3.367 \(\int \frac{1}{x^3 (-a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=116 \[ \frac{35 b^2}{4 a^4 \sqrt{b x-a}}-\frac{35 b^2}{12 a^3 (b x-a)^{3/2}}+\frac{35 b^2 \tan ^{-1}\left (\frac{\sqrt{b x-a}}{\sqrt{a}}\right )}{4 a^{9/2}}+\frac{7 b}{4 a^2 x (b x-a)^{3/2}}+\frac{1}{2 a x^2 (b x-a)^{3/2}} \]

[Out]

(-35*b^2)/(12*a^3*(-a + b*x)^(3/2)) + 1/(2*a*x^2*(-a + b*x)^(3/2)) + (7*b)/(4*a^2*x*(-a + b*x)^(3/2)) + (35*b^
2)/(4*a^4*Sqrt[-a + b*x]) + (35*b^2*ArcTan[Sqrt[-a + b*x]/Sqrt[a]])/(4*a^(9/2))

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Rubi [A]  time = 0.0331239, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {51, 63, 205} \[ \frac{35 b^2 \tan ^{-1}\left (\frac{\sqrt{b x-a}}{\sqrt{a}}\right )}{4 a^{9/2}}+\frac{35 \sqrt{b x-a}}{6 a^3 x^2}+\frac{14}{3 a^2 x^2 \sqrt{b x-a}}+\frac{35 b \sqrt{b x-a}}{4 a^4 x}-\frac{2}{3 a x^2 (b x-a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(-a + b*x)^(5/2)),x]

[Out]

-2/(3*a*x^2*(-a + b*x)^(3/2)) + 14/(3*a^2*x^2*Sqrt[-a + b*x]) + (35*Sqrt[-a + b*x])/(6*a^3*x^2) + (35*b*Sqrt[-
a + b*x])/(4*a^4*x) + (35*b^2*ArcTan[Sqrt[-a + b*x]/Sqrt[a]])/(4*a^(9/2))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{x^3 (-a+b x)^{5/2}} \, dx &=-\frac{2}{3 a x^2 (-a+b x)^{3/2}}-\frac{7 \int \frac{1}{x^3 (-a+b x)^{3/2}} \, dx}{3 a}\\ &=-\frac{2}{3 a x^2 (-a+b x)^{3/2}}+\frac{14}{3 a^2 x^2 \sqrt{-a+b x}}+\frac{35 \int \frac{1}{x^3 \sqrt{-a+b x}} \, dx}{3 a^2}\\ &=-\frac{2}{3 a x^2 (-a+b x)^{3/2}}+\frac{14}{3 a^2 x^2 \sqrt{-a+b x}}+\frac{35 \sqrt{-a+b x}}{6 a^3 x^2}+\frac{(35 b) \int \frac{1}{x^2 \sqrt{-a+b x}} \, dx}{4 a^3}\\ &=-\frac{2}{3 a x^2 (-a+b x)^{3/2}}+\frac{14}{3 a^2 x^2 \sqrt{-a+b x}}+\frac{35 \sqrt{-a+b x}}{6 a^3 x^2}+\frac{35 b \sqrt{-a+b x}}{4 a^4 x}+\frac{\left (35 b^2\right ) \int \frac{1}{x \sqrt{-a+b x}} \, dx}{8 a^4}\\ &=-\frac{2}{3 a x^2 (-a+b x)^{3/2}}+\frac{14}{3 a^2 x^2 \sqrt{-a+b x}}+\frac{35 \sqrt{-a+b x}}{6 a^3 x^2}+\frac{35 b \sqrt{-a+b x}}{4 a^4 x}+\frac{(35 b) \operatorname{Subst}\left (\int \frac{1}{\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{-a+b x}\right )}{4 a^4}\\ &=-\frac{2}{3 a x^2 (-a+b x)^{3/2}}+\frac{14}{3 a^2 x^2 \sqrt{-a+b x}}+\frac{35 \sqrt{-a+b x}}{6 a^3 x^2}+\frac{35 b \sqrt{-a+b x}}{4 a^4 x}+\frac{35 b^2 \tan ^{-1}\left (\frac{\sqrt{-a+b x}}{\sqrt{a}}\right )}{4 a^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0133301, size = 38, normalized size = 0.33 \[ -\frac{2 b^2 \, _2F_1\left (-\frac{3}{2},3;-\frac{1}{2};1-\frac{b x}{a}\right )}{3 a^3 (b x-a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(-a + b*x)^(5/2)),x]

[Out]

(-2*b^2*Hypergeometric2F1[-3/2, 3, -1/2, 1 - (b*x)/a])/(3*a^3*(-a + b*x)^(3/2))

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Maple [A]  time = 0.016, size = 92, normalized size = 0.8 \begin{align*} -{\frac{2\,{b}^{2}}{3\,{a}^{3}} \left ( bx-a \right ) ^{-{\frac{3}{2}}}}+6\,{\frac{{b}^{2}}{{a}^{4}\sqrt{bx-a}}}+{\frac{11}{4\,{a}^{4}{x}^{2}} \left ( bx-a \right ) ^{{\frac{3}{2}}}}+{\frac{13}{4\,{a}^{3}{x}^{2}}\sqrt{bx-a}}+{\frac{35\,{b}^{2}}{4}\arctan \left ({\sqrt{bx-a}{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*b^2/a^3/(b*x-a)^(3/2)+6*b^2/a^4/(b*x-a)^(1/2)+11/4/a^4/x^2*(b*x-a)^(3/2)+13/4/a^3/x^2*(b*x-a)^(1/2)+35/4*
b^2*arctan((b*x-a)^(1/2)/a^(1/2))/a^(9/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85456, size = 564, normalized size = 4.86 \begin{align*} \left [-\frac{105 \,{\left (b^{4} x^{4} - 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt{-a} \log \left (\frac{b x - 2 \, \sqrt{b x - a} \sqrt{-a} - 2 \, a}{x}\right ) - 2 \,{\left (105 \, a b^{3} x^{3} - 140 \, a^{2} b^{2} x^{2} + 21 \, a^{3} b x + 6 \, a^{4}\right )} \sqrt{b x - a}}{24 \,{\left (a^{5} b^{2} x^{4} - 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}}, \frac{105 \,{\left (b^{4} x^{4} - 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{b x - a}}{\sqrt{a}}\right ) +{\left (105 \, a b^{3} x^{3} - 140 \, a^{2} b^{2} x^{2} + 21 \, a^{3} b x + 6 \, a^{4}\right )} \sqrt{b x - a}}{12 \,{\left (a^{5} b^{2} x^{4} - 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x-a)^(5/2),x, algorithm="fricas")

[Out]

[-1/24*(105*(b^4*x^4 - 2*a*b^3*x^3 + a^2*b^2*x^2)*sqrt(-a)*log((b*x - 2*sqrt(b*x - a)*sqrt(-a) - 2*a)/x) - 2*(
105*a*b^3*x^3 - 140*a^2*b^2*x^2 + 21*a^3*b*x + 6*a^4)*sqrt(b*x - a))/(a^5*b^2*x^4 - 2*a^6*b*x^3 + a^7*x^2), 1/
12*(105*(b^4*x^4 - 2*a*b^3*x^3 + a^2*b^2*x^2)*sqrt(a)*arctan(sqrt(b*x - a)/sqrt(a)) + (105*a*b^3*x^3 - 140*a^2
*b^2*x^2 + 21*a^3*b*x + 6*a^4)*sqrt(b*x - a))/(a^5*b^2*x^4 - 2*a^6*b*x^3 + a^7*x^2)]

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Sympy [B]  time = 13.3606, size = 1112, normalized size = 9.59 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x-a)**(5/2),x)

[Out]

Piecewise((12*I*a**(89/2)*b**75*x**75/(24*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) - 1) - 24*a**(91/2)*b**
(153/2)*x**(157/2)*sqrt(a/(b*x) - 1)) + 42*I*a**(87/2)*b**76*x**76/(24*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/
(b*x) - 1) - 24*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) - 1)) - 280*I*a**(85/2)*b**77*x**77/(24*a**(93/2)
*b**(151/2)*x**(155/2)*sqrt(a/(b*x) - 1) - 24*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) - 1)) + 210*I*a**(8
3/2)*b**78*x**78/(24*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) - 1) - 24*a**(91/2)*b**(153/2)*x**(157/2)*sq
rt(a/(b*x) - 1)) + 210*I*a**42*b**(155/2)*x**(155/2)*sqrt(a/(b*x) - 1)*acosh(sqrt(a)/(sqrt(b)*sqrt(x)))/(24*a*
*(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) - 1) - 24*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) - 1)) - 105*
pi*a**42*b**(155/2)*x**(155/2)*sqrt(a/(b*x) - 1)/(24*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) - 1) - 24*a*
*(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) - 1)) - 210*I*a**41*b**(157/2)*x**(157/2)*sqrt(a/(b*x) - 1)*acosh(s
qrt(a)/(sqrt(b)*sqrt(x)))/(24*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) - 1) - 24*a**(91/2)*b**(153/2)*x**(
157/2)*sqrt(a/(b*x) - 1)) + 105*pi*a**41*b**(157/2)*x**(157/2)*sqrt(a/(b*x) - 1)/(24*a**(93/2)*b**(151/2)*x**(
155/2)*sqrt(a/(b*x) - 1) - 24*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) - 1)), Abs(a)/(Abs(b)*Abs(x)) > 1),
(-6*a**(89/2)*b**75*x**75/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(-a/(b*x) + 1) - 12*a**(91/2)*b**(153/2)*x*
*(157/2)*sqrt(-a/(b*x) + 1)) - 21*a**(87/2)*b**76*x**76/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(-a/(b*x) + 1)
- 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(-a/(b*x) + 1)) + 140*a**(85/2)*b**77*x**77/(12*a**(93/2)*b**(151/2)
*x**(155/2)*sqrt(-a/(b*x) + 1) - 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(-a/(b*x) + 1)) - 105*a**(83/2)*b**78*
x**78/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(-a/(b*x) + 1) - 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(-a/(b*x
) + 1)) - 105*a**42*b**(155/2)*x**(155/2)*sqrt(-a/(b*x) + 1)*asin(sqrt(a)/(sqrt(b)*sqrt(x)))/(12*a**(93/2)*b**
(151/2)*x**(155/2)*sqrt(-a/(b*x) + 1) - 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(-a/(b*x) + 1)) + 105*a**41*b**
(157/2)*x**(157/2)*sqrt(-a/(b*x) + 1)*asin(sqrt(a)/(sqrt(b)*sqrt(x)))/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt
(-a/(b*x) + 1) - 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(-a/(b*x) + 1)), True))

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Giac [A]  time = 1.2193, size = 131, normalized size = 1.13 \begin{align*} \frac{35 \, b^{2} \arctan \left (\frac{\sqrt{b x - a}}{\sqrt{a}}\right )}{4 \, a^{\frac{9}{2}}} + \frac{2 \,{\left (9 \,{\left (b x - a\right )} b^{2} - a b^{2}\right )}}{3 \,{\left (b x - a\right )}^{\frac{3}{2}} a^{4}} + \frac{11 \,{\left (b x - a\right )}^{\frac{3}{2}} b^{2} + 13 \, \sqrt{b x - a} a b^{2}}{4 \, a^{4} b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x-a)^(5/2),x, algorithm="giac")

[Out]

35/4*b^2*arctan(sqrt(b*x - a)/sqrt(a))/a^(9/2) + 2/3*(9*(b*x - a)*b^2 - a*b^2)/((b*x - a)^(3/2)*a^4) + 1/4*(11
*(b*x - a)^(3/2)*b^2 + 13*sqrt(b*x - a)*a*b^2)/(a^4*b^2*x^2)