3.292 \(\int x^3 (a+b x)^{3/2} \, dx\)

Optimal. Leaf size=72 \[ \frac{6 a^2 (a+b x)^{7/2}}{7 b^4}-\frac{2 a^3 (a+b x)^{5/2}}{5 b^4}+\frac{2 (a+b x)^{11/2}}{11 b^4}-\frac{2 a (a+b x)^{9/2}}{3 b^4} \]

[Out]

(-2*a^3*(a + b*x)^(5/2))/(5*b^4) + (6*a^2*(a + b*x)^(7/2))/(7*b^4) - (2*a*(a + b*x)^(9/2))/(3*b^4) + (2*(a + b
*x)^(11/2))/(11*b^4)

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Rubi [A]  time = 0.0188042, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {43} \[ \frac{6 a^2 (a+b x)^{7/2}}{7 b^4}-\frac{2 a^3 (a+b x)^{5/2}}{5 b^4}+\frac{2 (a+b x)^{11/2}}{11 b^4}-\frac{2 a (a+b x)^{9/2}}{3 b^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x)^(3/2),x]

[Out]

(-2*a^3*(a + b*x)^(5/2))/(5*b^4) + (6*a^2*(a + b*x)^(7/2))/(7*b^4) - (2*a*(a + b*x)^(9/2))/(3*b^4) + (2*(a + b
*x)^(11/2))/(11*b^4)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^3 (a+b x)^{3/2} \, dx &=\int \left (-\frac{a^3 (a+b x)^{3/2}}{b^3}+\frac{3 a^2 (a+b x)^{5/2}}{b^3}-\frac{3 a (a+b x)^{7/2}}{b^3}+\frac{(a+b x)^{9/2}}{b^3}\right ) \, dx\\ &=-\frac{2 a^3 (a+b x)^{5/2}}{5 b^4}+\frac{6 a^2 (a+b x)^{7/2}}{7 b^4}-\frac{2 a (a+b x)^{9/2}}{3 b^4}+\frac{2 (a+b x)^{11/2}}{11 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0390633, size = 46, normalized size = 0.64 \[ \frac{2 (a+b x)^{5/2} \left (40 a^2 b x-16 a^3-70 a b^2 x^2+105 b^3 x^3\right )}{1155 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x)^(3/2),x]

[Out]

(2*(a + b*x)^(5/2)*(-16*a^3 + 40*a^2*b*x - 70*a*b^2*x^2 + 105*b^3*x^3))/(1155*b^4)

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Maple [A]  time = 0.004, size = 43, normalized size = 0.6 \begin{align*} -{\frac{-210\,{b}^{3}{x}^{3}+140\,a{b}^{2}{x}^{2}-80\,{a}^{2}bx+32\,{a}^{3}}{1155\,{b}^{4}} \left ( bx+a \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x+a)^(3/2),x)

[Out]

-2/1155*(b*x+a)^(5/2)*(-105*b^3*x^3+70*a*b^2*x^2-40*a^2*b*x+16*a^3)/b^4

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Maxima [A]  time = 1.07472, size = 76, normalized size = 1.06 \begin{align*} \frac{2 \,{\left (b x + a\right )}^{\frac{11}{2}}}{11 \, b^{4}} - \frac{2 \,{\left (b x + a\right )}^{\frac{9}{2}} a}{3 \, b^{4}} + \frac{6 \,{\left (b x + a\right )}^{\frac{7}{2}} a^{2}}{7 \, b^{4}} - \frac{2 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{3}}{5 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/11*(b*x + a)^(11/2)/b^4 - 2/3*(b*x + a)^(9/2)*a/b^4 + 6/7*(b*x + a)^(7/2)*a^2/b^4 - 2/5*(b*x + a)^(5/2)*a^3/
b^4

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Fricas [A]  time = 1.50403, size = 147, normalized size = 2.04 \begin{align*} \frac{2 \,{\left (105 \, b^{5} x^{5} + 140 \, a b^{4} x^{4} + 5 \, a^{2} b^{3} x^{3} - 6 \, a^{3} b^{2} x^{2} + 8 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt{b x + a}}{1155 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1155*(105*b^5*x^5 + 140*a*b^4*x^4 + 5*a^2*b^3*x^3 - 6*a^3*b^2*x^2 + 8*a^4*b*x - 16*a^5)*sqrt(b*x + a)/b^4

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Sympy [B]  time = 3.73798, size = 1742, normalized size = 24.19 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b*x+a)**(3/2),x)

[Out]

-32*a**(51/2)*sqrt(1 + b*x/a)/(1155*a**20*b**4 + 6930*a**19*b**5*x + 17325*a**18*b**6*x**2 + 23100*a**17*b**7*
x**3 + 17325*a**16*b**8*x**4 + 6930*a**15*b**9*x**5 + 1155*a**14*b**10*x**6) + 32*a**(51/2)/(1155*a**20*b**4 +
6930*a**19*b**5*x + 17325*a**18*b**6*x**2 + 23100*a**17*b**7*x**3 + 17325*a**16*b**8*x**4 + 6930*a**15*b**9*x
**5 + 1155*a**14*b**10*x**6) - 176*a**(49/2)*b*x*sqrt(1 + b*x/a)/(1155*a**20*b**4 + 6930*a**19*b**5*x + 17325*
a**18*b**6*x**2 + 23100*a**17*b**7*x**3 + 17325*a**16*b**8*x**4 + 6930*a**15*b**9*x**5 + 1155*a**14*b**10*x**6
) + 192*a**(49/2)*b*x/(1155*a**20*b**4 + 6930*a**19*b**5*x + 17325*a**18*b**6*x**2 + 23100*a**17*b**7*x**3 + 1
7325*a**16*b**8*x**4 + 6930*a**15*b**9*x**5 + 1155*a**14*b**10*x**6) - 396*a**(47/2)*b**2*x**2*sqrt(1 + b*x/a)
/(1155*a**20*b**4 + 6930*a**19*b**5*x + 17325*a**18*b**6*x**2 + 23100*a**17*b**7*x**3 + 17325*a**16*b**8*x**4
+ 6930*a**15*b**9*x**5 + 1155*a**14*b**10*x**6) + 480*a**(47/2)*b**2*x**2/(1155*a**20*b**4 + 6930*a**19*b**5*x
+ 17325*a**18*b**6*x**2 + 23100*a**17*b**7*x**3 + 17325*a**16*b**8*x**4 + 6930*a**15*b**9*x**5 + 1155*a**14*b
**10*x**6) - 462*a**(45/2)*b**3*x**3*sqrt(1 + b*x/a)/(1155*a**20*b**4 + 6930*a**19*b**5*x + 17325*a**18*b**6*x
**2 + 23100*a**17*b**7*x**3 + 17325*a**16*b**8*x**4 + 6930*a**15*b**9*x**5 + 1155*a**14*b**10*x**6) + 640*a**(
45/2)*b**3*x**3/(1155*a**20*b**4 + 6930*a**19*b**5*x + 17325*a**18*b**6*x**2 + 23100*a**17*b**7*x**3 + 17325*a
**16*b**8*x**4 + 6930*a**15*b**9*x**5 + 1155*a**14*b**10*x**6) + 480*a**(43/2)*b**4*x**4/(1155*a**20*b**4 + 69
30*a**19*b**5*x + 17325*a**18*b**6*x**2 + 23100*a**17*b**7*x**3 + 17325*a**16*b**8*x**4 + 6930*a**15*b**9*x**5
+ 1155*a**14*b**10*x**6) + 1848*a**(41/2)*b**5*x**5*sqrt(1 + b*x/a)/(1155*a**20*b**4 + 6930*a**19*b**5*x + 17
325*a**18*b**6*x**2 + 23100*a**17*b**7*x**3 + 17325*a**16*b**8*x**4 + 6930*a**15*b**9*x**5 + 1155*a**14*b**10*
x**6) + 192*a**(41/2)*b**5*x**5/(1155*a**20*b**4 + 6930*a**19*b**5*x + 17325*a**18*b**6*x**2 + 23100*a**17*b**
7*x**3 + 17325*a**16*b**8*x**4 + 6930*a**15*b**9*x**5 + 1155*a**14*b**10*x**6) + 5544*a**(39/2)*b**6*x**6*sqrt
(1 + b*x/a)/(1155*a**20*b**4 + 6930*a**19*b**5*x + 17325*a**18*b**6*x**2 + 23100*a**17*b**7*x**3 + 17325*a**16
*b**8*x**4 + 6930*a**15*b**9*x**5 + 1155*a**14*b**10*x**6) + 32*a**(39/2)*b**6*x**6/(1155*a**20*b**4 + 6930*a*
*19*b**5*x + 17325*a**18*b**6*x**2 + 23100*a**17*b**7*x**3 + 17325*a**16*b**8*x**4 + 6930*a**15*b**9*x**5 + 11
55*a**14*b**10*x**6) + 8844*a**(37/2)*b**7*x**7*sqrt(1 + b*x/a)/(1155*a**20*b**4 + 6930*a**19*b**5*x + 17325*a
**18*b**6*x**2 + 23100*a**17*b**7*x**3 + 17325*a**16*b**8*x**4 + 6930*a**15*b**9*x**5 + 1155*a**14*b**10*x**6)
+ 8448*a**(35/2)*b**8*x**8*sqrt(1 + b*x/a)/(1155*a**20*b**4 + 6930*a**19*b**5*x + 17325*a**18*b**6*x**2 + 231
00*a**17*b**7*x**3 + 17325*a**16*b**8*x**4 + 6930*a**15*b**9*x**5 + 1155*a**14*b**10*x**6) + 4840*a**(33/2)*b*
*9*x**9*sqrt(1 + b*x/a)/(1155*a**20*b**4 + 6930*a**19*b**5*x + 17325*a**18*b**6*x**2 + 23100*a**17*b**7*x**3 +
17325*a**16*b**8*x**4 + 6930*a**15*b**9*x**5 + 1155*a**14*b**10*x**6) + 1540*a**(31/2)*b**10*x**10*sqrt(1 + b
*x/a)/(1155*a**20*b**4 + 6930*a**19*b**5*x + 17325*a**18*b**6*x**2 + 23100*a**17*b**7*x**3 + 17325*a**16*b**8*
x**4 + 6930*a**15*b**9*x**5 + 1155*a**14*b**10*x**6) + 210*a**(29/2)*b**11*x**11*sqrt(1 + b*x/a)/(1155*a**20*b
**4 + 6930*a**19*b**5*x + 17325*a**18*b**6*x**2 + 23100*a**17*b**7*x**3 + 17325*a**16*b**8*x**4 + 6930*a**15*b
**9*x**5 + 1155*a**14*b**10*x**6)

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Giac [B]  time = 1.17763, size = 157, normalized size = 2.18 \begin{align*} \frac{2 \,{\left (\frac{11 \,{\left (35 \,{\left (b x + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3}\right )} a}{b^{3}} + \frac{315 \,{\left (b x + a\right )}^{\frac{11}{2}} - 1540 \,{\left (b x + a\right )}^{\frac{9}{2}} a + 2970 \,{\left (b x + a\right )}^{\frac{7}{2}} a^{2} - 2772 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{3} + 1155 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{4}}{b^{3}}\right )}}{3465 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(11*(35*(b*x + a)^(9/2) - 135*(b*x + a)^(7/2)*a + 189*(b*x + a)^(5/2)*a^2 - 105*(b*x + a)^(3/2)*a^3)*a/
b^3 + (315*(b*x + a)^(11/2) - 1540*(b*x + a)^(9/2)*a + 2970*(b*x + a)^(7/2)*a^2 - 2772*(b*x + a)^(5/2)*a^3 + 1
155*(b*x + a)^(3/2)*a^4)/b^3)/b