3.250 \(\int x^5 (c \sqrt {a+b x^2})^{3/2} \, dx\)

Optimal. Leaf size=102 \[ \frac {2 a^2 \left (a+b x^2\right ) \left (c \sqrt {a+b x^2}\right )^{3/2}}{7 b^3}+\frac {2 \left (a+b x^2\right )^3 \left (c \sqrt {a+b x^2}\right )^{3/2}}{15 b^3}-\frac {4 a \left (a+b x^2\right )^2 \left (c \sqrt {a+b x^2}\right )^{3/2}}{11 b^3} \]

[Out]

2/7*a^2*(b*x^2+a)*(c*(b*x^2+a)^(1/2))^(3/2)/b^3-4/11*a*(b*x^2+a)^2*(c*(b*x^2+a)^(1/2))^(3/2)/b^3+2/15*(b*x^2+a
)^3*(c*(b*x^2+a)^(1/2))^(3/2)/b^3

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Rubi [A]  time = 0.16, antiderivative size = 113, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6720, 266, 43} \[ \frac {2 a^2 c \left (a+b x^2\right )^{3/2} \sqrt {c \sqrt {a+b x^2}}}{7 b^3}+\frac {2 c \left (a+b x^2\right )^{7/2} \sqrt {c \sqrt {a+b x^2}}}{15 b^3}-\frac {4 a c \left (a+b x^2\right )^{5/2} \sqrt {c \sqrt {a+b x^2}}}{11 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(c*Sqrt[a + b*x^2])^(3/2),x]

[Out]

(2*a^2*c*Sqrt[c*Sqrt[a + b*x^2]]*(a + b*x^2)^(3/2))/(7*b^3) - (4*a*c*Sqrt[c*Sqrt[a + b*x^2]]*(a + b*x^2)^(5/2)
)/(11*b^3) + (2*c*Sqrt[c*Sqrt[a + b*x^2]]*(a + b*x^2)^(7/2))/(15*b^3)

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])

Rule 266

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 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

\begin {align*} \int x^5 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx &=\frac {\left (c \sqrt {c \sqrt {a+b x^2}}\right ) \int x^5 \left (a+b x^2\right )^{3/4} \, dx}{\sqrt [4]{a+b x^2}}\\ &=\frac {\left (c \sqrt {c \sqrt {a+b x^2}}\right ) \operatorname {Subst}\left (\int x^2 (a+b x)^{3/4} \, dx,x,x^2\right )}{2 \sqrt [4]{a+b x^2}}\\ &=\frac {\left (c \sqrt {c \sqrt {a+b x^2}}\right ) \operatorname {Subst}\left (\int \left (\frac {a^2 (a+b x)^{3/4}}{b^2}-\frac {2 a (a+b x)^{7/4}}{b^2}+\frac {(a+b x)^{11/4}}{b^2}\right ) \, dx,x,x^2\right )}{2 \sqrt [4]{a+b x^2}}\\ &=\frac {2 a^2 c \sqrt {c \sqrt {a+b x^2}} \left (a+b x^2\right )^{3/2}}{7 b^3}-\frac {4 a c \sqrt {c \sqrt {a+b x^2}} \left (a+b x^2\right )^{5/2}}{11 b^3}+\frac {2 c \sqrt {c \sqrt {a+b x^2}} \left (a+b x^2\right )^{7/2}}{15 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 52, normalized size = 0.51 \[ \frac {2 \left (a+b x^2\right ) \left (32 a^2-56 a b x^2+77 b^2 x^4\right ) \left (c \sqrt {a+b x^2}\right )^{3/2}}{1155 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2)*(32*a^2 - 56*a*b*x^2 + 77*b^2*x^4))/(1155*b^3)

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fricas [A]  time = 0.49, size = 63, normalized size = 0.62 \[ \frac {2 \, {\left (77 \, b^{3} c x^{6} + 21 \, a b^{2} c x^{4} - 24 \, a^{2} b c x^{2} + 32 \, a^{3} c\right )} \sqrt {b x^{2} + a} \sqrt {\sqrt {b x^{2} + a} c}}{1155 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*(b*x^2+a)^(1/2))^(3/2),x, algorithm="fricas")

[Out]

2/1155*(77*b^3*c*x^6 + 21*a*b^2*c*x^4 - 24*a^2*b*c*x^2 + 32*a^3*c)*sqrt(b*x^2 + a)*sqrt(sqrt(b*x^2 + a)*c)/b^3

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giac [A]  time = 0.24, size = 109, normalized size = 1.07 \[ \frac {2 \, c^{\frac {3}{2}} {\left (\frac {5 \, {\left (21 \, {\left (b x^{2} + a\right )}^{\frac {11}{4}} - 66 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} a + 77 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a^{2}\right )} a}{b^{2}} + \frac {77 \, {\left (b x^{2} + a\right )}^{\frac {15}{4}} - 315 \, {\left (b x^{2} + a\right )}^{\frac {11}{4}} a + 495 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} a^{2} - 385 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a^{3}}{b^{2}}\right )}}{1155 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*(b*x^2+a)^(1/2))^(3/2),x, algorithm="giac")

[Out]

2/1155*c^(3/2)*(5*(21*(b*x^2 + a)^(11/4) - 66*(b*x^2 + a)^(7/4)*a + 77*(b*x^2 + a)^(3/4)*a^2)*a/b^2 + (77*(b*x
^2 + a)^(15/4) - 315*(b*x^2 + a)^(11/4)*a + 495*(b*x^2 + a)^(7/4)*a^2 - 385*(b*x^2 + a)^(3/4)*a^3)/b^2)/b

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maple [A]  time = 0.01, size = 47, normalized size = 0.46 \[ \frac {2 \left (b \,x^{2}+a \right ) \left (77 x^{4} b^{2}-56 a b \,x^{2}+32 a^{2}\right ) \left (\sqrt {b \,x^{2}+a}\, c \right )^{\frac {3}{2}}}{1155 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*((b*x^2+a)^(1/2)*c)^(3/2),x)

[Out]

2/1155*(b*x^2+a)*(77*b^2*x^4-56*a*b*x^2+32*a^2)*((b*x^2+a)^(1/2)*c)^(3/2)/b^3

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maxima [A]  time = 0.93, size = 64, normalized size = 0.63 \[ \frac {2 \, {\left (165 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {7}{2}} a^{2} c^{4} - 210 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {11}{2}} a c^{2} + 77 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {15}{2}}\right )}}{1155 \, b^{3} c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*(b*x^2+a)^(1/2))^(3/2),x, algorithm="maxima")

[Out]

2/1155*(165*(sqrt(b*x^2 + a)*c)^(7/2)*a^2*c^4 - 210*(sqrt(b*x^2 + a)*c)^(11/2)*a*c^2 + 77*(sqrt(b*x^2 + a)*c)^
(15/2))/(b^3*c^6)

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mupad [B]  time = 2.90, size = 88, normalized size = 0.86 \[ \sqrt {c\,\sqrt {b\,x^2+a}}\,\left (\frac {2\,c\,x^6\,\sqrt {b\,x^2+a}}{15}+\frac {64\,a^3\,c\,\sqrt {b\,x^2+a}}{1155\,b^3}+\frac {2\,a\,c\,x^4\,\sqrt {b\,x^2+a}}{55\,b}-\frac {16\,a^2\,c\,x^2\,\sqrt {b\,x^2+a}}{385\,b^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(c*(a + b*x^2)^(1/2))^(3/2),x)

[Out]

(c*(a + b*x^2)^(1/2))^(1/2)*((2*c*x^6*(a + b*x^2)^(1/2))/15 + (64*a^3*c*(a + b*x^2)^(1/2))/(1155*b^3) + (2*a*c
*x^4*(a + b*x^2)^(1/2))/(55*b) - (16*a^2*c*x^2*(a + b*x^2)^(1/2))/(385*b^2))

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sympy [A]  time = 88.61, size = 116, normalized size = 1.14 \[ \begin {cases} \frac {64 a^{3} c^{\frac {3}{2}} \left (a + b x^{2}\right )^{\frac {3}{4}}}{1155 b^{3}} - \frac {16 a^{2} c^{\frac {3}{2}} x^{2} \left (a + b x^{2}\right )^{\frac {3}{4}}}{385 b^{2}} + \frac {2 a c^{\frac {3}{2}} x^{4} \left (a + b x^{2}\right )^{\frac {3}{4}}}{55 b} + \frac {2 c^{\frac {3}{2}} x^{6} \left (a + b x^{2}\right )^{\frac {3}{4}}}{15} & \text {for}\: b \neq 0 \\\frac {x^{6} \left (\sqrt {a} c\right )^{\frac {3}{2}}}{6} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(c*(b*x**2+a)**(1/2))**(3/2),x)

[Out]

Piecewise((64*a**3*c**(3/2)*(a + b*x**2)**(3/4)/(1155*b**3) - 16*a**2*c**(3/2)*x**2*(a + b*x**2)**(3/4)/(385*b
**2) + 2*a*c**(3/2)*x**4*(a + b*x**2)**(3/4)/(55*b) + 2*c**(3/2)*x**6*(a + b*x**2)**(3/4)/15, Ne(b, 0)), (x**6
*(sqrt(a)*c)**(3/2)/6, True))

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