3.10.6 \(\int \frac {x^6 (a+b x)^n}{(c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=65 \[ \frac {x (a+b x)^{n+2}}{b^2 c^2 (n+2) \sqrt {c x^2}}-\frac {a x (a+b x)^{n+1}}{b^2 c^2 (n+1) \sqrt {c x^2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 65, 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 {x (a+b x)^{n+2}}{b^2 c^2 (n+2) \sqrt {c x^2}}-\frac {a x (a+b x)^{n+1}}{b^2 c^2 (n+1) \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^6*(a + b*x)^n)/(c*x^2)^(5/2),x]

[Out]

-((a*x*(a + b*x)^(1 + n))/(b^2*c^2*(1 + n)*Sqrt[c*x^2])) + (x*(a + b*x)^(2 + n))/(b^2*c^2*(2 + 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^6 (a+b x)^n}{\left (c x^2\right )^{5/2}} \, dx &=\frac {x \int x (a+b x)^n \, dx}{c^2 \sqrt {c x^2}}\\ &=\frac {x \int \left (-\frac {a (a+b x)^n}{b}+\frac {(a+b x)^{1+n}}{b}\right ) \, dx}{c^2 \sqrt {c x^2}}\\ &=-\frac {a x (a+b x)^{1+n}}{b^2 c^2 (1+n) \sqrt {c x^2}}+\frac {x (a+b x)^{2+n}}{b^2 c^2 (2+n) \sqrt {c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 46, normalized size = 0.71 \begin {gather*} \frac {x (a+b x)^{n+1} (b (n+1) x-a)}{b^2 c^2 (n+1) (n+2) \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(a + b*x)^n)/(c*x^2)^(5/2),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.25, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^6 (a+b x)^n}{\left (c x^2\right )^{5/2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^6*(a + b*x)^n)/(c*x^2)^(5/2),x]

[Out]

Defer[IntegrateAlgebraic][(x^6*(a + b*x)^n)/(c*x^2)^(5/2), x]

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fricas [A]  time = 0.92, size = 72, normalized size = 1.11 \begin {gather*} \frac {{\left (a b n x + {\left (b^{2} n + b^{2}\right )} x^{2} - a^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{{\left (b^{2} c^{3} n^{2} + 3 \, b^{2} c^{3} n + 2 \, b^{2} c^{3}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(b*x+a)^n/(c*x^2)^(5/2),x, algorithm="fricas")

[Out]

(a*b*n*x + (b^2*n + b^2)*x^2 - a^2)*sqrt(c*x^2)*(b*x + a)^n/((b^2*c^3*n^2 + 3*b^2*c^3*n + 2*b^2*c^3)*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^{6}}{\left (c x^{2}\right )^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(b*x+a)^n/(c*x^2)^(5/2),x, algorithm="giac")

[Out]

integrate((b*x + a)^n*x^6/(c*x^2)^(5/2), x)

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maple [A]  time = 0.00, size = 46, normalized size = 0.71 \begin {gather*} -\frac {\left (-x n b -b x +a \right ) x^{5} \left (b x +a \right )^{n +1}}{\left (c \,x^{2}\right )^{\frac {5}{2}} \left (n^{2}+3 n +2\right ) b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(b*x+a)^n/(c*x^2)^(5/2),x)

[Out]

-(b*x+a)^(n+1)*x^5*(-b*n*x-b*x+a)/(c*x^2)^(5/2)/b^2/(n^2+3*n+2)

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maxima [A]  time = 1.41, size = 45, normalized size = 0.69 \begin {gather*} \frac {{\left (b^{2} {\left (n + 1\right )} x^{2} + a b n x - a^{2}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{2} + 3 \, n + 2\right )} b^{2} c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(b*x+a)^n/(c*x^2)^(5/2),x, algorithm="maxima")

[Out]

(b^2*(n + 1)*x^2 + a*b*n*x - a^2)*(b*x + a)^n/((n^2 + 3*n + 2)*b^2*c^(5/2))

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mupad [B]  time = 0.29, size = 80, normalized size = 1.23 \begin {gather*} \frac {{\left (a+b\,x\right )}^n\,\left (\frac {x^3\,\left (n+1\right )}{c^2\,\left (n^2+3\,n+2\right )}-\frac {a^2\,x}{b^2\,c^2\,\left (n^2+3\,n+2\right )}+\frac {a\,n\,x^2}{b\,c^2\,\left (n^2+3\,n+2\right )}\right )}{\sqrt {c\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6*(a + b*x)^n)/(c*x^2)^(5/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {a^{n} x^{7}}{2 c^{\frac {5}{2}} \left (x^{2}\right )^{\frac {5}{2}}} & \text {for}\: b = 0 \\\int \frac {x^{6}}{\left (c x^{2}\right )^{\frac {5}{2}} \left (a + b x\right )^{2}}\, dx & \text {for}\: n = -2 \\\int \frac {x^{6}}{\left (c x^{2}\right )^{\frac {5}{2}} \left (a + b x\right )}\, dx & \text {for}\: n = -1 \\- \frac {a^{2} x^{5} \left (a + b x\right )^{n}}{b^{2} c^{\frac {5}{2}} n^{2} \left (x^{2}\right )^{\frac {5}{2}} + 3 b^{2} c^{\frac {5}{2}} n \left (x^{2}\right )^{\frac {5}{2}} + 2 b^{2} c^{\frac {5}{2}} \left (x^{2}\right )^{\frac {5}{2}}} + \frac {a b n x^{6} \left (a + b x\right )^{n}}{b^{2} c^{\frac {5}{2}} n^{2} \left (x^{2}\right )^{\frac {5}{2}} + 3 b^{2} c^{\frac {5}{2}} n \left (x^{2}\right )^{\frac {5}{2}} + 2 b^{2} c^{\frac {5}{2}} \left (x^{2}\right )^{\frac {5}{2}}} + \frac {b^{2} n x^{7} \left (a + b x\right )^{n}}{b^{2} c^{\frac {5}{2}} n^{2} \left (x^{2}\right )^{\frac {5}{2}} + 3 b^{2} c^{\frac {5}{2}} n \left (x^{2}\right )^{\frac {5}{2}} + 2 b^{2} c^{\frac {5}{2}} \left (x^{2}\right )^{\frac {5}{2}}} + \frac {b^{2} x^{7} \left (a + b x\right )^{n}}{b^{2} c^{\frac {5}{2}} n^{2} \left (x^{2}\right )^{\frac {5}{2}} + 3 b^{2} c^{\frac {5}{2}} n \left (x^{2}\right )^{\frac {5}{2}} + 2 b^{2} c^{\frac {5}{2}} \left (x^{2}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*(b*x+a)**n/(c*x**2)**(5/2),x)

[Out]

Piecewise((a**n*x**7/(2*c**(5/2)*(x**2)**(5/2)), Eq(b, 0)), (Integral(x**6/((c*x**2)**(5/2)*(a + b*x)**2), x),
Eq(n, -2)), (Integral(x**6/((c*x**2)**(5/2)*(a + b*x)), x), Eq(n, -1)), (-a**2*x**5*(a + b*x)**n/(b**2*c**(5/
2)*n**2*(x**2)**(5/2) + 3*b**2*c**(5/2)*n*(x**2)**(5/2) + 2*b**2*c**(5/2)*(x**2)**(5/2)) + a*b*n*x**6*(a + b*x
)**n/(b**2*c**(5/2)*n**2*(x**2)**(5/2) + 3*b**2*c**(5/2)*n*(x**2)**(5/2) + 2*b**2*c**(5/2)*(x**2)**(5/2)) + b*
*2*n*x**7*(a + b*x)**n/(b**2*c**(5/2)*n**2*(x**2)**(5/2) + 3*b**2*c**(5/2)*n*(x**2)**(5/2) + 2*b**2*c**(5/2)*(
x**2)**(5/2)) + b**2*x**7*(a + b*x)**n/(b**2*c**(5/2)*n**2*(x**2)**(5/2) + 3*b**2*c**(5/2)*n*(x**2)**(5/2) + 2
*b**2*c**(5/2)*(x**2)**(5/2)), True))

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