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

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

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Rubi [A]  time = 0.14, 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 = {788, 648} \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 648

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 788

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 = 0.04, size = 51, normalized size = 0.34 \begin {gather*} \frac {2 \sqrt {d+e x} (2 a e g+c d (g x-f))}{c^2 d^2 \sqrt {(d+e x) (a e+c d 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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IntegrateAlgebraic [A]  time = 1.36, size = 63, normalized size = 0.42 \begin {gather*} \frac {2 (d+e x)^{3/2} (a e+c d x) (g (a e+c d x)+a e g-c d f)}{c^2 d^2 ((d+e x) (a e+c d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.40, size = 96, normalized size = 0.64 \begin {gather*} \frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d g x - c d f + 2 \, a e g\right )} \sqrt {e x + d}}{c^{3} d^{3} e x^{2} + a c^{2} d^{3} e + {\left (c^{3} d^{4} + a c^{2} d^{2} e^{2}\right )} x} \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*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(c*d*g*x - c*d*f + 2*a*e*g)*sqrt(e*x + d)/(c^3*d^3*e*x^2 + a*c^2
*d^3*e + (c^3*d^4 + a*c^2*d^2*e^2)*x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 0.89Unable to transpose Error: Bad Argument Value

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maple [A]  time = 0.01, size = 66, normalized size = 0.44 \begin {gather*} \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}}}{\left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}} c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.57, size = 48, normalized size = 0.32 \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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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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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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