3.839 \(\int \frac{a+b x+c x^2}{(d+e x)^{5/2} \sqrt{f+g x}} \, dx\)
Optimal. Leaf size=160 \[ \frac{2 \sqrt{f+g x} \left (c \left (6 d e f-4 d^2 g\right )-e (-2 a e g-b d g+3 b e f)\right )}{3 e^2 \sqrt{d+e x} (e f-d g)^2}-\frac{2 \sqrt{f+g x} \left (a+\frac{d (c d-b e)}{e^2}\right )}{3 (d+e x)^{3/2} (e f-d g)}+\frac{2 c \tanh ^{-1}\left (\frac{\sqrt{g} \sqrt{d+e x}}{\sqrt{e} \sqrt{f+g x}}\right )}{e^{5/2} \sqrt{g}} \]
[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])
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Rubi [A] time = 0.180485, antiderivative size = 160, normalized size of antiderivative = 1.,
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} \[ \frac{2 \sqrt{f+g x} \left (c \left (6 d e f-4 d^2 g\right )-e (-2 a e g-b d g+3 b e f)\right )}{3 e^2 \sqrt{d+e x} (e f-d g)^2}-\frac{2 \sqrt{f+g x} \left (a+\frac{d (c d-b e)}{e^2}\right )}{3 (d+e x)^{3/2} (e f-d g)}+\frac{2 c \tanh ^{-1}\left (\frac{\sqrt{g} \sqrt{d+e x}}{\sqrt{e} \sqrt{f+g x}}\right )}{e^{5/2} \sqrt{g}} \]
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 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]
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 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])
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 = 0.2441, size = 173, normalized size = 1.08 \[ \frac{2 \sqrt{f+g x} \left (2 g (d+e x) \left (g (a g-b f)+c f^2\right )-(e f-d g) \left (g (a g-b f)+c f^2\right )+(f+g x) (2 c f-b g) (e f-d g)-\frac{c (e f-d g)^3 \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};\frac{g (d+e x)}{d g-e f}\right )}{e^2 \sqrt{\frac{e (f+g x)}{e f-d g}}}\right )}{3 g^2 (d+e x)^{3/2} (e f-d g)^2} \]
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))
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Maple [B] time = 0.327, size = 773, normalized size = 4.8 \begin{align*} \text{result too large to display} \end{align*}
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+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*x^2*c*d^2*e^2
*g^2-6*ln(1/2*(2*e*g*x+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*x^2*c*d*e^3*f*g+3*ln(1/2*(2
*e*g*x+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*x^2*c*e^4*f^2+6*ln(1/2*(2*e*g*x+2*((g*x+f)*
(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*x*c*d^3*e*g^2-12*ln(1/2*(2*e*g*x+2*((g*x+f)*(e*x+d))^(1/2)*(e
*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*x*c*d^2*e^2*f*g+6*ln(1/2*(2*e*g*x+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*
f)/(e*g)^(1/2))*x*c*d*e^3*f^2+3*ln(1/2*(2*e*g*x+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*c*
d^4*g^2-6*ln(1/2*(2*e*g*x+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*c*d^3*e*f*g+3*ln(1/2*(2*
e*g*x+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*c*d^2*e^2*f^2+4*x*a*e^3*g*((g*x+f)*(e*x+d))^
(1/2)*(e*g)^(1/2)+2*x*b*d*e^2*g*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)-6*x*b*e^3*f*((g*x+f)*(e*x+d))^(1/2)*(e*g)^
(1/2)-8*x*c*d^2*e*g*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+12*x*c*d*e^2*f*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+6*a
*d*e^2*g*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)-2*a*e^3*f*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)-4*b*d*e^2*f*((g*x+f
)*(e*x+d))^(1/2)*(e*g)^(1/2)-6*c*d^3*g*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+10*c*d^2*e*f*((g*x+f)*(e*x+d))^(1/2
)*(e*g)^(1/2))/(e*g)^(1/2)/(d*g-e*f)^2/((g*x+f)*(e*x+d))^(1/2)/e^2/(e*x+d)^(3/2)
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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)/(e*x+d)^(5/2)/(g*x+f)^(1/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 26.9215, size = 1671, normalized size = 10.44 \begin{align*} \text{result too large to display} \end{align*}
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)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b x + c x^{2}}{\left (d + e x\right )^{\frac{5}{2}} \sqrt{f + g x}}\, dx \end{align*}
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]
Integral((a + b*x + c*x**2)/((d + e*x)**(5/2)*sqrt(f + g*x)), x)
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Giac [B] time = 1.4474, size = 680, normalized size = 4.25 \begin{align*} -\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{align*}
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