3.3.28 \(\int x^5 (c (a+b x^2)^2)^{3/2} \, dx\) [228]

Optimal. Leaf size=143 \[ \frac {a^3 c x^6 \sqrt {c \left (a+b x^2\right )^2}}{6 \left (a+b x^2\right )}+\frac {3 a^2 b c x^8 \sqrt {c \left (a+b x^2\right )^2}}{8 \left (a+b x^2\right )}+\frac {3 a b^2 c x^{10} \sqrt {c \left (a+b x^2\right )^2}}{10 \left (a+b x^2\right )}+\frac {b^3 c x^{12} \sqrt {c \left (a+b x^2\right )^2}}{12 \left (a+b x^2\right )} \]

[Out]

1/6*a^3*c*x^6*(c*(b*x^2+a)^2)^(1/2)/(b*x^2+a)+3/8*a^2*b*c*x^8*(c*(b*x^2+a)^2)^(1/2)/(b*x^2+a)+3/10*a*b^2*c*x^1
0*(c*(b*x^2+a)^2)^(1/2)/(b*x^2+a)+1/12*b^3*c*x^12*(c*(b*x^2+a)^2)^(1/2)/(b*x^2+a)

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Rubi [A]
time = 0.05, antiderivative size = 143, 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 = {1973, 272, 45} \begin {gather*} \frac {a^3 c x^6 \sqrt {c \left (a+b x^2\right )^2}}{6 \left (a+b x^2\right )}+\frac {3 a^2 b c x^8 \sqrt {c \left (a+b x^2\right )^2}}{8 \left (a+b x^2\right )}+\frac {b^3 c x^{12} \sqrt {c \left (a+b x^2\right )^2}}{12 \left (a+b x^2\right )}+\frac {3 a b^2 c x^{10} \sqrt {c \left (a+b x^2\right )^2}}{10 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*(c*(a + b*x^2)^2)^(3/2),x]

[Out]

(a^3*c*x^6*Sqrt[c*(a + b*x^2)^2])/(6*(a + b*x^2)) + (3*a^2*b*c*x^8*Sqrt[c*(a + b*x^2)^2])/(8*(a + b*x^2)) + (3
*a*b^2*c*x^10*Sqrt[c*(a + b*x^2)^2])/(10*(a + b*x^2)) + (b^3*c*x^12*Sqrt[c*(a + b*x^2)^2])/(12*(a + b*x^2))

Rule 45

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])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1973

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/
a))^(p*q)], Int[u*(1 + b*(x^n/a))^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]

Rubi steps

\begin {align*} \int x^5 \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx &=\int x^5 \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2} \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int x^2 \left (a^2 c+2 a b c x+b^2 c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac {\sqrt {a^2 c+2 a b c x^2+b^2 c x^4} \text {Subst}\left (\int \left (\frac {a^2 \left (a b c+b^2 c x\right )^3}{b^2}-\frac {2 a \left (a b c+b^2 c x\right )^4}{b^3 c}+\frac {\left (a b c+b^2 c x\right )^5}{b^4 c^2}\right ) \, dx,x,x^2\right )}{2 b^2 c \left (a b c+b^2 c x^2\right )}\\ &=\frac {a^2 c \left (a+b x^2\right )^3 \sqrt {a^2 c+2 a b c x^2+b^2 c x^4}}{8 b^3}-\frac {a c \left (a+b x^2\right )^4 \sqrt {a^2 c+2 a b c x^2+b^2 c x^4}}{5 b^3}+\frac {c \left (a+b x^2\right )^5 \sqrt {a^2 c+2 a b c x^2+b^2 c x^4}}{12 b^3}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 63, normalized size = 0.44 \begin {gather*} \frac {x^6 \left (c \left (a+b x^2\right )^2\right )^{3/2} \left (20 a^3+45 a^2 b x^2+36 a b^2 x^4+10 b^3 x^6\right )}{120 \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*(c*(a + b*x^2)^2)^(3/2),x]

[Out]

(x^6*(c*(a + b*x^2)^2)^(3/2)*(20*a^3 + 45*a^2*b*x^2 + 36*a*b^2*x^4 + 10*b^3*x^6))/(120*(a + b*x^2)^3)

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Maple [A]
time = 0.06, size = 60, normalized size = 0.42

method	result	size
gosper	\(\frac {x^{6} \left (10 b^{3} x^{6}+36 a \,b^{2} x^{4}+45 a^{2} b \,x^{2}+20 a^{3}\right ) \left (c \left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{120 \left (b \,x^{2}+a \right )^{3}}\)	\(60\)
default	\(\frac {x^{6} \left (10 b^{3} x^{6}+36 a \,b^{2} x^{4}+45 a^{2} b \,x^{2}+20 a^{3}\right ) \left (c \left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{120 \left (b \,x^{2}+a \right )^{3}}\)	\(60\)
trager	\(\frac {c \,x^{6} \left (10 b^{3} x^{6}+36 a \,b^{2} x^{4}+45 a^{2} b \,x^{2}+20 a^{3}\right ) \sqrt {b^{2} c \,x^{4}+2 a b c \,x^{2}+a^{2} c}}{120 b \,x^{2}+120 a}\)	\(72\)
risch	\(\frac {a^{3} c \,x^{6} \sqrt {c \left (b \,x^{2}+a \right )^{2}}}{6 b \,x^{2}+6 a}+\frac {3 a^{2} b c \,x^{8} \sqrt {c \left (b \,x^{2}+a \right )^{2}}}{8 \left (b \,x^{2}+a \right )}+\frac {3 a \,b^{2} c \,x^{10} \sqrt {c \left (b \,x^{2}+a \right )^{2}}}{10 \left (b \,x^{2}+a \right )}+\frac {b^{3} c \,x^{12} \sqrt {c \left (b \,x^{2}+a \right )^{2}}}{12 b \,x^{2}+12 a}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(c*(b*x^2+a)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/120*x^6*(10*b^3*x^6+36*a*b^2*x^4+45*a^2*b*x^2+20*a^3)*(c*(b*x^2+a)^2)^(3/2)/(b*x^2+a)^3

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Maxima [A]
time = 0.29, size = 136, normalized size = 0.95 \begin {gather*} \frac {{\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )}^{\frac {3}{2}} a^{2} x^{2}}{8 \, b^{2}} + \frac {{\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )}^{\frac {3}{2}} a^{3}}{8 \, b^{3}} + \frac {{\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )}^{\frac {5}{2}} x^{2}}{12 \, b^{2} c} - \frac {7 \, {\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )}^{\frac {5}{2}} a}{60 \, b^{3} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*(b*x^2+a)^2)^(3/2),x, algorithm="maxima")

[Out]

1/8*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)^(3/2)*a^2*x^2/b^2 + 1/8*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)^(3/2)*a^3/b^3
+ 1/12*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)^(5/2)*x^2/(b^2*c) - 7/60*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)^(5/2)*a/(b
^3*c)

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Fricas [A]
time = 0.33, size = 74, normalized size = 0.52 \begin {gather*} \frac {{\left (10 \, b^{3} c x^{12} + 36 \, a b^{2} c x^{10} + 45 \, a^{2} b c x^{8} + 20 \, a^{3} c x^{6}\right )} \sqrt {b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}}{120 \, {\left (b x^{2} + a\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*(b*x^2+a)^2)^(3/2),x, algorithm="fricas")

[Out]

1/120*(10*b^3*c*x^12 + 36*a*b^2*c*x^10 + 45*a^2*b*c*x^8 + 20*a^3*c*x^6)*sqrt(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)/
(b*x^2 + a)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{5} \left (c \left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(c*(b*x**2+a)**2)**(3/2),x)

[Out]

Integral(x**5*(c*(a + b*x**2)**2)**(3/2), x)

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Giac [A]
time = 6.01, size = 72, normalized size = 0.50 \begin {gather*} \frac {1}{120} \, {\left (10 \, b^{3} x^{12} \mathrm {sgn}\left (b x^{2} + a\right ) + 36 \, a b^{2} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 45 \, a^{2} b x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 20 \, a^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right )\right )} c^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*(b*x^2+a)^2)^(3/2),x, algorithm="giac")

[Out]

1/120*(10*b^3*x^12*sgn(b*x^2 + a) + 36*a*b^2*x^10*sgn(b*x^2 + a) + 45*a^2*b*x^8*sgn(b*x^2 + a) + 20*a^3*x^6*sg
n(b*x^2 + a))*c^(3/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^5\,{\left (c\,{\left (b\,x^2+a\right )}^2\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(c*(a + b*x^2)^2)^(3/2),x)

[Out]

int(x^5*(c*(a + b*x^2)^2)^(3/2), x)

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