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

Optimal. Leaf size=61 \[ \frac {2 \sqrt {d+e x} \left (a e^2+c d^2\right )}{e^3}+\frac {2 c (d+e x)^{5/2}}{5 e^3}-\frac {4 c d (d+e x)^{3/2}}{3 e^3} \]

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Rubi [A]  time = 0.02, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {697} \begin {gather*} \frac {2 \sqrt {d+e x} \left (a e^2+c d^2\right )}{e^3}+\frac {2 c (d+e x)^{5/2}}{5 e^3}-\frac {4 c d (d+e x)^{3/2}}{3 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)/Sqrt[d + e*x],x]

[Out]

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

Rule 697

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

Rubi steps

\begin {align*} \int \frac {a+c x^2}{\sqrt {d+e x}} \, dx &=\int \left (\frac {c d^2+a e^2}{e^2 \sqrt {d+e x}}-\frac {2 c d \sqrt {d+e x}}{e^2}+\frac {c (d+e x)^{3/2}}{e^2}\right ) \, dx\\ &=\frac {2 \left (c d^2+a e^2\right ) \sqrt {d+e x}}{e^3}-\frac {4 c d (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 = 44, normalized size = 0.72 \begin {gather*} \frac {2 \sqrt {d+e x} \left (15 a e^2+c \left (8 d^2-4 d e x+3 e^2 x^2\right )\right )}{15 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 48, normalized size = 0.79 \begin {gather*} \frac {2 \sqrt {d+e x} \left (15 a e^2+15 c d^2-10 c d (d+e x)+3 c (d+e x)^2\right )}{15 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + c*x^2)/Sqrt[d + e*x],x]

[Out]

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

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fricas [A]  time = 0.39, size = 40, normalized size = 0.66 \begin {gather*} \frac {2 \, {\left (3 \, c e^{2} x^{2} - 4 \, c d e x + 8 \, c d^{2} + 15 \, a e^{2}\right )} \sqrt {e x + d}}{15 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 55, normalized size = 0.90 \begin {gather*} \frac {2}{15} \, {\left ({\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 )} + 15 \, \sqrt {x e + d} a\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 41, normalized size = 0.67 \begin {gather*} \frac {2 \sqrt {e x +d}\, \left (3 c \,e^{2} x^{2}-4 c d e x +15 a \,e^{2}+8 c \,d^{2}\right )}{15 e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.34, size = 53, normalized size = 0.87 \begin {gather*} \frac {2 \, {\left (15 \, \sqrt {e x + d} a + \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*sqrt(e*x + d)*a + (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 = 0.05, size = 44, normalized size = 0.72 \begin {gather*} \frac {2\,\sqrt {d+e\,x}\,\left (3\,c\,{\left (d+e\,x\right )}^2+15\,a\,e^2+15\,c\,d^2-10\,c\,d\,\left (d+e\,x\right )\right )}{15\,e^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)/(d + e*x)^(1/2),x)

[Out]

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

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sympy [A]  time = 7.46, size = 150, normalized size = 2.46 \begin {gather*} \begin {cases} \frac {- \frac {2 a d}{\sqrt {d + e x}} - 2 a \left (- \frac {d}{\sqrt {d + e x}} - \sqrt {d + e x}\right ) - \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 {a x + \frac {c x^{3}}{3}}{\sqrt {d}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-2*a*d/sqrt(d + e*x) - 2*a*(-d/sqrt(d + e*x) - sqrt(d + e*x)) - 2*c*d*(d**2/sqrt(d + e*x) + 2*d*sq
rt(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)), ((a*x + c*x**3/3)/sqrt(d), True))

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