3.2.97 \(\int \frac {1}{x^{7/2} \sqrt {a x^2+b x^3}} \, dx\)

Optimal. Leaf size=116 \[ \frac {32 b^3 \sqrt {a x^2+b x^3}}{35 a^4 x^{3/2}}-\frac {16 b^2 \sqrt {a x^2+b x^3}}{35 a^3 x^{5/2}}+\frac {12 b \sqrt {a x^2+b x^3}}{35 a^2 x^{7/2}}-\frac {2 \sqrt {a x^2+b x^3}}{7 a x^{9/2}} \]

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Rubi [A]  time = 0.16, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2016, 2014} \begin {gather*} \frac {32 b^3 \sqrt {a x^2+b x^3}}{35 a^4 x^{3/2}}-\frac {16 b^2 \sqrt {a x^2+b x^3}}{35 a^3 x^{5/2}}+\frac {12 b \sqrt {a x^2+b x^3}}{35 a^2 x^{7/2}}-\frac {2 \sqrt {a x^2+b x^3}}{7 a x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*Sqrt[a*x^2 + b*x^3]),x]

[Out]

(-2*Sqrt[a*x^2 + b*x^3])/(7*a*x^(9/2)) + (12*b*Sqrt[a*x^2 + b*x^3])/(35*a^2*x^(7/2)) - (16*b^2*Sqrt[a*x^2 + b*
x^3])/(35*a^3*x^(5/2)) + (32*b^3*Sqrt[a*x^2 + b*x^3])/(35*a^4*x^(3/2))

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps

\begin {align*} \int \frac {1}{x^{7/2} \sqrt {a x^2+b x^3}} \, dx &=-\frac {2 \sqrt {a x^2+b x^3}}{7 a x^{9/2}}-\frac {(6 b) \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx}{7 a}\\ &=-\frac {2 \sqrt {a x^2+b x^3}}{7 a x^{9/2}}+\frac {12 b \sqrt {a x^2+b x^3}}{35 a^2 x^{7/2}}+\frac {\left (24 b^2\right ) \int \frac {1}{x^{3/2} \sqrt {a x^2+b x^3}} \, dx}{35 a^2}\\ &=-\frac {2 \sqrt {a x^2+b x^3}}{7 a x^{9/2}}+\frac {12 b \sqrt {a x^2+b x^3}}{35 a^2 x^{7/2}}-\frac {16 b^2 \sqrt {a x^2+b x^3}}{35 a^3 x^{5/2}}-\frac {\left (16 b^3\right ) \int \frac {1}{\sqrt {x} \sqrt {a x^2+b x^3}} \, dx}{35 a^3}\\ &=-\frac {2 \sqrt {a x^2+b x^3}}{7 a x^{9/2}}+\frac {12 b \sqrt {a x^2+b x^3}}{35 a^2 x^{7/2}}-\frac {16 b^2 \sqrt {a x^2+b x^3}}{35 a^3 x^{5/2}}+\frac {32 b^3 \sqrt {a x^2+b x^3}}{35 a^4 x^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 55, normalized size = 0.47 \begin {gather*} \frac {2 \sqrt {x^2 (a+b x)} \left (-5 a^3+6 a^2 b x-8 a b^2 x^2+16 b^3 x^3\right )}{35 a^4 x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*Sqrt[a*x^2 + b*x^3]),x]

[Out]

(2*Sqrt[x^2*(a + b*x)]*(-5*a^3 + 6*a^2*b*x - 8*a*b^2*x^2 + 16*b^3*x^3))/(35*a^4*x^(9/2))

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IntegrateAlgebraic [A]  time = 0.14, size = 57, normalized size = 0.49 \begin {gather*} \frac {2 \sqrt {a x^2+b x^3} \left (-5 a^3+6 a^2 b x-8 a b^2 x^2+16 b^3 x^3\right )}{35 a^4 x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^(7/2)*Sqrt[a*x^2 + b*x^3]),x]

[Out]

(2*Sqrt[a*x^2 + b*x^3]*(-5*a^3 + 6*a^2*b*x - 8*a*b^2*x^2 + 16*b^3*x^3))/(35*a^4*x^(9/2))

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fricas [A]  time = 0.40, size = 51, normalized size = 0.44 \begin {gather*} \frac {2 \, {\left (16 \, b^{3} x^{3} - 8 \, a b^{2} x^{2} + 6 \, a^{2} b x - 5 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{35 \, a^{4} x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")

[Out]

2/35*(16*b^3*x^3 - 8*a*b^2*x^2 + 6*a^2*b*x - 5*a^3)*sqrt(b*x^3 + a*x^2)/(a^4*x^(9/2))

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giac [A]  time = 0.21, size = 103, normalized size = 0.89 \begin {gather*} \frac {64 \, {\left (35 \, {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{6} - 21 \, a {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} + 7 \, a^{2} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a^{3}\right )} b^{\frac {7}{2}}}{35 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="giac")

[Out]

64/35*(35*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^6 - 21*a*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 + 7*a^2*(sqrt(b)*sqrt
(x) - sqrt(b*x + a))^2 - a^3)*b^(7/2)/((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^7

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maple [A]  time = 0.04, size = 57, normalized size = 0.49 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 \sqrt {b \,x^{3}+a \,x^{2}}\, a^{4} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(b*x+a)*(-16*b^3*x^3+8*a*b^2*x^2-6*a^2*b*x+5*a^3)/x^(5/2)/a^4/(b*x^3+a*x^2)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b x^{3} + a x^{2}} x^{\frac {7}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*x^3 + a*x^2)*x^(7/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^{7/2}\,\sqrt {b\,x^3+a\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{\frac {7}{2}} \sqrt {x^{2} \left (a + b x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(7/2)/(b*x**3+a*x**2)**(1/2),x)

[Out]

Integral(1/(x**(7/2)*sqrt(x**2*(a + b*x))), x)

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