3.4.17 \(\int (d+e x)^{3/2} (b x+c x^2) \, dx\)

Optimal. Leaf size=68 \[ -\frac {2 (d+e x)^{7/2} (2 c d-b e)}{7 e^3}+\frac {2 d (d+e x)^{5/2} (c d-b e)}{5 e^3}+\frac {2 c (d+e x)^{9/2}}{9 e^3} \]

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Rubi [A]  time = 0.03, antiderivative size = 68, 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} \begin {gather*} -\frac {2 (d+e x)^{7/2} (2 c d-b e)}{7 e^3}+\frac {2 d (d+e x)^{5/2} (c d-b e)}{5 e^3}+\frac {2 c (d+e x)^{9/2}}{9 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(3/2)*(b*x + c*x^2),x]

[Out]

(2*d*(c*d - b*e)*(d + e*x)^(5/2))/(5*e^3) - (2*(2*c*d - b*e)*(d + e*x)^(7/2))/(7*e^3) + (2*c*(d + e*x)^(9/2))/
(9*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 (d+e x)^{3/2} \left (b x+c x^2\right ) \, dx &=\int \left (\frac {d (c d-b e) (d+e x)^{3/2}}{e^2}+\frac {(-2 c d+b e) (d+e x)^{5/2}}{e^2}+\frac {c (d+e x)^{7/2}}{e^2}\right ) \, dx\\ &=\frac {2 d (c d-b e) (d+e x)^{5/2}}{5 e^3}-\frac {2 (2 c d-b e) (d+e x)^{7/2}}{7 e^3}+\frac {2 c (d+e x)^{9/2}}{9 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 50, normalized size = 0.74 \begin {gather*} \frac {2 (d+e x)^{5/2} \left (9 b e (5 e x-2 d)+c \left (8 d^2-20 d e x+35 e^2 x^2\right )\right )}{315 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x)^(5/2)*(9*b*e*(-2*d + 5*e*x) + c*(8*d^2 - 20*d*e*x + 35*e^2*x^2)))/(315*e^3)

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IntegrateAlgebraic [A]  time = 0.04, size = 56, normalized size = 0.82 \begin {gather*} \frac {2 (d+e x)^{5/2} \left (45 b e (d+e x)-63 b d e+63 c d^2-90 c d (d+e x)+35 c (d+e x)^2\right )}{315 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^(3/2)*(b*x + c*x^2),x]

[Out]

(2*(d + e*x)^(5/2)*(63*c*d^2 - 63*b*d*e - 90*c*d*(d + e*x) + 45*b*e*(d + e*x) + 35*c*(d + e*x)^2))/(315*e^3)

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fricas [A]  time = 0.41, size = 95, normalized size = 1.40 \begin {gather*} \frac {2 \, {\left (35 \, c e^{4} x^{4} + 8 \, c d^{4} - 18 \, b d^{3} e + 5 \, {\left (10 \, c d e^{3} + 9 \, b e^{4}\right )} x^{3} + 3 \, {\left (c d^{2} e^{2} + 24 \, b d e^{3}\right )} x^{2} - {\left (4 \, c d^{3} e - 9 \, b d^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*c*e^4*x^4 + 8*c*d^4 - 18*b*d^3*e + 5*(10*c*d*e^3 + 9*b*e^4)*x^3 + 3*(c*d^2*e^2 + 24*b*d*e^3)*x^2 - (
4*c*d^3*e - 9*b*d^2*e^2)*x)*sqrt(e*x + d)/e^3

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giac [B]  time = 0.17, size = 289, normalized size = 4.25 \begin {gather*} \frac {2}{315} \, {\left (105 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} b d^{2} e^{\left (-1\right )} + 21 \, {\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^{2} e^{\left (-2\right )} + 42 \, {\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 )} b d e^{\left (-1\right )} + 18 \, {\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 d e^{\left (-2\right )} + 9 \, {\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 )} b e^{\left (-1\right )} + {\left (35 \, {\left (x e + d\right )}^{\frac {9}{2}} - 180 \, {\left (x e + d\right )}^{\frac {7}{2}} d + 378 \, {\left (x e + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (x e + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {x e + d} d^{4}\right )} c e^{\left (-2\right )}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(105*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*b*d^2*e^(-1) + 21*(3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)*d +
15*sqrt(x*e + d)*d^2)*c*d^2*e^(-2) + 42*(3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*b*d
*e^(-1) + 18*(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*d*e^
(-2) + 9*(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)*b*e^(-1) +
(35*(x*e + d)^(9/2) - 180*(x*e + d)^(7/2)*d + 378*(x*e + d)^(5/2)*d^2 - 420*(x*e + d)^(3/2)*d^3 + 315*sqrt(x*
e + d)*d^4)*c*e^(-2))*e^(-1)

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maple [A]  time = 0.05, size = 47, normalized size = 0.69 \begin {gather*} -\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (-35 c \,e^{2} x^{2}-45 b \,e^{2} x +20 c d e x +18 b d e -8 c \,d^{2}\right )}{315 e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)*(c*x^2+b*x),x)

[Out]

-2/315*(e*x+d)^(5/2)*(-35*c*e^2*x^2-45*b*e^2*x+20*c*d*e*x+18*b*d*e-8*c*d^2)/e^3

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maxima [A]  time = 1.32, size = 54, normalized size = 0.79 \begin {gather*} \frac {2 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} c - 45 \, {\left (2 \, c d - b e\right )} {\left (e x + d\right )}^{\frac {7}{2}} + 63 \, {\left (c d^{2} - b d e\right )} {\left (e x + d\right )}^{\frac {5}{2}}\right )}}{315 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*(e*x + d)^(9/2)*c - 45*(2*c*d - b*e)*(e*x + d)^(7/2) + 63*(c*d^2 - b*d*e)*(e*x + d)^(5/2))/e^3

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mupad [B]  time = 0.06, size = 52, normalized size = 0.76 \begin {gather*} \frac {2\,{\left (d+e\,x\right )}^{5/2}\,\left (35\,c\,{\left (d+e\,x\right )}^2+63\,c\,d^2+45\,b\,e\,\left (d+e\,x\right )-90\,c\,d\,\left (d+e\,x\right )-63\,b\,d\,e\right )}{315\,e^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)*(d + e*x)^(3/2),x)

[Out]

(2*(d + e*x)^(5/2)*(35*c*(d + e*x)^2 + 63*c*d^2 + 45*b*e*(d + e*x) - 90*c*d*(d + e*x) - 63*b*d*e))/(315*e^3)

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sympy [B]  time = 8.84, size = 178, normalized size = 2.62 \begin {gather*} \frac {2 b d \left (- \frac {d \left (d + e x\right )^{\frac {3}{2}}}{3} + \frac {\left (d + e x\right )^{\frac {5}{2}}}{5}\right )}{e^{2}} + \frac {2 b \left (\frac {d^{2} \left (d + e x\right )^{\frac {3}{2}}}{3} - \frac {2 d \left (d + e x\right )^{\frac {5}{2}}}{5} + \frac {\left (d + e x\right )^{\frac {7}{2}}}{7}\right )}{e^{2}} + \frac {2 c d \left (\frac {d^{2} \left (d + e x\right )^{\frac {3}{2}}}{3} - \frac {2 d \left (d + e x\right )^{\frac {5}{2}}}{5} + \frac {\left (d + e x\right )^{\frac {7}{2}}}{7}\right )}{e^{3}} + \frac {2 c \left (- \frac {d^{3} \left (d + e x\right )^{\frac {3}{2}}}{3} + \frac {3 d^{2} \left (d + e x\right )^{\frac {5}{2}}}{5} - \frac {3 d \left (d + e x\right )^{\frac {7}{2}}}{7} + \frac {\left (d + e x\right )^{\frac {9}{2}}}{9}\right )}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(3/2)*(c*x**2+b*x),x)

[Out]

2*b*d*(-d*(d + e*x)**(3/2)/3 + (d + e*x)**(5/2)/5)/e**2 + 2*b*(d**2*(d + e*x)**(3/2)/3 - 2*d*(d + e*x)**(5/2)/
5 + (d + e*x)**(7/2)/7)/e**2 + 2*c*d*(d**2*(d + e*x)**(3/2)/3 - 2*d*(d + e*x)**(5/2)/5 + (d + e*x)**(7/2)/7)/e
**3 + 2*c*(-d**3*(d + e*x)**(3/2)/3 + 3*d**2*(d + e*x)**(5/2)/5 - 3*d*(d + e*x)**(7/2)/7 + (d + e*x)**(9/2)/9)
/e**3

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