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

Optimal. Leaf size=63 \[ \frac {2 (d+e x)^{3/2} \left (a e^2+c d^2\right )}{3 e^3}+\frac {2 c (d+e x)^{7/2}}{7 e^3}-\frac {4 c d (d+e x)^{5/2}}{5 e^3} \]

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Rubi [A]  time = 0.02, antiderivative size = 63, 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 (d+e x)^{3/2} \left (a e^2+c d^2\right )}{3 e^3}+\frac {2 c (d+e x)^{7/2}}{7 e^3}-\frac {4 c d (d+e x)^{5/2}}{5 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d + e*x]*(a + c*x^2),x]

[Out]

(2*(c*d^2 + a*e^2)*(d + e*x)^(3/2))/(3*e^3) - (4*c*d*(d + e*x)^(5/2))/(5*e^3) + (2*c*(d + e*x)^(7/2))/(7*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 \sqrt {d+e x} \left (a+c x^2\right ) \, dx &=\int \left (\frac {\left (c d^2+a e^2\right ) \sqrt {d+e x}}{e^2}-\frac {2 c d (d+e x)^{3/2}}{e^2}+\frac {c (d+e x)^{5/2}}{e^2}\right ) \, dx\\ &=\frac {2 \left (c d^2+a e^2\right ) (d+e x)^{3/2}}{3 e^3}-\frac {4 c d (d+e x)^{5/2}}{5 e^3}+\frac {2 c (d+e x)^{7/2}}{7 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 44, normalized size = 0.70 \begin {gather*} \frac {2 (d+e x)^{3/2} \left (35 a e^2+c \left (8 d^2-12 d e x+15 e^2 x^2\right )\right )}{105 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x)^(3/2)*(35*a*e^2 + c*(8*d^2 - 12*d*e*x + 15*e^2*x^2)))/(105*e^3)

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IntegrateAlgebraic [A]  time = 0.03, size = 48, normalized size = 0.76 \begin {gather*} \frac {2 (d+e x)^{3/2} \left (35 a e^2+35 c d^2-42 c d (d+e x)+15 c (d+e x)^2\right )}{105 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x)^(3/2)*(35*c*d^2 + 35*a*e^2 - 42*c*d*(d + e*x) + 15*c*(d + e*x)^2))/(105*e^3)

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fricas [A]  time = 0.40, size = 62, normalized size = 0.98 \begin {gather*} \frac {2 \, {\left (15 \, c e^{3} x^{3} + 3 \, c d e^{2} x^{2} + 8 \, c d^{3} + 35 \, a d e^{2} - {\left (4 \, c d^{2} e - 35 \, a e^{3}\right )} x\right )} \sqrt {e x + d}}{105 \, 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]

2/105*(15*c*e^3*x^3 + 3*c*d*e^2*x^2 + 8*c*d^3 + 35*a*d*e^2 - (4*c*d^2*e - 35*a*e^3)*x)*sqrt(e*x + d)/e^3

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giac [B]  time = 0.18, size = 134, normalized size = 2.13 \begin {gather*} \frac {2}{105} \, {\left (7 \, {\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 d e^{\left (-2\right )} + 3 \, {\left (5 \, {\left (x e + d\right )}^{\frac {7}{2}} - 21 \, {\left (x e + d\right )}^{\frac {5}{2}} d + 35 \, {\left (x e + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {x e + d} d^{3}\right )} c e^{\left (-2\right )} + 105 \, \sqrt {x e + d} a d + 35 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} 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]

2/105*(7*(3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*c*d*e^(-2) + 3*(5*(x*e + d)^(7/2) -
21*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2 - 35*sqrt(x*e + d)*d^3)*c*e^(-2) + 105*sqrt(x*e + d)*a*d + 35*(
(x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*a)*e^(-1)

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maple [A]  time = 0.05, size = 41, normalized size = 0.65 \begin {gather*} \frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (15 c \,e^{2} x^{2}-12 c d e x +35 a \,e^{2}+8 c \,d^{2}\right )}{105 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]

2/105*(e*x+d)^(3/2)*(15*c*e^2*x^2-12*c*d*e*x+35*a*e^2+8*c*d^2)/e^3

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maxima [A]  time = 1.38, size = 47, normalized size = 0.75 \begin {gather*} \frac {2 \, {\left (15 \, {\left (e x + d\right )}^{\frac {7}{2}} c - 42 \, {\left (e x + d\right )}^{\frac {5}{2}} c d + 35 \, {\left (c d^{2} + a e^{2}\right )} {\left (e x + d\right )}^{\frac {3}{2}}\right )}}{105 \, 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="maxima")

[Out]

2/105*(15*(e*x + d)^(7/2)*c - 42*(e*x + d)^(5/2)*c*d + 35*(c*d^2 + a*e^2)*(e*x + d)^(3/2))/e^3

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mupad [B]  time = 0.05, size = 44, normalized size = 0.70 \begin {gather*} \frac {2\,{\left (d+e\,x\right )}^{3/2}\,\left (15\,c\,{\left (d+e\,x\right )}^2+35\,a\,e^2+35\,c\,d^2-42\,c\,d\,\left (d+e\,x\right )\right )}{105\,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)^(3/2)*(15*c*(d + e*x)^2 + 35*a*e^2 + 35*c*d^2 - 42*c*d*(d + e*x)))/(105*e^3)

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sympy [A]  time = 2.18, size = 61, normalized size = 0.97 \begin {gather*} \frac {2 \left (- \frac {2 c d \left (d + e x\right )^{\frac {5}{2}}}{5 e^{2}} + \frac {c \left (d + e x\right )^{\frac {7}{2}}}{7 e^{2}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (a e^{2} + c d^{2}\right )}{3 e^{2}}\right )}{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)

[Out]

2*(-2*c*d*(d + e*x)**(5/2)/(5*e**2) + c*(d + e*x)**(7/2)/(7*e**2) + (d + e*x)**(3/2)*(a*e**2 + c*d**2)/(3*e**2
))/e

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