3.840 \(\int \frac{a+b x+c x^2}{(d+e x)^{7/2} \sqrt{f+g x}} \, dx\)

Optimal. Leaf size=198 \[ \frac{2 \sqrt{f+g x} \left (2 e g (-4 a e g-b d g+5 b e f)-c \left (3 d^2 g^2-10 d e f g+15 e^2 f^2\right )\right )}{15 e^2 \sqrt{d+e x} (e f-d g)^3}-\frac{2 \sqrt{f+g x} \left (a+\frac{d (c d-b e)}{e^2}\right )}{5 (d+e x)^{5/2} (e f-d g)}+\frac{2 \sqrt{f+g x} (2 c d (5 e f-3 d g)-e (-4 a e g-b d g+5 b e f))}{15 e^2 (d+e x)^{3/2} (e f-d g)^2} \]

[Out]

(-2*(a + (d*(c*d - b*e))/e^2)*Sqrt[f + g*x])/(5*(e*f - d*g)*(d + e*x)^(5/2)) + (2*(2*c*d*(5*e*f - 3*d*g) - e*(
5*b*e*f - b*d*g - 4*a*e*g))*Sqrt[f + g*x])/(15*e^2*(e*f - d*g)^2*(d + e*x)^(3/2)) + (2*(2*e*g*(5*b*e*f - b*d*g
- 4*a*e*g) - c*(15*e^2*f^2 - 10*d*e*f*g + 3*d^2*g^2))*Sqrt[f + g*x])/(15*e^2*(e*f - d*g)^3*Sqrt[d + e*x])

________________________________________________________________________________________

Rubi [A]  time = 0.212091, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {949, 78, 37} \[ \frac{2 \sqrt{f+g x} \left (2 e g (-4 a e g-b d g+5 b e f)-c \left (3 d^2 g^2-10 d e f g+15 e^2 f^2\right )\right )}{15 e^2 \sqrt{d+e x} (e f-d g)^3}-\frac{2 \sqrt{f+g x} \left (a+\frac{d (c d-b e)}{e^2}\right )}{5 (d+e x)^{5/2} (e f-d g)}+\frac{2 \sqrt{f+g x} (2 c d (5 e f-3 d g)-e (-4 a e g-b d g+5 b e f))}{15 e^2 (d+e x)^{3/2} (e f-d g)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)/((d + e*x)^(7/2)*Sqrt[f + g*x]),x]

[Out]

(-2*(a + (d*(c*d - b*e))/e^2)*Sqrt[f + g*x])/(5*(e*f - d*g)*(d + e*x)^(5/2)) + (2*(2*c*d*(5*e*f - 3*d*g) - e*(
5*b*e*f - b*d*g - 4*a*e*g))*Sqrt[f + g*x])/(15*e^2*(e*f - d*g)^2*(d + e*x)^(3/2)) + (2*(2*e*g*(5*b*e*f - b*d*g
- 4*a*e*g) - c*(15*e^2*f^2 - 10*d*e*f*g + 3*d^2*g^2))*Sqrt[f + g*x])/(15*e^2*(e*f - d*g)^3*Sqrt[d + e*x])

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{a+b x+c x^2}{(d+e x)^{7/2} \sqrt{f+g x}} \, dx &=-\frac{2 \left (a+\frac{d (c d-b e)}{e^2}\right ) \sqrt{f+g x}}{5 (e f-d g) (d+e x)^{5/2}}-\frac{2 \int \frac{\frac{c d (5 e f-d g)-e (5 b e f-b d g-4 a e g)}{2 e^2}-\frac{5}{2} c \left (f-\frac{d g}{e}\right ) x}{(d+e x)^{5/2} \sqrt{f+g x}} \, dx}{5 (e f-d g)}\\ &=-\frac{2 \left (a+\frac{d (c d-b e)}{e^2}\right ) \sqrt{f+g x}}{5 (e f-d g) (d+e x)^{5/2}}+\frac{2 (2 c d (5 e f-3 d g)-e (5 b e f-b d g-4 a e g)) \sqrt{f+g x}}{15 e^2 (e f-d g)^2 (d+e x)^{3/2}}-\frac{\left (2 e g (5 b e f-b d g-4 a e g)-c \left (15 e^2 f^2-10 d e f g+3 d^2 g^2\right )\right ) \int \frac{1}{(d+e x)^{3/2} \sqrt{f+g x}} \, dx}{15 e^2 (e f-d g)^2}\\ &=-\frac{2 \left (a+\frac{d (c d-b e)}{e^2}\right ) \sqrt{f+g x}}{5 (e f-d g) (d+e x)^{5/2}}+\frac{2 (2 c d (5 e f-3 d g)-e (5 b e f-b d g-4 a e g)) \sqrt{f+g x}}{15 e^2 (e f-d g)^2 (d+e x)^{3/2}}+\frac{2 \left (2 e g (5 b e f-b d g-4 a e g)-c \left (15 e^2 f^2-10 d e f g+3 d^2 g^2\right )\right ) \sqrt{f+g x}}{15 e^2 (e f-d g)^3 \sqrt{d+e x}}\\ \end{align*}

Mathematica [A]  time = 0.239307, size = 178, normalized size = 0.9 \[ -\frac{2 \sqrt{f+g x} \left (a \left (15 d^2 g^2-10 d e g (f-2 g x)+e^2 \left (3 f^2-4 f g x+8 g^2 x^2\right )\right )+b \left (5 d^2 g (g x-2 f)+2 d e \left (f^2-13 f g x+g^2 x^2\right )+5 e^2 f x (f-2 g x)\right )+c \left (d^2 \left (8 f^2-4 f g x+3 g^2 x^2\right )+10 d e f x (2 f-g x)+15 e^2 f^2 x^2\right )\right )}{15 (d+e x)^{5/2} (e f-d g)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)/((d + e*x)^(7/2)*Sqrt[f + g*x]),x]

[Out]

(-2*Sqrt[f + g*x]*(b*(5*e^2*f*x*(f - 2*g*x) + 5*d^2*g*(-2*f + g*x) + 2*d*e*(f^2 - 13*f*g*x + g^2*x^2)) + c*(15
*e^2*f^2*x^2 + 10*d*e*f*x*(2*f - g*x) + d^2*(8*f^2 - 4*f*g*x + 3*g^2*x^2)) + a*(15*d^2*g^2 - 10*d*e*g*(f - 2*g
*x) + e^2*(3*f^2 - 4*f*g*x + 8*g^2*x^2))))/(15*(e*f - d*g)^3*(d + e*x)^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.057, size = 238, normalized size = 1.2 \begin{align*}{\frac{16\,a{e}^{2}{g}^{2}{x}^{2}+4\,bde{g}^{2}{x}^{2}-20\,b{e}^{2}fg{x}^{2}+6\,c{d}^{2}{g}^{2}{x}^{2}-20\,cdefg{x}^{2}+30\,c{e}^{2}{f}^{2}{x}^{2}+40\,ade{g}^{2}x-8\,a{e}^{2}fgx+10\,b{d}^{2}{g}^{2}x-52\,bdefgx+10\,b{e}^{2}{f}^{2}x-8\,c{d}^{2}fgx+40\,cde{f}^{2}x+30\,a{d}^{2}{g}^{2}-20\,adefg+6\,a{e}^{2}{f}^{2}-20\,b{d}^{2}fg+4\,bde{f}^{2}+16\,c{d}^{2}{f}^{2}}{15\,{g}^{3}{d}^{3}-45\,{d}^{2}ef{g}^{2}+45\,d{e}^{2}{f}^{2}g-15\,{e}^{3}{f}^{3}}\sqrt{gx+f} \left ( ex+d \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)/(e*x+d)^(7/2)/(g*x+f)^(1/2),x)

[Out]

2/15*(g*x+f)^(1/2)*(8*a*e^2*g^2*x^2+2*b*d*e*g^2*x^2-10*b*e^2*f*g*x^2+3*c*d^2*g^2*x^2-10*c*d*e*f*g*x^2+15*c*e^2
*f^2*x^2+20*a*d*e*g^2*x-4*a*e^2*f*g*x+5*b*d^2*g^2*x-26*b*d*e*f*g*x+5*b*e^2*f^2*x-4*c*d^2*f*g*x+20*c*d*e*f^2*x+
15*a*d^2*g^2-10*a*d*e*f*g+3*a*e^2*f^2-10*b*d^2*f*g+2*b*d*e*f^2+8*c*d^2*f^2)/(e*x+d)^(5/2)/(d^3*g^3-3*d^2*e*f*g
^2+3*d*e^2*f^2*g-e^3*f^3)

________________________________________________________________________________________

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)^(7/2)/(g*x+f)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 76.2546, size = 740, normalized size = 3.74 \begin{align*} -\frac{2 \,{\left (15 \, a d^{2} g^{2} +{\left (8 \, c d^{2} + 2 \, b d e + 3 \, a e^{2}\right )} f^{2} - 10 \,{\left (b d^{2} + a d e\right )} f g +{\left (15 \, c e^{2} f^{2} - 10 \,{\left (c d e + b e^{2}\right )} f g +{\left (3 \, c d^{2} + 2 \, b d e + 8 \, a e^{2}\right )} g^{2}\right )} x^{2} +{\left (5 \,{\left (4 \, c d e + b e^{2}\right )} f^{2} - 2 \,{\left (2 \, c d^{2} + 13 \, b d e + 2 \, a e^{2}\right )} f g + 5 \,{\left (b d^{2} + 4 \, a d e\right )} g^{2}\right )} x\right )} \sqrt{e x + d} \sqrt{g x + f}}{15 \,{\left (d^{3} e^{3} f^{3} - 3 \, d^{4} e^{2} f^{2} g + 3 \, d^{5} e f g^{2} - d^{6} g^{3} +{\left (e^{6} f^{3} - 3 \, d e^{5} f^{2} g + 3 \, d^{2} e^{4} f g^{2} - d^{3} e^{3} g^{3}\right )} x^{3} + 3 \,{\left (d e^{5} f^{3} - 3 \, d^{2} e^{4} f^{2} g + 3 \, d^{3} e^{3} f g^{2} - d^{4} e^{2} g^{3}\right )} x^{2} + 3 \,{\left (d^{2} e^{4} f^{3} - 3 \, d^{3} e^{3} f^{2} g + 3 \, d^{4} e^{2} f g^{2} - d^{5} e g^{3}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(e*x+d)^(7/2)/(g*x+f)^(1/2),x, algorithm="fricas")

[Out]

-2/15*(15*a*d^2*g^2 + (8*c*d^2 + 2*b*d*e + 3*a*e^2)*f^2 - 10*(b*d^2 + a*d*e)*f*g + (15*c*e^2*f^2 - 10*(c*d*e +
b*e^2)*f*g + (3*c*d^2 + 2*b*d*e + 8*a*e^2)*g^2)*x^2 + (5*(4*c*d*e + b*e^2)*f^2 - 2*(2*c*d^2 + 13*b*d*e + 2*a*
e^2)*f*g + 5*(b*d^2 + 4*a*d*e)*g^2)*x)*sqrt(e*x + d)*sqrt(g*x + f)/(d^3*e^3*f^3 - 3*d^4*e^2*f^2*g + 3*d^5*e*f*
g^2 - d^6*g^3 + (e^6*f^3 - 3*d*e^5*f^2*g + 3*d^2*e^4*f*g^2 - d^3*e^3*g^3)*x^3 + 3*(d*e^5*f^3 - 3*d^2*e^4*f^2*g
+ 3*d^3*e^3*f*g^2 - d^4*e^2*g^3)*x^2 + 3*(d^2*e^4*f^3 - 3*d^3*e^3*f^2*g + 3*d^4*e^2*f*g^2 - d^5*e*g^3)*x)

________________________________________________________________________________________

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)/(e*x+d)**(7/2)/(g*x+f)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.58423, size = 1458, normalized size = 7.36 \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)^(7/2)/(g*x+f)^(1/2),x, algorithm="giac")

[Out]

4/15*(3*c*d^4*g^(9/2)*e^(9/2) + 30*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^4*c*d
^2*g^(5/2)*e^(5/2) + 15*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^8*c*sqrt(g)*e^(1
/2) - 16*c*d^3*f*g^(7/2)*e^(11/2) + 2*b*d^3*g^(9/2)*e^(11/2) - 20*(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*f*g^(5/2)*e^(9/2) + 10*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e -
d*g*e + f*e^2))^2*b*d^2*g^(7/2)*e^(9/2) - 40*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*
e^2))^4*c*d*f*g^(3/2)*e^(7/2) - 10*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^4*b*d
*g^(5/2)*e^(7/2) - 60*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^6*c*f*sqrt(g)*e^(5
/2) + 30*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^6*b*g^(3/2)*e^(5/2) + 38*c*d^2*
f^2*g^(5/2)*e^(13/2) - 14*b*d^2*f*g^(7/2)*e^(13/2) + 8*a*d^2*g^(9/2)*e^(13/2) + 80*(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^2*g^(3/2)*e^(11/2) - 60*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sq
rt((x*e + d)*g*e - d*g*e + f*e^2))^2*b*d*f*g^(5/2)*e^(11/2) + 40*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e +
d)*g*e - d*g*e + f*e^2))^2*a*d*g^(7/2)*e^(11/2) + 90*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g
*e + f*e^2))^4*c*f^2*sqrt(g)*e^(9/2) - 70*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2)
)^4*b*f*g^(3/2)*e^(9/2) + 80*(sqrt(x*e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^4*a*g^(5/2)
*e^(9/2) - 40*c*d*f^3*g^(3/2)*e^(15/2) + 22*b*d*f^2*g^(5/2)*e^(15/2) - 16*a*d*f*g^(7/2)*e^(15/2) - 60*(sqrt(x*
e + d)*sqrt(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^2*c*f^3*sqrt(g)*e^(13/2) + 50*(sqrt(x*e + d)*sqr
t(g)*e^(1/2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^2*b*f^2*g^(3/2)*e^(13/2) - 40*(sqrt(x*e + d)*sqrt(g)*e^(1/
2) - sqrt((x*e + d)*g*e - d*g*e + f*e^2))^2*a*f*g^(5/2)*e^(13/2) + 15*c*f^4*sqrt(g)*e^(17/2) - 10*b*f^3*g^(3/2
)*e^(17/2) + 8*a*f^2*g^(5/2)*e^(17/2))*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)^5