3.9.82 \(\int x \sqrt {c x^2} (a+b x)^n \, dx\)

Optimal. Leaf size=96 \[ \frac {a^2 \sqrt {c x^2} (a+b x)^{n+1}}{b^3 (n+1) x}-\frac {2 a \sqrt {c x^2} (a+b x)^{n+2}}{b^3 (n+2) x}+\frac {\sqrt {c x^2} (a+b x)^{n+3}}{b^3 (n+3) x} \]

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Rubi [A]  time = 0.03, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {15, 43} \begin {gather*} \frac {a^2 \sqrt {c x^2} (a+b x)^{n+1}}{b^3 (n+1) x}-\frac {2 a \sqrt {c x^2} (a+b x)^{n+2}}{b^3 (n+2) x}+\frac {\sqrt {c x^2} (a+b x)^{n+3}}{b^3 (n+3) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[c*x^2]*(a + b*x)^n,x]

[Out]

(a^2*Sqrt[c*x^2]*(a + b*x)^(1 + n))/(b^3*(1 + n)*x) - (2*a*Sqrt[c*x^2]*(a + b*x)^(2 + n))/(b^3*(2 + n)*x) + (S
qrt[c*x^2]*(a + b*x)^(3 + n))/(b^3*(3 + n)*x)

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 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])

Rubi steps

\begin {align*} \int x \sqrt {c x^2} (a+b x)^n \, dx &=\frac {\sqrt {c x^2} \int x^2 (a+b x)^n \, dx}{x}\\ &=\frac {\sqrt {c x^2} \int \left (\frac {a^2 (a+b x)^n}{b^2}-\frac {2 a (a+b x)^{1+n}}{b^2}+\frac {(a+b x)^{2+n}}{b^2}\right ) \, dx}{x}\\ &=\frac {a^2 \sqrt {c x^2} (a+b x)^{1+n}}{b^3 (1+n) x}-\frac {2 a \sqrt {c x^2} (a+b x)^{2+n}}{b^3 (2+n) x}+\frac {\sqrt {c x^2} (a+b x)^{3+n}}{b^3 (3+n) x}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 68, normalized size = 0.71 \begin {gather*} \frac {c x (a+b x)^{n+1} \left (2 a^2-2 a b (n+1) x+b^2 \left (n^2+3 n+2\right ) x^2\right )}{b^3 (n+1) (n+2) (n+3) \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x*(a + b*x)^(1 + n)*(2*a^2 - 2*a*b*(1 + n)*x + b^2*(2 + 3*n + n^2)*x^2))/(b^3*(1 + n)*(2 + n)*(3 + n)*Sqrt[
c*x^2])

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IntegrateAlgebraic [F]  time = 0.20, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {c x^2} (a+b x)^n \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x*Sqrt[c*x^2]*(a + b*x)^n,x]

[Out]

Defer[IntegrateAlgebraic][x*Sqrt[c*x^2]*(a + b*x)^n, x]

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fricas [A]  time = 1.56, size = 106, normalized size = 1.10 \begin {gather*} -\frac {{\left (2 \, a^{2} b n x - {\left (b^{3} n^{2} + 3 \, b^{3} n + 2 \, b^{3}\right )} x^{3} - 2 \, a^{3} - {\left (a b^{2} n^{2} + a b^{2} n\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{{\left (b^{3} n^{3} + 6 \, b^{3} n^{2} + 11 \, b^{3} n + 6 \, b^{3}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^2*b*n*x - (b^3*n^2 + 3*b^3*n + 2*b^3)*x^3 - 2*a^3 - (a*b^2*n^2 + a*b^2*n)*x^2)*sqrt(c*x^2)*(b*x + a)^n/(
(b^3*n^3 + 6*b^3*n^2 + 11*b^3*n + 6*b^3)*x)

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giac [B]  time = 0.96, size = 200, normalized size = 2.08 \begin {gather*} -{\left (\frac {2 \, a^{3} a^{n} \mathrm {sgn}\relax (x)}{b^{3} n^{3} + 6 \, b^{3} n^{2} + 11 \, b^{3} n + 6 \, b^{3}} - \frac {{\left (b x + a\right )}^{n} b^{3} n^{2} x^{3} \mathrm {sgn}\relax (x) + {\left (b x + a\right )}^{n} a b^{2} n^{2} x^{2} \mathrm {sgn}\relax (x) + 3 \, {\left (b x + a\right )}^{n} b^{3} n x^{3} \mathrm {sgn}\relax (x) + {\left (b x + a\right )}^{n} a b^{2} n x^{2} \mathrm {sgn}\relax (x) + 2 \, {\left (b x + a\right )}^{n} b^{3} x^{3} \mathrm {sgn}\relax (x) - 2 \, {\left (b x + a\right )}^{n} a^{2} b n x \mathrm {sgn}\relax (x) + 2 \, {\left (b x + a\right )}^{n} a^{3} \mathrm {sgn}\relax (x)}{b^{3} n^{3} + 6 \, b^{3} n^{2} + 11 \, b^{3} n + 6 \, b^{3}}\right )} \sqrt {c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^3*a^n*sgn(x)/(b^3*n^3 + 6*b^3*n^2 + 11*b^3*n + 6*b^3) - ((b*x + a)^n*b^3*n^2*x^3*sgn(x) + (b*x + a)^n*a*
b^2*n^2*x^2*sgn(x) + 3*(b*x + a)^n*b^3*n*x^3*sgn(x) + (b*x + a)^n*a*b^2*n*x^2*sgn(x) + 2*(b*x + a)^n*b^3*x^3*s
gn(x) - 2*(b*x + a)^n*a^2*b*n*x*sgn(x) + 2*(b*x + a)^n*a^3*sgn(x))/(b^3*n^3 + 6*b^3*n^2 + 11*b^3*n + 6*b^3))*s
qrt(c)

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maple [A]  time = 0.01, size = 83, normalized size = 0.86 \begin {gather*} \frac {\left (b^{2} n^{2} x^{2}+3 b^{2} n \,x^{2}-2 a b n x +2 b^{2} x^{2}-2 a b x +2 a^{2}\right ) \sqrt {c \,x^{2}}\, \left (b x +a \right )^{n +1}}{\left (n^{3}+6 n^{2}+11 n +6\right ) b^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x+a)^(n+1)*(b^2*n^2*x^2+3*b^2*n*x^2-2*a*b*n*x+2*b^2*x^2-2*a*b*x+2*a^2)*(c*x^2)^(1/2)/x/b^3/(n^3+6*n^2+11*n+
6)

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maxima [A]  time = 1.41, size = 80, normalized size = 0.83 \begin {gather*} \frac {{\left ({\left (n^{2} + 3 \, n + 2\right )} b^{3} \sqrt {c} x^{3} + {\left (n^{2} + n\right )} a b^{2} \sqrt {c} x^{2} - 2 \, a^{2} b \sqrt {c} n x + 2 \, a^{3} \sqrt {c}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((n^2 + 3*n + 2)*b^3*sqrt(c)*x^3 + (n^2 + n)*a*b^2*sqrt(c)*x^2 - 2*a^2*b*sqrt(c)*n*x + 2*a^3*sqrt(c))*(b*x + a
)^n/((n^3 + 6*n^2 + 11*n + 6)*b^3)

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mupad [B]  time = 0.25, size = 142, normalized size = 1.48 \begin {gather*} \frac {{\left (a+b\,x\right )}^n\,\left (\frac {2\,a^3\,\sqrt {c\,x^2}}{b^3\,\left (n^3+6\,n^2+11\,n+6\right )}+\frac {x^3\,\sqrt {c\,x^2}\,\left (n^2+3\,n+2\right )}{n^3+6\,n^2+11\,n+6}-\frac {2\,a^2\,n\,x\,\sqrt {c\,x^2}}{b^2\,\left (n^3+6\,n^2+11\,n+6\right )}+\frac {a\,n\,x^2\,\sqrt {c\,x^2}\,\left (n+1\right )}{b\,\left (n^3+6\,n^2+11\,n+6\right )}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*x^2)^(1/2)*(a + b*x)^n,x)

[Out]

((a + b*x)^n*((2*a^3*(c*x^2)^(1/2))/(b^3*(11*n + 6*n^2 + n^3 + 6)) + (x^3*(c*x^2)^(1/2)*(3*n + n^2 + 2))/(11*n
+ 6*n^2 + n^3 + 6) - (2*a^2*n*x*(c*x^2)^(1/2))/(b^2*(11*n + 6*n^2 + n^3 + 6)) + (a*n*x^2*(c*x^2)^(1/2)*(n + 1
))/(b*(11*n + 6*n^2 + n^3 + 6))))/x

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {a^{n} \sqrt {c} x^{2} \sqrt {x^{2}}}{3} & \text {for}\: b = 0 \\\int \frac {x \sqrt {c x^{2}}}{\left (a + b x\right )^{3}}\, dx & \text {for}\: n = -3 \\\int \frac {x \sqrt {c x^{2}}}{\left (a + b x\right )^{2}}\, dx & \text {for}\: n = -2 \\\int \frac {x \sqrt {c x^{2}}}{a + b x}\, dx & \text {for}\: n = -1 \\\frac {2 a^{3} \sqrt {c} \left (a + b x\right )^{n} \sqrt {x^{2}}}{b^{3} n^{3} x + 6 b^{3} n^{2} x + 11 b^{3} n x + 6 b^{3} x} - \frac {2 a^{2} b \sqrt {c} n x \left (a + b x\right )^{n} \sqrt {x^{2}}}{b^{3} n^{3} x + 6 b^{3} n^{2} x + 11 b^{3} n x + 6 b^{3} x} + \frac {a b^{2} \sqrt {c} n^{2} x^{2} \left (a + b x\right )^{n} \sqrt {x^{2}}}{b^{3} n^{3} x + 6 b^{3} n^{2} x + 11 b^{3} n x + 6 b^{3} x} + \frac {a b^{2} \sqrt {c} n x^{2} \left (a + b x\right )^{n} \sqrt {x^{2}}}{b^{3} n^{3} x + 6 b^{3} n^{2} x + 11 b^{3} n x + 6 b^{3} x} + \frac {b^{3} \sqrt {c} n^{2} x^{3} \left (a + b x\right )^{n} \sqrt {x^{2}}}{b^{3} n^{3} x + 6 b^{3} n^{2} x + 11 b^{3} n x + 6 b^{3} x} + \frac {3 b^{3} \sqrt {c} n x^{3} \left (a + b x\right )^{n} \sqrt {x^{2}}}{b^{3} n^{3} x + 6 b^{3} n^{2} x + 11 b^{3} n x + 6 b^{3} x} + \frac {2 b^{3} \sqrt {c} x^{3} \left (a + b x\right )^{n} \sqrt {x^{2}}}{b^{3} n^{3} x + 6 b^{3} n^{2} x + 11 b^{3} n x + 6 b^{3} x} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**n*sqrt(c)*x**2*sqrt(x**2)/3, Eq(b, 0)), (Integral(x*sqrt(c*x**2)/(a + b*x)**3, x), Eq(n, -3)), (
Integral(x*sqrt(c*x**2)/(a + b*x)**2, x), Eq(n, -2)), (Integral(x*sqrt(c*x**2)/(a + b*x), x), Eq(n, -1)), (2*a
**3*sqrt(c)*(a + b*x)**n*sqrt(x**2)/(b**3*n**3*x + 6*b**3*n**2*x + 11*b**3*n*x + 6*b**3*x) - 2*a**2*b*sqrt(c)*
n*x*(a + b*x)**n*sqrt(x**2)/(b**3*n**3*x + 6*b**3*n**2*x + 11*b**3*n*x + 6*b**3*x) + a*b**2*sqrt(c)*n**2*x**2*
(a + b*x)**n*sqrt(x**2)/(b**3*n**3*x + 6*b**3*n**2*x + 11*b**3*n*x + 6*b**3*x) + a*b**2*sqrt(c)*n*x**2*(a + b*
x)**n*sqrt(x**2)/(b**3*n**3*x + 6*b**3*n**2*x + 11*b**3*n*x + 6*b**3*x) + b**3*sqrt(c)*n**2*x**3*(a + b*x)**n*
sqrt(x**2)/(b**3*n**3*x + 6*b**3*n**2*x + 11*b**3*n*x + 6*b**3*x) + 3*b**3*sqrt(c)*n*x**3*(a + b*x)**n*sqrt(x*
*2)/(b**3*n**3*x + 6*b**3*n**2*x + 11*b**3*n*x + 6*b**3*x) + 2*b**3*sqrt(c)*x**3*(a + b*x)**n*sqrt(x**2)/(b**3
*n**3*x + 6*b**3*n**2*x + 11*b**3*n*x + 6*b**3*x), True))

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