3.798 \(\int \frac{(a^2-b^2 x^2)^{3/2}}{(a+b x)^8} \, dx\)

Optimal. Leaf size=133 \[ -\frac{2 \left (a^2-b^2 x^2\right )^{5/2}}{1155 a^4 b (a+b x)^5}-\frac{2 \left (a^2-b^2 x^2\right )^{5/2}}{231 a^3 b (a+b x)^6}-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{33 a^2 b (a+b x)^7}-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{11 a b (a+b x)^8} \]

[Out]

-(a^2 - b^2*x^2)^(5/2)/(11*a*b*(a + b*x)^8) - (a^2 - b^2*x^2)^(5/2)/(33*a^2*b*(a + b*x)^7) - (2*(a^2 - b^2*x^2
)^(5/2))/(231*a^3*b*(a + b*x)^6) - (2*(a^2 - b^2*x^2)^(5/2))/(1155*a^4*b*(a + b*x)^5)

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Rubi [A]  time = 0.0554768, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {659, 651} \[ -\frac{2 \left (a^2-b^2 x^2\right )^{5/2}}{1155 a^4 b (a+b x)^5}-\frac{2 \left (a^2-b^2 x^2\right )^{5/2}}{231 a^3 b (a+b x)^6}-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{33 a^2 b (a+b x)^7}-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{11 a b (a+b x)^8} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 - b^2*x^2)^(3/2)/(a + b*x)^8,x]

[Out]

-(a^2 - b^2*x^2)^(5/2)/(11*a*b*(a + b*x)^8) - (a^2 - b^2*x^2)^(5/2)/(33*a^2*b*(a + b*x)^7) - (2*(a^2 - b^2*x^2
)^(5/2))/(231*a^3*b*(a + b*x)^6) - (2*(a^2 - b^2*x^2)^(5/2))/(1155*a^4*b*(a + b*x)^5)

Rule 659

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*(m + p + 1)), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2
], 0]

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)^8} \, dx &=-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{11 a b (a+b x)^8}+\frac{3 \int \frac{\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^7} \, dx}{11 a}\\ &=-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{11 a b (a+b x)^8}-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{33 a^2 b (a+b x)^7}+\frac{2 \int \frac{\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^6} \, dx}{33 a^2}\\ &=-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{11 a b (a+b x)^8}-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{33 a^2 b (a+b x)^7}-\frac{2 \left (a^2-b^2 x^2\right )^{5/2}}{231 a^3 b (a+b x)^6}+\frac{2 \int \frac{\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^5} \, dx}{231 a^3}\\ &=-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{11 a b (a+b x)^8}-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{33 a^2 b (a+b x)^7}-\frac{2 \left (a^2-b^2 x^2\right )^{5/2}}{231 a^3 b (a+b x)^6}-\frac{2 \left (a^2-b^2 x^2\right )^{5/2}}{1155 a^4 b (a+b x)^5}\\ \end{align*}

Mathematica [A]  time = 0.0575, size = 71, normalized size = 0.53 \[ -\frac{(a-b x)^2 \sqrt{a^2-b^2 x^2} \left (61 a^2 b x+152 a^3+16 a b^2 x^2+2 b^3 x^3\right )}{1155 a^4 b (a+b x)^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 - b^2*x^2)^(3/2)/(a + b*x)^8,x]

[Out]

-((a - b*x)^2*Sqrt[a^2 - b^2*x^2]*(152*a^3 + 61*a^2*b*x + 16*a*b^2*x^2 + 2*b^3*x^3))/(1155*a^4*b*(a + b*x)^6)

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Maple [A]  time = 0.045, size = 66, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,{b}^{3}{x}^{3}+16\,a{b}^{2}{x}^{2}+61\,x{a}^{2}b+152\,{a}^{3} \right ) \left ( -bx+a \right ) }{1155\, \left ( bx+a \right ) ^{7}{a}^{4}b} \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)^8,x)

[Out]

-1/1155*(-b*x+a)*(2*b^3*x^3+16*a*b^2*x^2+61*a^2*b*x+152*a^3)*(-b^2*x^2+a^2)^(3/2)/(b*x+a)^7/a^4/b

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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)^8,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.14769, size = 448, normalized size = 3.37 \begin{align*} -\frac{152 \, b^{6} x^{6} + 912 \, a b^{5} x^{5} + 2280 \, a^{2} b^{4} x^{4} + 3040 \, a^{3} b^{3} x^{3} + 2280 \, a^{4} b^{2} x^{2} + 912 \, a^{5} b x + 152 \, a^{6} +{\left (2 \, b^{5} x^{5} + 12 \, a b^{4} x^{4} + 31 \, a^{2} b^{3} x^{3} + 46 \, a^{3} b^{2} x^{2} - 243 \, a^{4} b x + 152 \, a^{5}\right )} \sqrt{-b^{2} x^{2} + a^{2}}}{1155 \,{\left (a^{4} b^{7} x^{6} + 6 \, a^{5} b^{6} x^{5} + 15 \, a^{6} b^{5} x^{4} + 20 \, a^{7} b^{4} x^{3} + 15 \, a^{8} b^{3} x^{2} + 6 \, a^{9} b^{2} x + a^{10} 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)^8,x, algorithm="fricas")

[Out]

-1/1155*(152*b^6*x^6 + 912*a*b^5*x^5 + 2280*a^2*b^4*x^4 + 3040*a^3*b^3*x^3 + 2280*a^4*b^2*x^2 + 912*a^5*b*x +
152*a^6 + (2*b^5*x^5 + 12*a*b^4*x^4 + 31*a^2*b^3*x^3 + 46*a^3*b^2*x^2 - 243*a^4*b*x + 152*a^5)*sqrt(-b^2*x^2 +
a^2))/(a^4*b^7*x^6 + 6*a^5*b^6*x^5 + 15*a^6*b^5*x^4 + 20*a^7*b^4*x^3 + 15*a^8*b^3*x^2 + 6*a^9*b^2*x + a^10*b)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**8,x)

[Out]

Timed out

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Giac [B]  time = 1.24842, size = 474, normalized size = 3.56 \begin{align*} \frac{2 \,{\left (\frac{517 \,{\left (a b + \sqrt{-b^{2} x^{2} + a^{2}}{\left | b \right |}\right )}}{b^{2} x} + \frac{4895 \,{\left (a b + \sqrt{-b^{2} x^{2} + a^{2}}{\left | b \right |}\right )}^{2}}{b^{4} x^{2}} + \frac{11220 \,{\left (a b + \sqrt{-b^{2} x^{2} + a^{2}}{\left | b \right |}\right )}^{3}}{b^{6} x^{3}} + \frac{27060 \,{\left (a b + \sqrt{-b^{2} x^{2} + a^{2}}{\left | b \right |}\right )}^{4}}{b^{8} x^{4}} + \frac{32802 \,{\left (a b + \sqrt{-b^{2} x^{2} + a^{2}}{\left | b \right |}\right )}^{5}}{b^{10} x^{5}} + \frac{37422 \,{\left (a b + \sqrt{-b^{2} x^{2} + a^{2}}{\left | b \right |}\right )}^{6}}{b^{12} x^{6}} + \frac{23100 \,{\left (a b + \sqrt{-b^{2} x^{2} + a^{2}}{\left | b \right |}\right )}^{7}}{b^{14} x^{7}} + \frac{13860 \,{\left (a b + \sqrt{-b^{2} x^{2} + a^{2}}{\left | b \right |}\right )}^{8}}{b^{16} x^{8}} + \frac{3465 \,{\left (a b + \sqrt{-b^{2} x^{2} + a^{2}}{\left | b \right |}\right )}^{9}}{b^{18} x^{9}} + \frac{1155 \,{\left (a b + \sqrt{-b^{2} x^{2} + a^{2}}{\left | b \right |}\right )}^{10}}{b^{20} x^{10}} + 152\right )}}{1155 \, a^{4}{\left (\frac{a b + \sqrt{-b^{2} x^{2} + a^{2}}{\left | b \right |}}{b^{2} x} + 1\right )}^{11}{\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)^8,x, algorithm="giac")

[Out]

2/1155*(517*(a*b + sqrt(-b^2*x^2 + a^2)*abs(b))/(b^2*x) + 4895*(a*b + sqrt(-b^2*x^2 + a^2)*abs(b))^2/(b^4*x^2)
+ 11220*(a*b + sqrt(-b^2*x^2 + a^2)*abs(b))^3/(b^6*x^3) + 27060*(a*b + sqrt(-b^2*x^2 + a^2)*abs(b))^4/(b^8*x^
4) + 32802*(a*b + sqrt(-b^2*x^2 + a^2)*abs(b))^5/(b^10*x^5) + 37422*(a*b + sqrt(-b^2*x^2 + a^2)*abs(b))^6/(b^1
2*x^6) + 23100*(a*b + sqrt(-b^2*x^2 + a^2)*abs(b))^7/(b^14*x^7) + 13860*(a*b + sqrt(-b^2*x^2 + a^2)*abs(b))^8/
(b^16*x^8) + 3465*(a*b + sqrt(-b^2*x^2 + a^2)*abs(b))^9/(b^18*x^9) + 1155*(a*b + sqrt(-b^2*x^2 + a^2)*abs(b))^
10/(b^20*x^10) + 152)/(a^4*((a*b + sqrt(-b^2*x^2 + a^2)*abs(b))/(b^2*x) + 1)^11*abs(b))