3.3.52 \(\int x (c \sqrt {a+b x^2})^{3/2} \, dx\) [252]

Optimal. Leaf size=36 \[ \frac {2 c \sqrt {c \sqrt {a+b x^2}} \left (a+b x^2\right )^{3/2}}{7 b} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1605, 15, 30} \begin {gather*} \frac {2 c \left (a+b x^2\right )^{3/2} \sqrt {c \sqrt {a+b x^2}}}{7 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(c*Sqrt[a + b*x^2])^(3/2),x]

[Out]

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

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Mathematica [A]
time = 0.01, size = 31, normalized size = 0.86 \begin {gather*} \frac {2 \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right )}{7 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.04, size = 26, normalized size = 0.72

method	result	size
gosper	\(\frac {2 \left (b \,x^{2}+a \right ) \left (c \sqrt {b \,x^{2}+a}\right )^{\frac {3}{2}}}{7 b}\)	\(26\)
derivativedivides	\(\frac {2 \left (b \,x^{2}+a \right ) \left (c \sqrt {b \,x^{2}+a}\right )^{\frac {3}{2}}}{7 b}\)	\(26\)
default	\(\frac {2 \left (b \,x^{2}+a \right ) \left (c \sqrt {b \,x^{2}+a}\right )^{\frac {3}{2}}}{7 b}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 25, normalized size = 0.69 \begin {gather*} \frac {2 \, {\left (b x^{2} + a\right )} \left (\sqrt {b x^{2} + a} c\right )^{\frac {3}{2}}}{7 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 37, normalized size = 1.03 \begin {gather*} \frac {2 \, {\left (b c x^{2} + a c\right )} \sqrt {b x^{2} + a} \sqrt {\sqrt {b x^{2} + a} c}}{7 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(b*c*x^2 + a*c)*sqrt(b*x^2 + a)*sqrt(sqrt(b*x^2 + a)*c)/b

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Sympy [A]
time = 1.70, size = 58, normalized size = 1.61 \begin {gather*} \begin {cases} \frac {2 a \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{7 b} + \frac {2 x^{2} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{7} & \text {for}\: b \neq 0 \\\frac {x^{2} \left (\sqrt {a} c\right )^{\frac {3}{2}}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*a*(c*sqrt(a + b*x**2))**(3/2)/(7*b) + 2*x**2*(c*sqrt(a + b*x**2))**(3/2)/7, Ne(b, 0)), (x**2*(sqr
t(a)*c)**(3/2)/2, True))

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Giac [A]
time = 4.10, size = 17, normalized size = 0.47 \begin {gather*} \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} c^{\frac {3}{2}}}{7 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(b*x^2 + a)^(7/4)*c^(3/2)/b

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Mupad [B]
time = 2.77, size = 28, normalized size = 0.78 \begin {gather*} \frac {2\,c\,{\left (b\,x^2+a\right )}^{3/2}\,\sqrt {c\,\sqrt {b\,x^2+a}}}{7\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*(a + b*x^2)^(1/2))^(3/2),x)

[Out]

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

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