### 3.795 $$\int \frac{(a^2-b^2 x^2)^{3/2}}{(a+b x)^5} \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0090839, antiderivative size = 33, normalized size of antiderivative = 1., 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} $-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{5 a b (a+b x)^5}$

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*}

Mathematica [A]  time = 0.0455292, size = 41, normalized size = 1.24 $-\frac{(a-b x)^2 \sqrt{a^2-b^2 x^2}}{5 a b (a+b x)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 36, normalized size = 1.1 \begin{align*} -{\frac{-bx+a}{5\, \left ( bx+a \right ) ^{4}ba} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 2.13154, size = 198, normalized size = 6. \begin{align*} -\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{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}

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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Giac [C]  time = 1.29834, size = 139, normalized size = 4.21 \begin{align*} -\frac{1}{15} \,{\left (-\frac{3 i \, \mathrm{sgn}\left (\frac{1}{b x + a}\right ) \mathrm{sgn}\left (b\right )}{a b^{2}} - \frac{5 \,{\left (\frac{2 \, a}{b x + a} - 1\right )}^{\frac{3}{2}} \mathrm{sgn}\left (\frac{1}{b x + a}\right ) \mathrm{sgn}\left (b\right ) -{\left (3 \,{\left (\frac{2 \, a}{b x + a} - 1\right )}^{\frac{5}{2}} + 5 \,{\left (\frac{2 \, a}{b x + a} - 1\right )}^{\frac{3}{2}}\right )} \mathrm{sgn}\left (\frac{1}{b x + a}\right ) \mathrm{sgn}\left (b\right )}{a b^{2}}\right )}{\left | b \right |} \end{align*}

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