3.6.81 \(\int \frac {a+b x+c x^2}{(d+e x)^{3/2} \sqrt {f+g x}} \, dx\)

Optimal. Leaf size=129 \[ -\frac {2 \sqrt {f+g x} \left (a+\frac {d (c d-b e)}{e^2}\right )}{\sqrt {d+e x} (e f-d g)}-\frac {(-2 b e g+3 c d g+c e f) \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{e^{5/2} g^{3/2}}+\frac {c \sqrt {d+e x} \sqrt {f+g x}}{e^2 g} \]

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Rubi [A]  time = 0.13, antiderivative size = 129, 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, 80, 63, 217, 206} \begin {gather*} -\frac {2 \sqrt {f+g x} \left (a+\frac {d (c d-b e)}{e^2}\right )}{\sqrt {d+e x} (e f-d g)}-\frac {(-2 b e g+3 c d g+c e f) \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{e^{5/2} g^{3/2}}+\frac {c \sqrt {d+e x} \sqrt {f+g x}}{e^2 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
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 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)^{3/2} \sqrt {f+g x}} \, dx &=-\frac {2 \left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{(e f-d g) \sqrt {d+e x}}-\frac {2 \int \frac {\frac {(c d-b e) (e f-d g)}{2 e^2}-\frac {c (e f-d g) x}{2 e}}{\sqrt {d+e x} \sqrt {f+g x}} \, dx}{e f-d g}\\ &=-\frac {2 \left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{(e f-d g) \sqrt {d+e x}}+\frac {c \sqrt {d+e x} \sqrt {f+g x}}{e^2 g}-\frac {(c e f+3 c d g-2 b e g) \int \frac {1}{\sqrt {d+e x} \sqrt {f+g x}} \, dx}{2 e^2 g}\\ &=-\frac {2 \left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{(e f-d g) \sqrt {d+e x}}+\frac {c \sqrt {d+e x} \sqrt {f+g x}}{e^2 g}-\frac {(c e f+3 c d g-2 b e g) \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 g}\\ &=-\frac {2 \left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{(e f-d g) \sqrt {d+e x}}+\frac {c \sqrt {d+e x} \sqrt {f+g x}}{e^2 g}-\frac {(c e f+3 c d g-2 b e g) \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 g}\\ &=-\frac {2 \left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{(e f-d g) \sqrt {d+e x}}+\frac {c \sqrt {d+e x} \sqrt {f+g x}}{e^2 g}-\frac {(c e f+3 c d g-2 b e g) \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{e^{5/2} g^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.59, size = 222, normalized size = 1.72 \begin {gather*} -\frac {2 \sqrt {f+g x} \left (e \sqrt {e f-d g} \sqrt {\frac {e (f+g x)}{e f-d g}} \left (g^2 (a e-b d)+c f (2 d g-e f)\right )+e \sqrt {g} \sqrt {d+e x} (2 c f-b g) (e f-d g) \sinh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e f-d g}}\right )+c (e f-d g)^{5/2} \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};\frac {g (d+e x)}{d g-e f}\right )\right )}{e^2 g^2 \sqrt {d+e x} (e f-d g)^{3/2} \sqrt {\frac {e (f+g x)}{e f-d g}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[f + g*x]*(e*Sqrt[e*f - d*g]*((-(b*d) + a*e)*g^2 + c*f*(-(e*f) + 2*d*g))*Sqrt[(e*(f + g*x))/(e*f - d*g
)] + e*Sqrt[g]*(2*c*f - b*g)*(e*f - d*g)*Sqrt[d + e*x]*ArcSinh[(Sqrt[g]*Sqrt[d + e*x])/Sqrt[e*f - d*g]] + c*(e
*f - d*g)^(5/2)*Hypergeometric2F1[-3/2, -1/2, 1/2, (g*(d + e*x))/(-(e*f) + d*g)]))/(e^2*g^2*(e*f - d*g)^(3/2)*
Sqrt[d + e*x]*Sqrt[(e*(f + g*x))/(e*f - d*g)])

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IntegrateAlgebraic [A]  time = 0.34, size = 216, normalized size = 1.67 \begin {gather*} \frac {(2 b e g-3 c d g-c e f) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {g} \sqrt {d+e x}}\right )}{e^{5/2} g^{3/2}}-\frac {\sqrt {f+g x} \left (\frac {2 a e^3 g (f+g x)}{d+e x}-2 a e^2 g^2-\frac {2 b d e^2 g (f+g x)}{d+e x}+2 b d e g^2+\frac {2 c d^2 e g (f+g x)}{d+e x}-3 c d^2 g^2+2 c d e f g-c e^2 f^2\right )}{e^2 g \sqrt {d+e x} (e f-d g) \left (\frac {e (f+g x)}{d+e x}-g\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.44, size = 588, normalized size = 4.56 \begin {gather*} \left [-\frac {{\left (c d e^{2} f^{2} + 2 \, {\left (c d^{2} e - b d e^{2}\right )} f g - {\left (3 \, c d^{3} - 2 \, b d^{2} e\right )} g^{2} + {\left (c e^{3} f^{2} + 2 \, {\left (c d e^{2} - b e^{3}\right )} f g - {\left (3 \, c d^{2} e - 2 \, b d e^{2}\right )} 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 (c d e^{2} f g - {\left (3 \, c d^{2} e - 2 \, b d e^{2} + 2 \, a e^{3}\right )} g^{2} + {\left (c e^{3} f g - c d e^{2} g^{2}\right )} x\right )} \sqrt {e x + d} \sqrt {g x + f}}{4 \, {\left (d e^{4} f g^{2} - d^{2} e^{3} g^{3} + {\left (e^{5} f g^{2} - d e^{4} g^{3}\right )} x\right )}}, \frac {{\left (c d e^{2} f^{2} + 2 \, {\left (c d^{2} e - b d e^{2}\right )} f g - {\left (3 \, c d^{3} - 2 \, b d^{2} e\right )} g^{2} + {\left (c e^{3} f^{2} + 2 \, {\left (c d e^{2} - b e^{3}\right )} f g - {\left (3 \, c d^{2} e - 2 \, b d e^{2}\right )} 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 (c d e^{2} f g - {\left (3 \, c d^{2} e - 2 \, b d e^{2} + 2 \, a e^{3}\right )} g^{2} + {\left (c e^{3} f g - c d e^{2} g^{2}\right )} x\right )} \sqrt {e x + d} \sqrt {g x + f}}{2 \, {\left (d e^{4} f g^{2} - d^{2} e^{3} g^{3} + {\left (e^{5} f g^{2} - d 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)^(3/2)/(g*x+f)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.39, size = 201, normalized size = 1.56 \begin {gather*} \frac {\sqrt {{\left (x e + d\right )} g e - d g e + f e^{2}} \sqrt {x e + d} c e^{\left (-3\right )}}{g} + \frac {4 \, {\left (c d^{2} \sqrt {g} e^{\frac {1}{2}} - b d \sqrt {g} e^{\frac {3}{2}} + a \sqrt {g} e^{\frac {5}{2}}\right )} e^{\left (-2\right )}}{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}} + \frac {{\left (3 \, c d g^{\frac {3}{2}} e^{\frac {1}{2}} + c f \sqrt {g} e^{\frac {3}{2}} - 2 \, b g^{\frac {3}{2}} e^{\frac {3}{2}}\right )} e^{\left (-3\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 )}{2 \, g^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt((x*e + d)*g*e - d*g*e + f*e^2)*sqrt(x*e + d)*c*e^(-3)/g + 4*(c*d^2*sqrt(g)*e^(1/2) - b*d*sqrt(g)*e^(3/2)
+ a*sqrt(g)*e^(5/2))*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) + 1/2*(3*c*d*g^(3/2)*e^(1/2) + c*f*sqrt(g)*e^(3/2) - 2*b*g^(3/2)*e^(3/2))*e^(-3)*log((sqrt(x*e + d)*sq
rt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^2)/g^2

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maple [B]  time = 0.03, size = 697, normalized size = 5.40 \begin {gather*} \frac {\sqrt {g x +f}\, \left (2 b d \,e^{2} 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 )-2 b \,e^{3} 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 )-3 c \,d^{2} 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 )+2 c d \,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 )+c \,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 )+2 b \,d^{2} e \,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 )-2 b d \,e^{2} 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^{3} 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 )+2 c \,d^{2} 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 )+c d \,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 )+2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, c d e g x -2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, c \,e^{2} f x +4 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, a \,e^{2} g -4 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, b d e g +6 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, c \,d^{2} g -2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, c d e f \right )}{2 \sqrt {e g}\, \left (d g -e f \right ) \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e x +d}\, e^{2} g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(3/2)/(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?

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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 )}^{3/2}} \,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)^(3/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b x + c x^{2}}{\left (d + e x\right )^{\frac {3}{2}} \sqrt {f + g x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x + c*x**2)/((d + e*x)**(3/2)*sqrt(f + g*x)), x)

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