3.9.40 \(\int \frac {a+b x+c x^2}{(d+e x)^{7/2} \sqrt {f+g x}} \, dx\) [840]

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

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 198, normalized size of antiderivative = 1.00, 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 = {963, 79, 37} \begin {gather*} \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} \end {gather*}

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 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]

Rule 79

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] || L
tQ[p, n]))))

Rule 963

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)^{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*}

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Mathematica [A]
time = 0.18, size = 177, normalized size = 0.89 \begin {gather*} -\frac {2 \sqrt {f+g x} \left (15 c f^2-15 b f g+15 a g^2-\frac {10 c d f (f+g x)}{d+e x}+\frac {5 b e f (f+g x)}{d+e x}+\frac {5 b d g (f+g x)}{d+e x}-\frac {10 a e g (f+g x)}{d+e x}+\frac {3 c d^2 (f+g x)^2}{(d+e x)^2}-\frac {3 b d e (f+g x)^2}{(d+e x)^2}+\frac {3 a e^2 (f+g x)^2}{(d+e x)^2}\right )}{15 (e f-d g)^3 \sqrt {d+e x}} \end {gather*}

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

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Maple [A]
time = 0.11, size = 210, normalized size = 1.06

method	result	size
default	\(\frac {2 \sqrt {g x +f}\, \left (8 a \,e^{2} g^{2} x^{2}+2 b d e \,g^{2} x^{2}-10 b \,e^{2} f g \,x^{2}+3 c \,d^{2} g^{2} x^{2}-10 c d e f g \,x^{2}+15 c \,e^{2} f^{2} x^{2}+20 a d e \,g^{2} x -4 a \,e^{2} f g x +5 b \,d^{2} g^{2} x -26 b d e f g x +5 b \,e^{2} f^{2} x -4 c \,d^{2} f g x +20 c d e \,f^{2} x +15 a \,d^{2} g^{2}-10 a d e f g +3 a \,e^{2} f^{2}-10 b \,d^{2} f g +2 b d e \,f^{2}+8 c \,d^{2} f^{2}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} \left (d g -e f \right )^{3}}\)	\(210\)
gosper	\(\frac {2 \sqrt {g x +f}\, \left (8 a \,e^{2} g^{2} x^{2}+2 b d e \,g^{2} x^{2}-10 b \,e^{2} f g \,x^{2}+3 c \,d^{2} g^{2} x^{2}-10 c d e f g \,x^{2}+15 c \,e^{2} f^{2} x^{2}+20 a d e \,g^{2} x -4 a \,e^{2} f g x +5 b \,d^{2} g^{2} x -26 b d e f g x +5 b \,e^{2} f^{2} x -4 c \,d^{2} f g x +20 c d e \,f^{2} x +15 a \,d^{2} g^{2}-10 a d e f g +3 a \,e^{2} f^{2}-10 b \,d^{2} f g +2 b d e \,f^{2}+8 c \,d^{2} f^{2}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} \left (g^{3} d^{3}-3 d^{2} e f \,g^{2}+3 d \,e^{2} f^{2} g -e^{3} f^{3}\right )}\)	\(238\)

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,method=_RETURNVERBOSE)

[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*g-e*f)^3

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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)^(7/2)/(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(d*g-%e*f>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 19.89, size = 360, normalized size = 1.82 \begin {gather*} \frac {2 \, {\left (3 \, c d^{2} g^{2} x^{2} + 8 \, c d^{2} f^{2} - 10 \, b d^{2} f g + 15 \, a d^{2} g^{2} - {\left (4 \, c d^{2} f g - 5 \, b d^{2} g^{2}\right )} x + {\left (3 \, a f^{2} + {\left (15 \, c f^{2} - 10 \, b f g + 8 \, a g^{2}\right )} x^{2} + {\left (5 \, b f^{2} - 4 \, a f g\right )} x\right )} e^{2} + 2 \, {\left (b d f^{2} - 5 \, a d f g - {\left (5 \, c d f g - b d g^{2}\right )} x^{2} + {\left (10 \, c d f^{2} - 13 \, b d f g + 10 \, a d g^{2}\right )} x\right )} e\right )} \sqrt {g x + f} \sqrt {x e + d}}{15 \, {\left (d^{6} g^{3} - f^{3} x^{3} e^{6} + 3 \, {\left (d f^{2} g x^{3} - d f^{3} x^{2}\right )} e^{5} - 3 \, {\left (d^{2} f g^{2} x^{3} - 3 \, d^{2} f^{2} g x^{2} + d^{2} f^{3} x\right )} e^{4} + {\left (d^{3} g^{3} x^{3} - 9 \, d^{3} f g^{2} x^{2} + 9 \, d^{3} f^{2} g x - d^{3} f^{3}\right )} e^{3} + 3 \, {\left (d^{4} g^{3} x^{2} - 3 \, d^{4} f g^{2} x + d^{4} f^{2} g\right )} e^{2} + 3 \, {\left (d^{5} g^{3} x - d^{5} f g^{2}\right )} e\right )}} \end {gather*}

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*(3*c*d^2*g^2*x^2 + 8*c*d^2*f^2 - 10*b*d^2*f*g + 15*a*d^2*g^2 - (4*c*d^2*f*g - 5*b*d^2*g^2)*x + (3*a*f^2 +
(15*c*f^2 - 10*b*f*g + 8*a*g^2)*x^2 + (5*b*f^2 - 4*a*f*g)*x)*e^2 + 2*(b*d*f^2 - 5*a*d*f*g - (5*c*d*f*g - b*d*
g^2)*x^2 + (10*c*d*f^2 - 13*b*d*f*g + 10*a*d*g^2)*x)*e)*sqrt(g*x + f)*sqrt(x*e + d)/(d^6*g^3 - f^3*x^3*e^6 + 3
*(d*f^2*g*x^3 - d*f^3*x^2)*e^5 - 3*(d^2*f*g^2*x^3 - 3*d^2*f^2*g*x^2 + d^2*f^3*x)*e^4 + (d^3*g^3*x^3 - 9*d^3*f*
g^2*x^2 + 9*d^3*f^2*g*x - d^3*f^3)*e^3 + 3*(d^4*g^3*x^2 - 3*d^4*f*g^2*x + d^4*f^2*g)*e^2 + 3*(d^5*g^3*x - d^5*
f*g^2)*e)

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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 {7}{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)**(7/2)/(g*x+f)**(1/2),x)

[Out]

Integral((a + b*x + c*x**2)/((d + e*x)**(7/2)*sqrt(f + g*x)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 452 vs. \(2 (188) = 376\).
time = 6.15, size = 452, normalized size = 2.28 \begin {gather*} \frac {2 \, {\left ({\left (g x + f\right )} {\left (\frac {{\left (3 \, c d^{2} g^{8} e^{2} - 10 \, c d f g^{7} e^{3} + 2 \, b d g^{8} e^{3} + 15 \, c f^{2} g^{6} e^{4} - 10 \, b f g^{7} e^{4} + 8 \, a g^{8} e^{4}\right )} {\left (g x + f\right )}}{d^{3} g^{5} {\left | g \right |} e^{2} - 3 \, d^{2} f g^{4} {\left | g \right |} e^{3} + 3 \, d f^{2} g^{3} {\left | g \right |} e^{4} - f^{3} g^{2} {\left | g \right |} e^{5}} - \frac {5 \, {\left (2 \, c d^{2} f g^{8} e^{2} - b d^{2} g^{9} e^{2} - 8 \, c d f^{2} g^{7} e^{3} + 6 \, b d f g^{8} e^{3} - 4 \, a d g^{9} e^{3} + 6 \, c f^{3} g^{6} e^{4} - 5 \, b f^{2} g^{7} e^{4} + 4 \, a f g^{8} e^{4}\right )}}{d^{3} g^{5} {\left | g \right |} e^{2} - 3 \, d^{2} f g^{4} {\left | g \right |} e^{3} + 3 \, d f^{2} g^{3} {\left | g \right |} e^{4} - f^{3} g^{2} {\left | g \right |} e^{5}}\right )} + \frac {15 \, {\left (c d^{2} f^{2} g^{8} e^{2} - b d^{2} f g^{9} e^{2} + a d^{2} g^{10} e^{2} - 2 \, c d f^{3} g^{7} e^{3} + 2 \, b d f^{2} g^{8} e^{3} - 2 \, a d f g^{9} e^{3} + c f^{4} g^{6} e^{4} - b f^{3} g^{7} e^{4} + a f^{2} g^{8} e^{4}\right )}}{d^{3} g^{5} {\left | g \right |} e^{2} - 3 \, d^{2} f g^{4} {\left | g \right |} e^{3} + 3 \, d f^{2} g^{3} {\left | g \right |} e^{4} - f^{3} g^{2} {\left | g \right |} e^{5}}\right )} \sqrt {g x + f}}{15 \, {\left (d g^{2} + {\left (g x + f\right )} g e - f g e\right )}^{\frac {5}{2}}} \end {gather*}

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]

2/15*((g*x + f)*((3*c*d^2*g^8*e^2 - 10*c*d*f*g^7*e^3 + 2*b*d*g^8*e^3 + 15*c*f^2*g^6*e^4 - 10*b*f*g^7*e^4 + 8*a
*g^8*e^4)*(g*x + f)/(d^3*g^5*abs(g)*e^2 - 3*d^2*f*g^4*abs(g)*e^3 + 3*d*f^2*g^3*abs(g)*e^4 - f^3*g^2*abs(g)*e^5
) - 5*(2*c*d^2*f*g^8*e^2 - b*d^2*g^9*e^2 - 8*c*d*f^2*g^7*e^3 + 6*b*d*f*g^8*e^3 - 4*a*d*g^9*e^3 + 6*c*f^3*g^6*e
^4 - 5*b*f^2*g^7*e^4 + 4*a*f*g^8*e^4)/(d^3*g^5*abs(g)*e^2 - 3*d^2*f*g^4*abs(g)*e^3 + 3*d*f^2*g^3*abs(g)*e^4 -
f^3*g^2*abs(g)*e^5)) + 15*(c*d^2*f^2*g^8*e^2 - b*d^2*f*g^9*e^2 + a*d^2*g^10*e^2 - 2*c*d*f^3*g^7*e^3 + 2*b*d*f^
2*g^8*e^3 - 2*a*d*f*g^9*e^3 + c*f^4*g^6*e^4 - b*f^3*g^7*e^4 + a*f^2*g^8*e^4)/(d^3*g^5*abs(g)*e^2 - 3*d^2*f*g^4
*abs(g)*e^3 + 3*d*f^2*g^3*abs(g)*e^4 - f^3*g^2*abs(g)*e^5))*sqrt(g*x + f)/(d*g^2 + (g*x + f)*g*e - f*g*e)^(5/2
)

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Mupad [B]
time = 4.30, size = 260, normalized size = 1.31 \begin {gather*} \frac {\sqrt {f+g\,x}\,\left (\frac {16\,c\,d^2\,f^2-20\,b\,d^2\,f\,g+30\,a\,d^2\,g^2+4\,b\,d\,e\,f^2-20\,a\,d\,e\,f\,g+6\,a\,e^2\,f^2}{15\,e^2\,{\left (d\,g-e\,f\right )}^3}+\frac {x\,\left (-8\,c\,d^2\,f\,g+10\,b\,d^2\,g^2+40\,c\,d\,e\,f^2-52\,b\,d\,e\,f\,g+40\,a\,d\,e\,g^2+10\,b\,e^2\,f^2-8\,a\,e^2\,f\,g\right )}{15\,e^2\,{\left (d\,g-e\,f\right )}^3}+\frac {x^2\,\left (6\,c\,d^2\,g^2-20\,c\,d\,e\,f\,g+4\,b\,d\,e\,g^2+30\,c\,e^2\,f^2-20\,b\,e^2\,f\,g+16\,a\,e^2\,g^2\right )}{15\,e^2\,{\left (d\,g-e\,f\right )}^3}\right )}{x^2\,\sqrt {d+e\,x}+\frac {d^2\,\sqrt {d+e\,x}}{e^2}+\frac {2\,d\,x\,\sqrt {d+e\,x}}{e}} \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)^(7/2)),x)

[Out]

((f + g*x)^(1/2)*((30*a*d^2*g^2 + 6*a*e^2*f^2 + 16*c*d^2*f^2 + 4*b*d*e*f^2 - 20*b*d^2*f*g - 20*a*d*e*f*g)/(15*
e^2*(d*g - e*f)^3) + (x*(10*b*d^2*g^2 + 10*b*e^2*f^2 + 40*a*d*e*g^2 + 40*c*d*e*f^2 - 8*a*e^2*f*g - 8*c*d^2*f*g
- 52*b*d*e*f*g))/(15*e^2*(d*g - e*f)^3) + (x^2*(16*a*e^2*g^2 + 6*c*d^2*g^2 + 30*c*e^2*f^2 + 4*b*d*e*g^2 - 20*
b*e^2*f*g - 20*c*d*e*f*g))/(15*e^2*(d*g - e*f)^3)))/(x^2*(d + e*x)^(1/2) + (d^2*(d + e*x)^(1/2))/e^2 + (2*d*x*
(d + e*x)^(1/2))/e)

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