3.3.38 \(\int x (c (a+b x^2)^3)^{3/2} \, dx\) [238]

Optimal. Leaf size=32 \[ \frac {c \left (a+b x^2\right )^4 \sqrt {c \left (a+b x^2\right )^3}}{11 b} \]

[Out]

1/11*c*(b*x^2+a)^4*(c*(b*x^2+a)^3)^(1/2)/b

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Rubi [A]
time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1605, 15, 30} \begin {gather*} \frac {c \left (a+b x^2\right )^4 \sqrt {c \left (a+b x^2\right )^3}}{11 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(c*(a + b*x^2)^3)^(3/2),x]

[Out]

(c*(a + b*x^2)^4*Sqrt[c*(a + b*x^2)^3])/(11*b)

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 1605

Int[((a_.) + (b_.)*(Pq_)^(n_.))^(p_.)*(Qr_), x_Symbol] :> With[{q = Expon[Pq, x], r = Expon[Qr, x]}, Dist[Coef
f[Qr, x, r]/(q*Coeff[Pq, x, q]), Subst[Int[(a + b*x^n)^p, x], x, Pq], x] /; EqQ[r, q - 1] && EqQ[Coeff[Qr, x,
r]*D[Pq, x], q*Coeff[Pq, x, q]*Qr]] /; FreeQ[{a, b, n, p}, x] && PolyQ[Pq, x] && PolyQ[Qr, x]

Rubi steps

\begin {align*} \int x \left (c \left (a+b x^2\right )^3\right )^{3/2} \, dx &=\frac {\text {Subst}\left (\int \left (c x^3\right )^{3/2} \, dx,x,a+b x^2\right )}{2 b}\\ &=\frac {\left (c \sqrt {c \left (a+b x^2\right )^3}\right ) \text {Subst}\left (\int x^{9/2} \, dx,x,a+b x^2\right )}{2 b \left (a+b x^2\right )^{3/2}}\\ &=\frac {c \left (a+b x^2\right )^4 \sqrt {c \left (a+b x^2\right )^3}}{11 b}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 29, normalized size = 0.91 \begin {gather*} \frac {\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^3\right )^{3/2}}{11 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)*(c*(a + b*x^2)^3)^(3/2))/(11*b)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(137\) vs. \(2(28)=56\).
time = 0.04, size = 138, normalized size = 4.31

method	result	size
gosper	\(\frac {\left (b \,x^{2}+a \right ) \left (c \left (b \,x^{2}+a \right )^{3}\right )^{\frac {3}{2}}}{11 b}\)	\(26\)
risch	\(\frac {c \sqrt {c \left (b \,x^{2}+a \right )^{3}}\, \left (b^{5} x^{10}+5 b^{4} a \,x^{8}+10 a^{2} b^{3} x^{6}+10 b^{2} a^{3} x^{4}+5 b \,a^{4} x^{2}+a^{5}\right )}{11 \left (b \,x^{2}+a \right ) b}\)	\(80\)
trager	\(\frac {c \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right ) \sqrt {b^{3} c \,x^{6}+3 a \,b^{2} c \,x^{4}+3 a^{2} b c \,x^{2}+a^{3} c}}{11 b}\)	\(83\)
default	\(-\frac {\left (c \left (b \,x^{2}+a \right )^{3}\right )^{\frac {3}{2}} \left (-5 x^{6} \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} b^{3}-15 \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} a \,b^{2} x^{4}-15 \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} a^{2} b \,x^{2}+6 \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} a^{3}-11 a^{3} \left (c \left (b \,x^{2}+a \right )\right )^{\frac {5}{2}}\right )}{55 b \left (b \,x^{2}+a \right )^{3} \left (c \left (b \,x^{2}+a \right )\right )^{\frac {3}{2}} c}\)	\(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*(b*x^2+a)^3)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/55*(c*(b*x^2+a)^3)^(3/2)/b*(-5*x^6*(b*c*x^2+a*c)^(5/2)*b^3-15*(b*c*x^2+a*c)^(5/2)*a*b^2*x^4-15*(b*c*x^2+a*c
)^(5/2)*a^2*b*x^2+6*(b*c*x^2+a*c)^(5/2)*a^3-11*a^3*(c*(b*x^2+a))^(5/2))/(b*x^2+a)^3/(c*(b*x^2+a))^(3/2)/c

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (28) = 56\).
time = 0.29, size = 70, normalized size = 2.19 \begin {gather*} \frac {{\left (b^{4} c^{\frac {3}{2}} x^{8} + 4 \, a b^{3} c^{\frac {3}{2}} x^{6} + 6 \, a^{2} b^{2} c^{\frac {3}{2}} x^{4} + 4 \, a^{3} b c^{\frac {3}{2}} x^{2} + a^{4} c^{\frac {3}{2}}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}}}{11 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*(b^4*c^(3/2)*x^8 + 4*a*b^3*c^(3/2)*x^6 + 6*a^2*b^2*c^(3/2)*x^4 + 4*a^3*b*c^(3/2)*x^2 + a^4*c^(3/2))*(b*x^
2 + a)^(3/2)/b

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (28) = 56\).
time = 0.36, size = 87, normalized size = 2.72 \begin {gather*} \frac {{\left (b^{4} c x^{8} + 4 \, a b^{3} c x^{6} + 6 \, a^{2} b^{2} c x^{4} + 4 \, a^{3} b c x^{2} + a^{4} c\right )} \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{11 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*(b^4*c*x^8 + 4*a*b^3*c*x^6 + 6*a^2*b^2*c*x^4 + 4*a^3*b*c*x^2 + a^4*c)*sqrt(b^3*c*x^6 + 3*a*b^2*c*x^4 + 3*
a^2*b*c*x^2 + a^3*c)/b

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (c \left (a + b x^{2}\right )^{3}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(c*(a + b*x**2)**3)**(3/2), x)

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Giac [A]
time = 4.30, size = 28, normalized size = 0.88 \begin {gather*} \frac {{\left (b c x^{2} + a c\right )}^{\frac {11}{2}} \mathrm {sgn}\left (b x^{2} + a\right )}{11 \, b c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*(b*c*x^2 + a*c)^(11/2)*sgn(b*x^2 + a)/(b*c^4)

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Mupad [B]
time = 2.71, size = 62, normalized size = 1.94 \begin {gather*} \sqrt {c\,{\left (b\,x^2+a\right )}^3}\,\left (\frac {a^4\,c}{11\,b}+\frac {4\,a^3\,c\,x^2}{11}+\frac {b^3\,c\,x^8}{11}+\frac {6\,a^2\,b\,c\,x^4}{11}+\frac {4\,a\,b^2\,c\,x^6}{11}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*(a + b*x^2)^3)^(3/2),x)

[Out]

(c*(a + b*x^2)^3)^(1/2)*((a^4*c)/(11*b) + (4*a^3*c*x^2)/11 + (b^3*c*x^8)/11 + (6*a^2*b*c*x^4)/11 + (4*a*b^2*c*
x^6)/11)

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