3.7.36 \(\int \frac {(a^2-b^2 x^2)^{3/2}}{(a+b x)^5} \, dx\)

Optimal. Leaf size=33 \[ -\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 a b (a+b x)^5} \]

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Rubi [A]  time = 0.01, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {651} \begin {gather*} -\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 a b (a+b x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 - b^2*x^2)^(3/2)/(a + b*x)^5,x]

[Out]

-(a^2 - b^2*x^2)^(5/2)/(5*a*b*(a + b*x)^5)

Rule 651

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rubi steps

\begin {align*} \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^5} \, dx &=-\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 a b (a+b x)^5}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 41, normalized size = 1.24 \begin {gather*} -\frac {(a-b x)^2 \sqrt {a^2-b^2 x^2}}{5 a b (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 - b^2*x^2)^(3/2)/(a + b*x)^5,x]

[Out]

-1/5*((a - b*x)^2*Sqrt[a^2 - b^2*x^2])/(a*b*(a + b*x)^3)

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IntegrateAlgebraic [A]  time = 0.45, size = 52, normalized size = 1.58 \begin {gather*} \frac {\sqrt {a^2-b^2 x^2} \left (-a^2+2 a b x-b^2 x^2\right )}{5 a b (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a^2 - b^2*x^2)^(3/2)/(a + b*x)^5,x]

[Out]

(Sqrt[a^2 - b^2*x^2]*(-a^2 + 2*a*b*x - b^2*x^2))/(5*a*b*(a + b*x)^3)

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fricas [B]  time = 0.40, size = 96, normalized size = 2.91 \begin {gather*} -\frac {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3} + {\left (b^{2} x^{2} - 2 \, a b x + a^{2}\right )} \sqrt {-b^{2} x^{2} + a^{2}}}{5 \, {\left (a b^{4} x^{3} + 3 \, a^{2} b^{3} x^{2} + 3 \, a^{3} b^{2} x + a^{4} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*x^2+a^2)^(3/2)/(b*x+a)^5,x, algorithm="fricas")

[Out]

-1/5*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3 + (b^2*x^2 - 2*a*b*x + a^2)*sqrt(-b^2*x^2 + a^2))/(a*b^4*x^3 + 3
*a^2*b^3*x^2 + 3*a^3*b^2*x + a^4*b)

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giac [C]  time = 0.23, size = 160, normalized size = 4.85 \begin {gather*} -\frac {1}{15} \, {\left (-\frac {3 i \, \mathrm {sgn}\left (\frac {1}{b x + a}\right ) \mathrm {sgn}\relax (b)}{a b^{2}} + \frac {{\left (3 \, {\left (\frac {2 \, a}{b x + a} - 1\right )}^{\frac {5}{2}} + 10 \, {\left (\frac {2 \, a}{b x + a} - 1\right )}^{\frac {3}{2}} + 15 \, \sqrt {\frac {2 \, a}{b x + a} - 1}\right )} \mathrm {sgn}\left (\frac {1}{b x + a}\right ) \mathrm {sgn}\relax (b) - 10 \, {\left ({\left (\frac {2 \, a}{b x + a} - 1\right )}^{\frac {3}{2}} + 3 \, \sqrt {\frac {2 \, a}{b x + a} - 1}\right )} \mathrm {sgn}\left (\frac {1}{b x + a}\right ) \mathrm {sgn}\relax (b) + 15 \, \sqrt {\frac {2 \, a}{b x + a} - 1} \mathrm {sgn}\left (\frac {1}{b x + a}\right ) \mathrm {sgn}\relax (b)}{a b^{2}}\right )} {\left | b \right |} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*x^2+a^2)^(3/2)/(b*x+a)^5,x, algorithm="giac")

[Out]

-1/15*(-3*I*sgn(1/(b*x + a))*sgn(b)/(a*b^2) + ((3*(2*a/(b*x + a) - 1)^(5/2) + 10*(2*a/(b*x + a) - 1)^(3/2) + 1
5*sqrt(2*a/(b*x + a) - 1))*sgn(1/(b*x + a))*sgn(b) - 10*((2*a/(b*x + a) - 1)^(3/2) + 3*sqrt(2*a/(b*x + a) - 1)
)*sgn(1/(b*x + a))*sgn(b) + 15*sqrt(2*a/(b*x + a) - 1)*sgn(1/(b*x + a))*sgn(b))/(a*b^2))*abs(b)

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maple [A]  time = 0.04, size = 36, normalized size = 1.09 \begin {gather*} -\frac {\left (-b x +a \right ) \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}}}{5 \left (b x +a \right )^{4} a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b^2*x^2+a^2)^(3/2)/(b*x+a)^5,x)

[Out]

-1/5*(-b*x+a)/(b*x+a)^4/b/a*(-b^2*x^2+a^2)^(3/2)

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maxima [B]  time = 1.52, size = 179, normalized size = 5.42 \begin {gather*} -\frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}}}{b^{5} x^{4} + 4 \, a b^{4} x^{3} + 6 \, a^{2} b^{3} x^{2} + 4 \, a^{3} b^{2} x + a^{4} b} + \frac {6 \, \sqrt {-b^{2} x^{2} + a^{2}} a}{5 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} - \frac {\sqrt {-b^{2} x^{2} + a^{2}}}{5 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} - \frac {\sqrt {-b^{2} x^{2} + a^{2}}}{5 \, {\left (a b^{2} x + a^{2} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*x^2+a^2)^(3/2)/(b*x+a)^5,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.80, size = 37, normalized size = 1.12 \begin {gather*} -\frac {\sqrt {a^2-b^2\,x^2}\,{\left (a-b\,x\right )}^2}{5\,a\,b\,{\left (a+b\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 - b^2*x^2)^(3/2)/(a + b*x)^5,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (- a + b x\right ) \left (a + b x\right )\right )^{\frac {3}{2}}}{\left (a + b x\right )^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b**2*x**2+a**2)**(3/2)/(b*x+a)**5,x)

[Out]

Integral((-(-a + b*x)*(a + b*x))**(3/2)/(a + b*x)**5, x)

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