3.3.51 \(\int x^3 (c \sqrt {a+b x^2})^{3/2} \, dx\) [251]

Optimal. Leaf size=66 \[ -\frac {2 a \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right )}{7 b^2}+\frac {2 \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right )^2}{11 b^2} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 66, normalized size of antiderivative = 1.00, 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 = {1973, 272, 45} \begin {gather*} \frac {2 \left (a+b x^2\right )^2 \left (c \sqrt {a+b x^2}\right )^{3/2}}{11 b^2}-\frac {2 a \left (a+b x^2\right ) \left (c \sqrt {a+b x^2}\right )^{3/2}}{7 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(c*Sqrt[a + b*x^2])^(3/2),x]

[Out]

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

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Mathematica [A]
time = 0.03, size = 41, normalized size = 0.62 \begin {gather*} \frac {2 \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right ) \left (-4 a+7 b x^2\right )}{77 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2)*(-4*a + 7*b*x^2))/(77*b^2)

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Maple [A]
time = 0.01, size = 36, normalized size = 0.55

method	result	size
gosper	\(-\frac {2 \left (b \,x^{2}+a \right ) \left (-7 b \,x^{2}+4 a \right ) \left (c \sqrt {b \,x^{2}+a}\right )^{\frac {3}{2}}}{77 b^{2}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c*(b*x^2+a)^(1/2))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/77*(b*x^2+a)*(-7*b*x^2+4*a)*(c*(b*x^2+a)^(1/2))^(3/2)/b^2

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Maxima [A]
time = 0.28, size = 43, normalized size = 0.65 \begin {gather*} -\frac {2 \, {\left (11 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {7}{2}} a c^{2} - 7 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {11}{2}}\right )}}{77 \, b^{2} c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/77*(11*(sqrt(b*x^2 + a)*c)^(7/2)*a*c^2 - 7*(sqrt(b*x^2 + a)*c)^(11/2))/(b^2*c^4)

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Fricas [A]
time = 0.38, size = 51, normalized size = 0.77 \begin {gather*} \frac {2 \, {\left (7 \, b^{2} c x^{4} + 3 \, a b c x^{2} - 4 \, a^{2} c\right )} \sqrt {b x^{2} + a} \sqrt {\sqrt {b x^{2} + a} c}}{77 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/77*(7*b^2*c*x^4 + 3*a*b*c*x^2 - 4*a^2*c)*sqrt(b*x^2 + a)*sqrt(sqrt(b*x^2 + a)*c)/b^2

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Sympy [A]
time = 3.50, size = 87, normalized size = 1.32 \begin {gather*} \begin {cases} - \frac {8 a^{2} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{77 b^{2}} + \frac {6 a x^{2} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{77 b} + \frac {2 x^{4} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{11} & \text {for}\: b \neq 0 \\\frac {x^{4} \left (\sqrt {a} c\right )^{\frac {3}{2}}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-8*a**2*(c*sqrt(a + b*x**2))**(3/2)/(77*b**2) + 6*a*x**2*(c*sqrt(a + b*x**2))**(3/2)/(77*b) + 2*x**
4*(c*sqrt(a + b*x**2))**(3/2)/11, Ne(b, 0)), (x**4*(sqrt(a)*c)**(3/2)/4, True))

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Giac [A]
time = 4.55, size = 81, normalized size = 1.23 \begin {gather*} \frac {2 \, {\left (\frac {11 \, {\left (3 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} - 7 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a\right )} a}{b} + \frac {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}}{b}\right )} c^{\frac {3}{2}}}{231 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/231*(11*(3*(b*x^2 + a)^(7/4) - 7*(b*x^2 + a)^(3/4)*a)*a/b + (21*(b*x^2 + a)^(11/4) - 66*(b*x^2 + a)^(7/4)*a
+ 77*(b*x^2 + a)^(3/4)*a^2)/b)*c^(3/2)/b

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Mupad [B]
time = 2.89, size = 67, normalized size = 1.02 \begin {gather*} \sqrt {c\,\sqrt {b\,x^2+a}}\,\left (\frac {2\,c\,x^4\,\sqrt {b\,x^2+a}}{11}-\frac {8\,a^2\,c\,\sqrt {b\,x^2+a}}{77\,b^2}+\frac {6\,a\,c\,x^2\,\sqrt {b\,x^2+a}}{77\,b}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*(a + b*x^2)^(1/2))^(1/2)*((2*c*x^4*(a + b*x^2)^(1/2))/11 - (8*a^2*c*(a + b*x^2)^(1/2))/(77*b^2) + (6*a*c*x^
2*(a + b*x^2)^(1/2))/(77*b))

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