3.341 \(\int \frac {b x+c x^2}{\sqrt {d+e x}} \, dx\)

Optimal. Leaf size=66 \[ -\frac {2 (d+e x)^{3/2} (2 c d-b e)}{3 e^3}+\frac {2 d \sqrt {d+e x} (c d-b e)}{e^3}+\frac {2 c (d+e x)^{5/2}}{5 e^3} \]

[Out]

-2/3*(-b*e+2*c*d)*(e*x+d)^(3/2)/e^3+2/5*c*(e*x+d)^(5/2)/e^3+2*d*(-b*e+c*d)*(e*x+d)^(1/2)/e^3

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Rubi [A]  time = 0.03, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {698} \[ -\frac {2 (d+e x)^{3/2} (2 c d-b e)}{3 e^3}+\frac {2 d \sqrt {d+e x} (c d-b e)}{e^3}+\frac {2 c (d+e x)^{5/2}}{5 e^3} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)/Sqrt[d + e*x],x]

[Out]

(2*d*(c*d - b*e)*Sqrt[d + e*x])/e^3 - (2*(2*c*d - b*e)*(d + e*x)^(3/2))/(3*e^3) + (2*c*(d + e*x)^(5/2))/(5*e^3
)

Rule 698

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

Rubi steps

\begin {align*} \int \frac {b x+c x^2}{\sqrt {d+e x}} \, dx &=\int \left (\frac {d (c d-b e)}{e^2 \sqrt {d+e x}}+\frac {(-2 c d+b e) \sqrt {d+e x}}{e^2}+\frac {c (d+e x)^{3/2}}{e^2}\right ) \, dx\\ &=\frac {2 d (c d-b e) \sqrt {d+e x}}{e^3}-\frac {2 (2 c d-b e) (d+e x)^{3/2}}{3 e^3}+\frac {2 c (d+e x)^{5/2}}{5 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 49, normalized size = 0.74 \[ \frac {2 \sqrt {d+e x} \left (5 b e (e x-2 d)+c \left (8 d^2-4 d e x+3 e^2 x^2\right )\right )}{15 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.88, size = 48, normalized size = 0.73 \[ \frac {2 \, {\left (3 \, c e^{2} x^{2} + 8 \, c d^{2} - 10 \, b d e - {\left (4 \, c d e - 5 \, b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 69, normalized size = 1.05 \[ \frac {2}{15} \, {\left (5 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} b e^{\left (-1\right )} + {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} - 10 \, {\left (x e + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {x e + d} d^{2}\right )} c e^{\left (-2\right )}\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

2/15*(5*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*b*e^(-1) + (3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)*d + 15*sqrt(x
*e + d)*d^2)*c*e^(-2))*e^(-1)

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maple [A]  time = 0.04, size = 47, normalized size = 0.71 \[ -\frac {2 \left (-3 c \,e^{2} x^{2}-5 b \,e^{2} x +4 c d e x +10 b d e -8 c \,d^{2}\right ) \sqrt {e x +d}}{15 e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)/(e*x+d)^(1/2),x)

[Out]

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

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maxima [A]  time = 1.37, size = 67, normalized size = 1.02 \[ \frac {2 \, {\left (\frac {5 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} b}{e} + \frac {{\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} c}{e^{2}}\right )}}{15 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

2/15*(5*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*b/e + (3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x +
d)*d^2)*c/e^2)/e

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mupad [B]  time = 0.06, size = 52, normalized size = 0.79 \[ \frac {2\,\sqrt {d+e\,x}\,\left (3\,c\,{\left (d+e\,x\right )}^2+15\,c\,d^2+5\,b\,e\,\left (d+e\,x\right )-10\,c\,d\,\left (d+e\,x\right )-15\,b\,d\,e\right )}{15\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)/(d + e*x)^(1/2),x)

[Out]

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

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sympy [A]  time = 10.65, size = 182, normalized size = 2.76 \[ \begin {cases} \frac {- \frac {2 b d \left (- \frac {d}{\sqrt {d + e x}} - \sqrt {d + e x}\right )}{e} - \frac {2 b \left (\frac {d^{2}}{\sqrt {d + e x}} + 2 d \sqrt {d + e x} - \frac {\left (d + e x\right )^{\frac {3}{2}}}{3}\right )}{e} - \frac {2 c d \left (\frac {d^{2}}{\sqrt {d + e x}} + 2 d \sqrt {d + e x} - \frac {\left (d + e x\right )^{\frac {3}{2}}}{3}\right )}{e^{2}} - \frac {2 c \left (- \frac {d^{3}}{\sqrt {d + e x}} - 3 d^{2} \sqrt {d + e x} + d \left (d + e x\right )^{\frac {3}{2}} - \frac {\left (d + e x\right )^{\frac {5}{2}}}{5}\right )}{e^{2}}}{e} & \text {for}\: e \neq 0 \\\frac {\frac {b x^{2}}{2} + \frac {c x^{3}}{3}}{\sqrt {d}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)/(e*x+d)**(1/2),x)

[Out]

Piecewise(((-2*b*d*(-d/sqrt(d + e*x) - sqrt(d + e*x))/e - 2*b*(d**2/sqrt(d + e*x) + 2*d*sqrt(d + e*x) - (d + e
*x)**(3/2)/3)/e - 2*c*d*(d**2/sqrt(d + e*x) + 2*d*sqrt(d + e*x) - (d + e*x)**(3/2)/3)/e**2 - 2*c*(-d**3/sqrt(d
+ e*x) - 3*d**2*sqrt(d + e*x) + d*(d + e*x)**(3/2) - (d + e*x)**(5/2)/5)/e**2)/e, Ne(e, 0)), ((b*x**2/2 + c*x
**3/3)/sqrt(d), True))

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