### 3.592 $$\int (d+e x)^{3/2} (a+c x^2) \, dx$$

Optimal. Leaf size=63 $\frac{2 (d+e x)^{5/2} \left (a e^2+c d^2\right )}{5 e^3}+\frac{2 c (d+e x)^{9/2}}{9 e^3}-\frac{4 c d (d+e x)^{7/2}}{7 e^3}$

[Out]

(2*(c*d^2 + a*e^2)*(d + e*x)^(5/2))/(5*e^3) - (4*c*d*(d + e*x)^(7/2))/(7*e^3) + (2*c*(d + e*x)^(9/2))/(9*e^3)

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Rubi [A]  time = 0.0210846, antiderivative size = 63, normalized size of antiderivative = 1., 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} $\frac{2 (d+e x)^{5/2} \left (a e^2+c d^2\right )}{5 e^3}+\frac{2 c (d+e x)^{9/2}}{9 e^3}-\frac{4 c d (d+e x)^{7/2}}{7 e^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^(3/2)*(a + c*x^2),x]

[Out]

(2*(c*d^2 + a*e^2)*(d + e*x)^(5/2))/(5*e^3) - (4*c*d*(d + e*x)^(7/2))/(7*e^3) + (2*c*(d + e*x)^(9/2))/(9*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 (d+e x)^{3/2} \left (a+c x^2\right ) \, dx &=\int \left (\frac{\left (c d^2+a e^2\right ) (d+e x)^{3/2}}{e^2}-\frac{2 c d (d+e x)^{5/2}}{e^2}+\frac{c (d+e x)^{7/2}}{e^2}\right ) \, dx\\ &=\frac{2 \left (c d^2+a e^2\right ) (d+e x)^{5/2}}{5 e^3}-\frac{4 c d (d+e x)^{7/2}}{7 e^3}+\frac{2 c (d+e x)^{9/2}}{9 e^3}\\ \end{align*}

Mathematica [A]  time = 0.0370441, size = 44, normalized size = 0.7 $\frac{2 (d+e x)^{5/2} \left (63 a e^2+c \left (8 d^2-20 d e x+35 e^2 x^2\right )\right )}{315 e^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(d + e*x)^(5/2)*(63*a*e^2 + c*(8*d^2 - 20*d*e*x + 35*e^2*x^2)))/(315*e^3)

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Maple [A]  time = 0.043, size = 41, normalized size = 0.7 \begin{align*}{\frac{70\,c{e}^{2}{x}^{2}-40\,cdex+126\,a{e}^{2}+16\,c{d}^{2}}{315\,{e}^{3}} \left ( ex+d \right ) ^{{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(e*x+d)^(5/2)*(35*c*e^2*x^2-20*c*d*e*x+63*a*e^2+8*c*d^2)/e^3

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Maxima [A]  time = 1.19216, size = 63, normalized size = 1. \begin{align*} \frac{2 \,{\left (35 \,{\left (e x + d\right )}^{\frac{9}{2}} c - 90 \,{\left (e x + d\right )}^{\frac{7}{2}} c d + 63 \,{\left (c d^{2} + a e^{2}\right )}{\left (e x + d\right )}^{\frac{5}{2}}\right )}}{315 \, e^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*(e*x + d)^(9/2)*c - 90*(e*x + d)^(7/2)*c*d + 63*(c*d^2 + a*e^2)*(e*x + d)^(5/2))/e^3

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Fricas [A]  time = 1.71276, size = 194, normalized size = 3.08 \begin{align*} \frac{2 \,{\left (35 \, c e^{4} x^{4} + 50 \, c d e^{3} x^{3} + 8 \, c d^{4} + 63 \, a d^{2} e^{2} + 3 \,{\left (c d^{2} e^{2} + 21 \, a e^{4}\right )} x^{2} - 2 \,{\left (2 \, c d^{3} e - 63 \, a d e^{3}\right )} x\right )} \sqrt{e x + d}}{315 \, e^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*c*e^4*x^4 + 50*c*d*e^3*x^3 + 8*c*d^4 + 63*a*d^2*e^2 + 3*(c*d^2*e^2 + 21*a*e^4)*x^2 - 2*(2*c*d^3*e -
63*a*d*e^3)*x)*sqrt(e*x + d)/e^3

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Sympy [A]  time = 5.60879, size = 155, normalized size = 2.46 \begin{align*} a d \left (\begin{cases} \sqrt{d} x & \text{for}\: e = 0 \\\frac{2 \left (d + e x\right )^{\frac{3}{2}}}{3 e} & \text{otherwise} \end{cases}\right ) + \frac{2 a \left (- \frac{d \left (d + e x\right )^{\frac{3}{2}}}{3} + \frac{\left (d + e x\right )^{\frac{5}{2}}}{5}\right )}{e} + \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{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d*Piecewise((sqrt(d)*x, Eq(e, 0)), (2*(d + e*x)**(3/2)/(3*e), True)) + 2*a*(-d*(d + e*x)**(3/2)/3 + (d + e*x
)**(5/2)/5)/e + 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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Giac [B]  time = 1.41944, size = 182, normalized size = 2.89 \begin{align*} \frac{2}{315} \,{\left (3 \,{\left (15 \,{\left (x e + d\right )}^{\frac{7}{2}} - 42 \,{\left (x e + d\right )}^{\frac{5}{2}} d + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{2}\right )} c d e^{\left (-2\right )} + 105 \,{\left (x e + d\right )}^{\frac{3}{2}} a d +{\left (35 \,{\left (x e + d\right )}^{\frac{9}{2}} - 135 \,{\left (x e + d\right )}^{\frac{7}{2}} d + 189 \,{\left (x e + d\right )}^{\frac{5}{2}} d^{2} - 105 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{3}\right )} c e^{\left (-2\right )} + 21 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} a\right )} e^{\left (-1\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(3*(15*(x*e + d)^(7/2) - 42*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*c*d*e^(-2) + 105*(x*e + d)^(3/2)
*a*d + (35*(x*e + d)^(9/2) - 135*(x*e + d)^(7/2)*d + 189*(x*e + d)^(5/2)*d^2 - 105*(x*e + d)^(3/2)*d^3)*c*e^(-
2) + 21*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*a)*e^(-1)