3.2.32 \(\int x^4 (b x^2+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=106 \[ -\frac {16 b^3 \left (b x^2+c x^4\right )^{5/2}}{1155 c^4 x^5}+\frac {8 b^2 \left (b x^2+c x^4\right )^{5/2}}{231 c^3 x^3}-\frac {2 b \left (b x^2+c x^4\right )^{5/2}}{33 c^2 x}+\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c} \]

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Rubi [A]  time = 0.20, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2016, 2002, 2014} \begin {gather*} -\frac {16 b^3 \left (b x^2+c x^4\right )^{5/2}}{1155 c^4 x^5}+\frac {8 b^2 \left (b x^2+c x^4\right )^{5/2}}{231 c^3 x^3}-\frac {2 b \left (b x^2+c x^4\right )^{5/2}}{33 c^2 x}+\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(-16*b^3*(b*x^2 + c*x^4)^(5/2))/(1155*c^4*x^5) + (8*b^2*(b*x^2 + c*x^4)^(5/2))/(231*c^3*x^3) - (2*b*(b*x^2 + c
*x^4)^(5/2))/(33*c^2*x) + (x*(b*x^2 + c*x^4)^(5/2))/(11*c)

Rule 2002

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j -
1)), x] - Dist[(b*(n*p + n - j + 1))/(a*(j*p + 1)), Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j,
n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 0]

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps

\begin {align*} \int x^4 \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {(6 b) \int x^2 \left (b x^2+c x^4\right )^{3/2} \, dx}{11 c}\\ &=-\frac {2 b \left (b x^2+c x^4\right )^{5/2}}{33 c^2 x}+\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}+\frac {\left (8 b^2\right ) \int \left (b x^2+c x^4\right )^{3/2} \, dx}{33 c^2}\\ &=\frac {8 b^2 \left (b x^2+c x^4\right )^{5/2}}{231 c^3 x^3}-\frac {2 b \left (b x^2+c x^4\right )^{5/2}}{33 c^2 x}+\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {\left (16 b^3\right ) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^2} \, dx}{231 c^3}\\ &=-\frac {16 b^3 \left (b x^2+c x^4\right )^{5/2}}{1155 c^4 x^5}+\frac {8 b^2 \left (b x^2+c x^4\right )^{5/2}}{231 c^3 x^3}-\frac {2 b \left (b x^2+c x^4\right )^{5/2}}{33 c^2 x}+\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 64, normalized size = 0.60 \begin {gather*} \frac {x \left (b+c x^2\right )^3 \left (-16 b^3+40 b^2 c x^2-70 b c^2 x^4+105 c^3 x^6\right )}{1155 c^4 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x*(b + c*x^2)^3*(-16*b^3 + 40*b^2*c*x^2 - 70*b*c^2*x^4 + 105*c^3*x^6))/(1155*c^4*Sqrt[x^2*(b + c*x^2)])

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IntegrateAlgebraic [A]  time = 0.34, size = 79, normalized size = 0.75 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-16 b^5+8 b^4 c x^2-6 b^3 c^2 x^4+5 b^2 c^3 x^6+140 b c^4 x^8+105 c^5 x^{10}\right )}{1155 c^4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^4*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-16*b^5 + 8*b^4*c*x^2 - 6*b^3*c^2*x^4 + 5*b^2*c^3*x^6 + 140*b*c^4*x^8 + 105*c^5*x^10))/(
1155*c^4*x)

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fricas [A]  time = 0.76, size = 75, normalized size = 0.71 \begin {gather*} \frac {{\left (105 \, c^{5} x^{10} + 140 \, b c^{4} x^{8} + 5 \, b^{2} c^{3} x^{6} - 6 \, b^{3} c^{2} x^{4} + 8 \, b^{4} c x^{2} - 16 \, b^{5}\right )} \sqrt {c x^{4} + b x^{2}}}{1155 \, c^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

1/1155*(105*c^5*x^10 + 140*b*c^4*x^8 + 5*b^2*c^3*x^6 - 6*b^3*c^2*x^4 + 8*b^4*c*x^2 - 16*b^5)*sqrt(c*x^4 + b*x^
2)/(c^4*x)

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giac [A]  time = 0.16, size = 76, normalized size = 0.72 \begin {gather*} \frac {16 \, b^{\frac {11}{2}} \mathrm {sgn}\relax (x)}{1155 \, c^{4}} + \frac {105 \, {\left (c x^{2} + b\right )}^{\frac {11}{2}} \mathrm {sgn}\relax (x) - 385 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} b \mathrm {sgn}\relax (x) + 495 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} b^{2} \mathrm {sgn}\relax (x) - 231 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} b^{3} \mathrm {sgn}\relax (x)}{1155 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

16/1155*b^(11/2)*sgn(x)/c^4 + 1/1155*(105*(c*x^2 + b)^(11/2)*sgn(x) - 385*(c*x^2 + b)^(9/2)*b*sgn(x) + 495*(c*
x^2 + b)^(7/2)*b^2*sgn(x) - 231*(c*x^2 + b)^(5/2)*b^3*sgn(x))/c^4

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maple [A]  time = 0.01, size = 61, normalized size = 0.58 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-105 c^{3} x^{6}+70 b \,c^{2} x^{4}-40 b^{2} c \,x^{2}+16 b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{1155 c^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(c*x^4+b*x^2)^(3/2),x)

[Out]

-1/1155*(c*x^2+b)*(-105*c^3*x^6+70*b*c^2*x^4-40*b^2*c*x^2+16*b^3)*(c*x^4+b*x^2)^(3/2)/c^4/x^3

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maxima [A]  time = 1.50, size = 68, normalized size = 0.64 \begin {gather*} \frac {{\left (105 \, c^{5} x^{10} + 140 \, b c^{4} x^{8} + 5 \, b^{2} c^{3} x^{6} - 6 \, b^{3} c^{2} x^{4} + 8 \, b^{4} c x^{2} - 16 \, b^{5}\right )} \sqrt {c x^{2} + b}}{1155 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

1/1155*(105*c^5*x^10 + 140*b*c^4*x^8 + 5*b^2*c^3*x^6 - 6*b^3*c^2*x^4 + 8*b^4*c*x^2 - 16*b^5)*sqrt(c*x^2 + b)/c
^4

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mupad [B]  time = 4.30, size = 62, normalized size = 0.58 \begin {gather*} -\frac {{\left (c\,x^2+b\right )}^2\,\sqrt {c\,x^4+b\,x^2}\,\left (16\,b^3-40\,b^2\,c\,x^2+70\,b\,c^2\,x^4-105\,c^3\,x^6\right )}{1155\,c^4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x^2 + c*x^4)^(3/2),x)

[Out]

-((b + c*x^2)^2*(b*x^2 + c*x^4)^(1/2)*(16*b^3 - 105*c^3*x^6 - 40*b^2*c*x^2 + 70*b*c^2*x^4))/(1155*c^4*x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{4} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**4*(x**2*(b + c*x**2))**(3/2), x)

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