3.7.67 \(\int \frac {(d+e x)^{3/2} (f+g x)}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\) [667]

Optimal. Leaf size=150 \[ -\frac {2 (c d f-a e g) (d+e x)^{3/2}}{c d \left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {2 \left (2 a e^2 g-c d (e f+d g)\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^2 d^2 \left (c d^2-a e^2\right ) \sqrt {d+e x}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {802, 662} \begin {gather*} -\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (2 a e^2 g-c d (d g+e f)\right )}{c^2 d^2 \sqrt {d+e x} \left (c d^2-a e^2\right )}-\frac {2 (d+e x)^{3/2} (c d f-a e g)}{c d \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rule 802

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

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 51, normalized size = 0.34 \begin {gather*} \frac {2 \sqrt {d+e x} (2 a e g+c d (-f+g x))}{c^2 d^2 \sqrt {(a e+c d x) (d+e x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.14, size = 58, normalized size = 0.39

method	result	size
default	\(\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (c d g x +2 a e g -c d f \right )}{\sqrt {e x +d}\, \left (c d x +a e \right ) c^{2} d^{2}}\)	\(58\)
gosper	\(\frac {2 \left (c d x +a e \right ) \left (c d g x +2 a e g -c d f \right ) \left (e x +d \right )^{\frac {3}{2}}}{c^{2} d^{2} \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)*(g*x+f)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.31, size = 51, normalized size = 0.34 \begin {gather*} -\frac {2 \, f}{\sqrt {c d x + a e} c d} + \frac {2 \, {\left (c d x + 2 \, a e\right )} g}{\sqrt {c d x + a e} c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*f/(sqrt(c*d*x + a*e)*c*d) + 2*(c*d*x + 2*a*e)*g/(sqrt(c*d*x + a*e)*c^2*d^2)

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Fricas [A]
time = 1.88, size = 98, normalized size = 0.65 \begin {gather*} \frac {2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (c d g x - c d f + 2 \, a g e\right )} \sqrt {x e + d}}{c^{3} d^{4} x + a c^{2} d^{2} x e^{2} + {\left (c^{3} d^{3} x^{2} + a c^{2} d^{3}\right )} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.53, size = 127, normalized size = 0.85 \begin {gather*} \frac {2 \, \sqrt {{\left (x e + d\right )} c d e - c d^{2} e + a e^{3}} g e^{\left (-1\right )}}{c^{2} d^{2}} + \frac {2 \, {\left (c d^{2} g + c d f e - 2 \, a g e^{2}\right )}}{\sqrt {-c d^{2} e + a e^{3}} c^{2} d^{2}} - \frac {2 \, {\left (c d f e - a g e^{2}\right )}}{\sqrt {{\left (x e + d\right )} c d e - c d^{2} e + a e^{3}} c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.37, size = 118, normalized size = 0.79 \begin {gather*} \frac {\left (\frac {\left (4\,a\,e\,g-2\,c\,d\,f\right )\,\sqrt {d+e\,x}}{c^3\,d^3\,e}+\frac {2\,g\,x\,\sqrt {d+e\,x}}{c^2\,d^2\,e}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{\frac {a}{c}+x^2+\frac {x\,\left (c^3\,d^4+a\,c^2\,d^2\,e^2\right )}{c^3\,d^3\,e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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