])

________________________________________________________________________________________

fricas [B]  time = 3.55, size = 792, normalized size = 4.95 \begin {gather*} \left [\frac {3 \, {\left (c d^{2} e^{2} f^{2} - 2 \, c d^{3} e f g + c d^{4} g^{2} + {\left (c e^{4} f^{2} - 2 \, c d e^{3} f g + c d^{2} e^{2} g^{2}\right )} x^{2} + 2 \, {\left (c d e^{3} f^{2} - 2 \, c d^{2} e^{2} f g + c d^{3} e g^{2}\right )} x\right )} \sqrt {e g} \log \left (8 \, e^{2} g^{2} x^{2} + e^{2} f^{2} + 6 \, d e f g + d^{2} g^{2} + 4 \, {\left (2 \, e g x + e f + d g\right )} \sqrt {e g} \sqrt {e x + d} \sqrt {g x + f} + 8 \, {\left (e^{2} f g + d e g^{2}\right )} x\right ) + 4 \, {\left ({\left (5 \, c d^{2} e^{2} - 2 \, b d e^{3} - a e^{4}\right )} f g - 3 \, {\left (c d^{3} e - a d e^{3}\right )} g^{2} + {\left (3 \, {\left (2 \, c d e^{3} - b e^{4}\right )} f g - {\left (4 \, c d^{2} e^{2} - b d e^{3} - 2 \, a e^{4}\right )} g^{2}\right )} x\right )} \sqrt {e x + d} \sqrt {g x + f}}{6 \, {\left (d^{2} e^{5} f^{2} g - 2 \, d^{3} e^{4} f g^{2} + d^{4} e^{3} g^{3} + {\left (e^{7} f^{2} g - 2 \, d e^{6} f g^{2} + d^{2} e^{5} g^{3}\right )} x^{2} + 2 \, {\left (d e^{6} f^{2} g - 2 \, d^{2} e^{5} f g^{2} + d^{3} e^{4} g^{3}\right )} x\right )}}, -\frac {3 \, {\left (c d^{2} e^{2} f^{2} - 2 \, c d^{3} e f g + c d^{4} g^{2} + {\left (c e^{4} f^{2} - 2 \, c d e^{3} f g + c d^{2} e^{2} g^{2}\right )} x^{2} + 2 \, {\left (c d e^{3} f^{2} - 2 \, c d^{2} e^{2} f g + c d^{3} e g^{2}\right )} x\right )} \sqrt {-e g} \arctan \left (\frac {{\left (2 \, e g x + e f + d g\right )} \sqrt {-e g} \sqrt {e x + d} \sqrt {g x + f}}{2 \, {\left (e^{2} g^{2} x^{2} + d e f g + {\left (e^{2} f g + d e g^{2}\right )} x\right )}}\right ) - 2 \, {\left ({\left (5 \, c d^{2} e^{2} - 2 \, b d e^{3} - a e^{4}\right )} f g - 3 \, {\left (c d^{3} e - a d e^{3}\right )} g^{2} + {\left (3 \, {\left (2 \, c d e^{3} - b e^{4}\right )} f g - {\left (4 \, c d^{2} e^{2} - b d e^{3} - 2 \, a e^{4}\right )} g^{2}\right )} x\right )} \sqrt {e x + d} \sqrt {g x + f}}{3 \, {\left (d^{2} e^{5} f^{2} g - 2 \, d^{3} e^{4} f g^{2} + d^{4} e^{3} g^{3} + {\left (e^{7} f^{2} g - 2 \, d e^{6} f g^{2} + d^{2} e^{5} g^{3}\right )} x^{2} + 2 \, {\left (d e^{6} f^{2} g - 2 \, d^{2} e^{5} f g^{2} + d^{3} e^{4} g^{3}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(c*d^2*e^2*f^2 - 2*c*d^3*e*f*g + c*d^4*g^2 + (c*e^4*f^2 - 2*c*d*e^3*f*g + c*d^2*e^2*g^2)*x^2 + 2*(c*d*
e^3*f^2 - 2*c*d^2*e^2*f*g + c*d^3*e*g^2)*x)*sqrt(e*g)*log(8*e^2*g^2*x^2 + e^2*f^2 + 6*d*e*f*g + d^2*g^2 + 4*(2
*e*g*x + e*f + d*g)*sqrt(e*g)*sqrt(e*x + d)*sqrt(g*x + f) + 8*(e^2*f*g + d*e*g^2)*x) + 4*((5*c*d^2*e^2 - 2*b*d
*e^3 - a*e^4)*f*g - 3*(c*d^3*e - a*d*e^3)*g^2 + (3*(2*c*d*e^3 - b*e^4)*f*g - (4*c*d^2*e^2 - b*d*e^3 - 2*a*e^4)
*g^2)*x)*sqrt(e*x + d)*sqrt(g*x + f))/(d^2*e^5*f^2*g - 2*d^3*e^4*f*g^2 + d^4*e^3*g^3 + (e^7*f^2*g - 2*d*e^6*f*
g^2 + d^2*e^5*g^3)*x^2 + 2*(d*e^6*f^2*g - 2*d^2*e^5*f*g^2 + d^3*e^4*g^3)*x), -1/3*(3*(c*d^2*e^2*f^2 - 2*c*d^3*
e*f*g + c*d^4*g^2 + (c*e^4*f^2 - 2*c*d*e^3*f*g + c*d^2*e^2*g^2)*x^2 + 2*(c*d*e^3*f^2 - 2*c*d^2*e^2*f*g + c*d^3
*e*g^2)*x)*sqrt(-e*g)*arctan(1/2*(2*e*g*x + e*f + d*g)*sqrt(-e*g)*sqrt(e*x + d)*sqrt(g*x + f)/(e^2*g^2*x^2 + d
*e*f*g + (e^2*f*g + d*e*g^2)*x)) - 2*((5*c*d^2*e^2 - 2*b*d*e^3 - a*e^4)*f*g - 3*(c*d^3*e - a*d*e^3)*g^2 + (3*(
2*c*d*e^3 - b*e^4)*f*g - (4*c*d^2*e^2 - b*d*e^3 - 2*a*e^4)*g^2)*x)*sqrt(e*x + d)*sqrt(g*x + f))/(d^2*e^5*f^2*g
- 2*d^3*e^4*f*g^2 + d^4*e^3*g^3 + (e^7*f^2*g - 2*d*e^6*f*g^2 + d^2*e^5*g^3)*x^2 + 2*(d*e^6*f^2*g - 2*d^2*e^5*
f*g^2 + d^3*e^4*g^3)*x)]

________________________________________________________________________________________

giac [B]  time = 0.68, size = 504, normalized size = 3.15 \begin {gather*} -\frac {c e^{\left (-\frac {5}{2}\right )} \log \left ({\left (\sqrt {x e + d} \sqrt {g} e^{\frac {1}{2}} - \sqrt {{\left (x e + d\right )} g e - d g e + f e^{2}}\right )}^{2}\right )}{\sqrt {g}} - \frac {4 \, {\left (4 \, c d^{3} g^{\frac {5}{2}} e^{\frac {5}{2}} + 6 \, {\left (\sqrt {x e + d} \sqrt {g} e^{\frac {1}{2}} - \sqrt {{\left (x e + d\right )} g e - d g e + f e^{2}}\right )}^{2} c d^{2} g^{\frac {3}{2}} e^{\frac {3}{2}} + 6 \, {\left (\sqrt {x e + d} \sqrt {g} e^{\frac {1}{2}} - \sqrt {{\left (x e + d\right )} g e - d g e + f e^{2}}\right )}^{4} c d \sqrt {g} e^{\frac {1}{2}} - 10 \, c d^{2} f g^{\frac {3}{2}} e^{\frac {7}{2}} - b d^{2} g^{\frac {5}{2}} e^{\frac {7}{2}} - 12 \, {\left (\sqrt {x e + d} \sqrt {g} e^{\frac {1}{2}} - \sqrt {{\left (x e + d\right )} g e - d g e + f e^{2}}\right )}^{2} c d f \sqrt {g} e^{\frac {5}{2}} - 3 \, {\left (\sqrt {x e + d} \sqrt {g} e^{\frac {1}{2}} - \sqrt {{\left (x e + d\right )} g e - d g e + f e^{2}}\right )}^{4} b \sqrt {g} e^{\frac {3}{2}} + 6 \, c d f^{2} \sqrt {g} e^{\frac {9}{2}} + 4 \, b d f g^{\frac {3}{2}} e^{\frac {9}{2}} - 2 \, a d g^{\frac {5}{2}} e^{\frac {9}{2}} + 6 \, {\left (\sqrt {x e + d} \sqrt {g} e^{\frac {1}{2}} - \sqrt {{\left (x e + d\right )} g e - d g e + f e^{2}}\right )}^{2} b f \sqrt {g} e^{\frac {7}{2}} - 6 \, {\left (\sqrt {x e + d} \sqrt {g} e^{\frac {1}{2}} - \sqrt {{\left (x e + d\right )} g e - d g e + f e^{2}}\right )}^{2} a g^{\frac {3}{2}} e^{\frac {7}{2}} - 3 \, b f^{2} \sqrt {g} e^{\frac {11}{2}} + 2 \, a f g^{\frac {3}{2}} e^{\frac {11}{2}}\right )} e^{\left (-2\right )}}{3 \, {\left (d g e + {\left (\sqrt {x e + d} \sqrt {g} e^{\frac {1}{2}} - \sqrt {{\left (x e + d\right )} g e - d g e + f e^{2}}\right )}^{2} - f e^{2}\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*e^(-5/2)*log((sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^2)/sqrt(g) - 4/3*(4*c*d^
3*g^(5/2)*e^(5/2) + 6*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^2*c*d^2*g^(3/2)*e^
(3/2) + 6*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^4*c*d*sqrt(g)*e^(1/2) - 10*c*d
^2*f*g^(3/2)*e^(7/2) - b*d^2*g^(5/2)*e^(7/2) - 12*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e
+ f*e^2))^2*c*d*f*sqrt(g)*e^(5/2) - 3*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^4*
b*sqrt(g)*e^(3/2) + 6*c*d*f^2*sqrt(g)*e^(9/2) + 4*b*d*f*g^(3/2)*e^(9/2) - 2*a*d*g^(5/2)*e^(9/2) + 6*(sqrt(x*e
+ d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^2*b*f*sqrt(g)*e^(7/2) - 6*(sqrt(x*e + d)*sqrt(g)*e
^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^2*a*g^(3/2)*e^(7/2) - 3*b*f^2*sqrt(g)*e^(11/2) + 2*a*f*g^(3/2)*e
^(11/2))*e^(-2)/(d*g*e + (sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^2 - f*e^2)^3

________________________________________________________________________________________

maple [B]  time = 0.03, size = 773, normalized size = 4.83 \begin {gather*} \frac {\sqrt {g x +f}\, \left (3 c \,d^{2} e^{2} g^{2} x^{2} \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )-6 c d \,e^{3} f g \,x^{2} \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )+3 c \,e^{4} f^{2} x^{2} \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )+6 c \,d^{3} e \,g^{2} x \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )-12 c \,d^{2} e^{2} f g x \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )+6 c d \,e^{3} f^{2} x \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )+3 c \,d^{4} g^{2} \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )-6 c \,d^{3} e f g \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )+3 c \,d^{2} e^{2} f^{2} \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )+4 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, a \,e^{3} g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, b d \,e^{2} g x -6 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, b \,e^{3} f x -8 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, c \,d^{2} e g x +12 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, c d \,e^{2} f x +6 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, a d \,e^{2} g -2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, a \,e^{3} f -4 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, b d \,e^{2} f -6 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, c \,d^{3} g +10 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, c \,d^{2} e f \right )}{3 \sqrt {e g}\, \left (d g -e f \right )^{2} \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \left (e x +d \right )^{\frac {3}{2}} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)/(e*x+d)^(5/2)/(g*x+f)^(1/2),x)

[Out]

1/3*(g*x+f)^(1/2)*(3*ln(1/2*(2*e*g*x+d*g+e*f+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2))/(e*g)^(1/2))*x^2*c*d^2*e^2
*g^2-6*ln(1/2*(2*e*g*x+d*g+e*f+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2))/(e*g)^(1/2))*x^2*c*d*e^3*f*g+3*ln(1/2*(2
*e*g*x+d*g+e*f+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2))/(e*g)^(1/2))*x^2*c*e^4*f^2+6*ln(1/2*(2*e*g*x+d*g+e*f+2*(
(e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2))/(e*g)^(1/2))*x*c*d^3*e*g^2-12*ln(1/2*(2*e*g*x+d*g+e*f+2*((e*x+d)*(g*x+f))^
(1/2)*(e*g)^(1/2))/(e*g)^(1/2))*x*c*d^2*e^2*f*g+6*ln(1/2*(2*e*g*x+d*g+e*f+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2
))/(e*g)^(1/2))*x*c*d*e^3*f^2+3*ln(1/2*(2*e*g*x+d*g+e*f+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2))/(e*g)^(1/2))*c*
d^4*g^2-6*ln(1/2*(2*e*g*x+d*g+e*f+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2))/(e*g)^(1/2))*c*d^3*e*f*g+3*ln(1/2*(2*
e*g*x+d*g+e*f+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2))/(e*g)^(1/2))*c*d^2*e^2*f^2+4*x*a*e^3*g*((e*x+d)*(g*x+f))^
(1/2)*(e*g)^(1/2)+2*x*b*d*e^2*g*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)-6*x*b*e^3*f*((e*x+d)*(g*x+f))^(1/2)*(e*g)^
(1/2)-8*x*c*d^2*e*g*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)+12*x*c*d*e^2*f*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)+6*a
*d*e^2*g*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)-2*a*e^3*f*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)-4*b*d*e^2*f*((e*x+d
)*(g*x+f))^(1/2)*(e*g)^(1/2)-6*c*d^3*g*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)+10*c*d^2*e*f*((e*x+d)*(g*x+f))^(1/2
)*(e*g)^(1/2))/(e*g)^(1/2)/(d*g-e*f)^2/((e*x+d)*(g*x+f))^(1/2)/e^2/(e*x+d)^(3/2)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(e*x+d)^(5/2)/(g*x+f)^(1/2),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(g>0)', see `assume?` for more
details)Is g positive or negative?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {c\,x^2+b\,x+a}{\sqrt {f+g\,x}\,{\left (d+e\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)/((f + g*x)^(1/2)*(d + e*x)^(5/2)),x)

[Out]

int((a + b*x + c*x^2)/((f + g*x)^(1/2)*(d + e*x)^(5/2)), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)/(e*x+d)**(5/2)/(g*x+f)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________