3.9.96 \(\int \frac {x^4 (a+b x)^n}{\sqrt {c x^2}} \, dx\)

Optimal. Leaf size=123 \[ -\frac {a^3 x (a+b x)^{n+1}}{b^4 (n+1) \sqrt {c x^2}}+\frac {3 a^2 x (a+b x)^{n+2}}{b^4 (n+2) \sqrt {c x^2}}-\frac {3 a x (a+b x)^{n+3}}{b^4 (n+3) \sqrt {c x^2}}+\frac {x (a+b x)^{n+4}}{b^4 (n+4) \sqrt {c x^2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 43} \begin {gather*} -\frac {a^3 x (a+b x)^{n+1}}{b^4 (n+1) \sqrt {c x^2}}+\frac {3 a^2 x (a+b x)^{n+2}}{b^4 (n+2) \sqrt {c x^2}}-\frac {3 a x (a+b x)^{n+3}}{b^4 (n+3) \sqrt {c x^2}}+\frac {x (a+b x)^{n+4}}{b^4 (n+4) \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*x)^n)/Sqrt[c*x^2],x]

[Out]

-((a^3*x*(a + b*x)^(1 + n))/(b^4*(1 + n)*Sqrt[c*x^2])) + (3*a^2*x*(a + b*x)^(2 + n))/(b^4*(2 + n)*Sqrt[c*x^2])
- (3*a*x*(a + b*x)^(3 + n))/(b^4*(3 + n)*Sqrt[c*x^2]) + (x*(a + b*x)^(4 + n))/(b^4*(4 + n)*Sqrt[c*x^2])

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 \frac {x^4 (a+b x)^n}{\sqrt {c x^2}} \, dx &=\frac {x \int x^3 (a+b x)^n \, dx}{\sqrt {c x^2}}\\ &=\frac {x \int \left (-\frac {a^3 (a+b x)^n}{b^3}+\frac {3 a^2 (a+b x)^{1+n}}{b^3}-\frac {3 a (a+b x)^{2+n}}{b^3}+\frac {(a+b x)^{3+n}}{b^3}\right ) \, dx}{\sqrt {c x^2}}\\ &=-\frac {a^3 x (a+b x)^{1+n}}{b^4 (1+n) \sqrt {c x^2}}+\frac {3 a^2 x (a+b x)^{2+n}}{b^4 (2+n) \sqrt {c x^2}}-\frac {3 a x (a+b x)^{3+n}}{b^4 (3+n) \sqrt {c x^2}}+\frac {x (a+b x)^{4+n}}{b^4 (4+n) \sqrt {c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 96, normalized size = 0.78 \begin {gather*} \frac {x (a+b x)^{n+1} \left (-6 a^3+6 a^2 b (n+1) x-3 a b^2 \left (n^2+3 n+2\right ) x^2+b^3 \left (n^3+6 n^2+11 n+6\right ) x^3\right )}{b^4 (n+1) (n+2) (n+3) (n+4) \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*x)^n)/Sqrt[c*x^2],x]

[Out]

(x*(a + b*x)^(1 + n)*(-6*a^3 + 6*a^2*b*(1 + n)*x - 3*a*b^2*(2 + 3*n + n^2)*x^2 + b^3*(6 + 11*n + 6*n^2 + n^3)*
x^3))/(b^4*(1 + n)*(2 + n)*(3 + n)*(4 + n)*Sqrt[c*x^2])

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IntegrateAlgebraic [F]  time = 0.22, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^4 (a+b x)^n}{\sqrt {c x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^4*(a + b*x)^n)/Sqrt[c*x^2],x]

[Out]

Defer[IntegrateAlgebraic][(x^4*(a + b*x)^n)/Sqrt[c*x^2], x]

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fricas [A]  time = 1.06, size = 158, normalized size = 1.28 \begin {gather*} \frac {{\left (6 \, a^{3} b n x + {\left (b^{4} n^{3} + 6 \, b^{4} n^{2} + 11 \, b^{4} n + 6 \, b^{4}\right )} x^{4} - 6 \, a^{4} + {\left (a b^{3} n^{3} + 3 \, a b^{3} n^{2} + 2 \, a b^{3} n\right )} x^{3} - 3 \, {\left (a^{2} b^{2} n^{2} + a^{2} b^{2} n\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{{\left (b^{4} c n^{4} + 10 \, b^{4} c n^{3} + 35 \, b^{4} c n^{2} + 50 \, b^{4} c n + 24 \, b^{4} c\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x+a)^n/(c*x^2)^(1/2),x, algorithm="fricas")

[Out]

(6*a^3*b*n*x + (b^4*n^3 + 6*b^4*n^2 + 11*b^4*n + 6*b^4)*x^4 - 6*a^4 + (a*b^3*n^3 + 3*a*b^3*n^2 + 2*a*b^3*n)*x^
3 - 3*(a^2*b^2*n^2 + a^2*b^2*n)*x^2)*sqrt(c*x^2)*(b*x + a)^n/((b^4*c*n^4 + 10*b^4*c*n^3 + 35*b^4*c*n^2 + 50*b^
4*c*n + 24*b^4*c)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x + a\right )}^{n} x^{4}}{\sqrt {c x^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x+a)^n/(c*x^2)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x + a)^n*x^4/sqrt(c*x^2), x)

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maple [A]  time = 0.01, size = 134, normalized size = 1.09 \begin {gather*} -\frac {\left (-b^{3} n^{3} x^{3}-6 b^{3} n^{2} x^{3}+3 a \,b^{2} n^{2} x^{2}-11 b^{3} n \,x^{3}+9 a \,b^{2} n \,x^{2}-6 b^{3} x^{3}-6 a^{2} b n x +6 a \,b^{2} x^{2}-6 a^{2} b x +6 a^{3}\right ) x \left (b x +a \right )^{n +1}}{\sqrt {c \,x^{2}}\, \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right ) b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x+a)^n/(c*x^2)^(1/2),x)

[Out]

-(b*x+a)^(n+1)*x*(-b^3*n^3*x^3-6*b^3*n^2*x^3+3*a*b^2*n^2*x^2-11*b^3*n*x^3+9*a*b^2*n*x^2-6*b^3*x^3-6*a^2*b*n*x+
6*a*b^2*x^2-6*a^2*b*x+6*a^3)/(c*x^2)^(1/2)/b^4/(n^4+10*n^3+35*n^2+50*n+24)

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maxima [A]  time = 1.45, size = 104, normalized size = 0.85 \begin {gather*} \frac {{\left ({\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{4} x^{4} + {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a b^{3} x^{3} - 3 \, {\left (n^{2} + n\right )} a^{2} b^{2} x^{2} + 6 \, a^{3} b n x - 6 \, a^{4}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} b^{4} \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x+a)^n/(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

((n^3 + 6*n^2 + 11*n + 6)*b^4*x^4 + (n^3 + 3*n^2 + 2*n)*a*b^3*x^3 - 3*(n^2 + n)*a^2*b^2*x^2 + 6*a^3*b*n*x - 6*
a^4)*(b*x + a)^n/((n^4 + 10*n^3 + 35*n^2 + 50*n + 24)*b^4*sqrt(c))

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mupad [B]  time = 0.37, size = 186, normalized size = 1.51 \begin {gather*} \frac {{\left (a+b\,x\right )}^n\,\left (\frac {x^5\,\left (n^3+6\,n^2+11\,n+6\right )}{n^4+10\,n^3+35\,n^2+50\,n+24}-\frac {6\,a^4\,x}{b^4\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {6\,a^3\,n\,x^2}{b^3\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {a\,n\,x^4\,\left (n^2+3\,n+2\right )}{b\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}-\frac {3\,a^2\,n\,x^3\,\left (n+1\right )}{b^2\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}\right )}{\sqrt {c\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*x)^n)/(c*x^2)^(1/2),x)

[Out]

((a + b*x)^n*((x^5*(11*n + 6*n^2 + n^3 + 6))/(50*n + 35*n^2 + 10*n^3 + n^4 + 24) - (6*a^4*x)/(b^4*(50*n + 35*n
^2 + 10*n^3 + n^4 + 24)) + (6*a^3*n*x^2)/(b^3*(50*n + 35*n^2 + 10*n^3 + n^4 + 24)) + (a*n*x^4*(3*n + n^2 + 2))
/(b*(50*n + 35*n^2 + 10*n^3 + n^4 + 24)) - (3*a^2*n*x^3*(n + 1))/(b^2*(50*n + 35*n^2 + 10*n^3 + n^4 + 24))))/(
c*x^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b*x+a)**n/(c*x**2)**(1/2),x)

[Out]

Piecewise((a**n*x**5/(4*sqrt(c)*sqrt(x**2)), Eq(b, 0)), (Integral(x**4/(sqrt(c*x**2)*(a + b*x)**4), x), Eq(n,
-4)), (Integral(x**4/(sqrt(c*x**2)*(a + b*x)**3), x), Eq(n, -3)), (Integral(x**4/(sqrt(c*x**2)*(a + b*x)**2),
x), Eq(n, -2)), (Integral(x**4/(sqrt(c*x**2)*(a + b*x)), x), Eq(n, -1)), (-6*a**4*x*(a + b*x)**n/(b**4*sqrt(c)
*n**4*sqrt(x**2) + 10*b**4*sqrt(c)*n**3*sqrt(x**2) + 35*b**4*sqrt(c)*n**2*sqrt(x**2) + 50*b**4*sqrt(c)*n*sqrt(
x**2) + 24*b**4*sqrt(c)*sqrt(x**2)) + 6*a**3*b*n*x**2*(a + b*x)**n/(b**4*sqrt(c)*n**4*sqrt(x**2) + 10*b**4*sqr
t(c)*n**3*sqrt(x**2) + 35*b**4*sqrt(c)*n**2*sqrt(x**2) + 50*b**4*sqrt(c)*n*sqrt(x**2) + 24*b**4*sqrt(c)*sqrt(x
**2)) - 3*a**2*b**2*n**2*x**3*(a + b*x)**n/(b**4*sqrt(c)*n**4*sqrt(x**2) + 10*b**4*sqrt(c)*n**3*sqrt(x**2) + 3
5*b**4*sqrt(c)*n**2*sqrt(x**2) + 50*b**4*sqrt(c)*n*sqrt(x**2) + 24*b**4*sqrt(c)*sqrt(x**2)) - 3*a**2*b**2*n*x*
*3*(a + b*x)**n/(b**4*sqrt(c)*n**4*sqrt(x**2) + 10*b**4*sqrt(c)*n**3*sqrt(x**2) + 35*b**4*sqrt(c)*n**2*sqrt(x*
*2) + 50*b**4*sqrt(c)*n*sqrt(x**2) + 24*b**4*sqrt(c)*sqrt(x**2)) + a*b**3*n**3*x**4*(a + b*x)**n/(b**4*sqrt(c)
*n**4*sqrt(x**2) + 10*b**4*sqrt(c)*n**3*sqrt(x**2) + 35*b**4*sqrt(c)*n**2*sqrt(x**2) + 50*b**4*sqrt(c)*n*sqrt(
x**2) + 24*b**4*sqrt(c)*sqrt(x**2)) + 3*a*b**3*n**2*x**4*(a + b*x)**n/(b**4*sqrt(c)*n**4*sqrt(x**2) + 10*b**4*
sqrt(c)*n**3*sqrt(x**2) + 35*b**4*sqrt(c)*n**2*sqrt(x**2) + 50*b**4*sqrt(c)*n*sqrt(x**2) + 24*b**4*sqrt(c)*sqr
t(x**2)) + 2*a*b**3*n*x**4*(a + b*x)**n/(b**4*sqrt(c)*n**4*sqrt(x**2) + 10*b**4*sqrt(c)*n**3*sqrt(x**2) + 35*b
**4*sqrt(c)*n**2*sqrt(x**2) + 50*b**4*sqrt(c)*n*sqrt(x**2) + 24*b**4*sqrt(c)*sqrt(x**2)) + b**4*n**3*x**5*(a +
b*x)**n/(b**4*sqrt(c)*n**4*sqrt(x**2) + 10*b**4*sqrt(c)*n**3*sqrt(x**2) + 35*b**4*sqrt(c)*n**2*sqrt(x**2) + 5
0*b**4*sqrt(c)*n*sqrt(x**2) + 24*b**4*sqrt(c)*sqrt(x**2)) + 6*b**4*n**2*x**5*(a + b*x)**n/(b**4*sqrt(c)*n**4*s
qrt(x**2) + 10*b**4*sqrt(c)*n**3*sqrt(x**2) + 35*b**4*sqrt(c)*n**2*sqrt(x**2) + 50*b**4*sqrt(c)*n*sqrt(x**2) +
24*b**4*sqrt(c)*sqrt(x**2)) + 11*b**4*n*x**5*(a + b*x)**n/(b**4*sqrt(c)*n**4*sqrt(x**2) + 10*b**4*sqrt(c)*n**
3*sqrt(x**2) + 35*b**4*sqrt(c)*n**2*sqrt(x**2) + 50*b**4*sqrt(c)*n*sqrt(x**2) + 24*b**4*sqrt(c)*sqrt(x**2)) +
6*b**4*x**5*(a + b*x)**n/(b**4*sqrt(c)*n**4*sqrt(x**2) + 10*b**4*sqrt(c)*n**3*sqrt(x**2) + 35*b**4*sqrt(c)*n**
2*sqrt(x**2) + 50*b**4*sqrt(c)*n*sqrt(x**2) + 24*b**4*sqrt(c)*sqrt(x**2)), True))

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