3.482 \(\int \frac{x^{7/2}}{(-a+b x)^3} \, dx\)

Optimal. Leaf size=97 \[ -\frac{35 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{9/2}}+\frac{7 x^{5/2}}{4 b^2 (a-b x)}+\frac{35 a \sqrt{x}}{4 b^4}-\frac{x^{7/2}}{2 b (a-b x)^2}+\frac{35 x^{3/2}}{12 b^3} \]

[Out]

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

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Rubi [A]  time = 0.0301179, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {47, 50, 63, 208} \[ -\frac{35 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{9/2}}+\frac{7 x^{5/2}}{4 b^2 (a-b x)}+\frac{35 a \sqrt{x}}{4 b^4}-\frac{x^{7/2}}{2 b (a-b x)^2}+\frac{35 x^{3/2}}{12 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/(-a + b*x)^3,x]

[Out]

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

Rule 47

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 208

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

Rubi steps

\begin{align*} \int \frac{x^{7/2}}{(-a+b x)^3} \, dx &=-\frac{x^{7/2}}{2 b (a-b x)^2}+\frac{7 \int \frac{x^{5/2}}{(-a+b x)^2} \, dx}{4 b}\\ &=-\frac{x^{7/2}}{2 b (a-b x)^2}+\frac{7 x^{5/2}}{4 b^2 (a-b x)}+\frac{35 \int \frac{x^{3/2}}{-a+b x} \, dx}{8 b^2}\\ &=\frac{35 x^{3/2}}{12 b^3}-\frac{x^{7/2}}{2 b (a-b x)^2}+\frac{7 x^{5/2}}{4 b^2 (a-b x)}+\frac{(35 a) \int \frac{\sqrt{x}}{-a+b x} \, dx}{8 b^3}\\ &=\frac{35 a \sqrt{x}}{4 b^4}+\frac{35 x^{3/2}}{12 b^3}-\frac{x^{7/2}}{2 b (a-b x)^2}+\frac{7 x^{5/2}}{4 b^2 (a-b x)}+\frac{\left (35 a^2\right ) \int \frac{1}{\sqrt{x} (-a+b x)} \, dx}{8 b^4}\\ &=\frac{35 a \sqrt{x}}{4 b^4}+\frac{35 x^{3/2}}{12 b^3}-\frac{x^{7/2}}{2 b (a-b x)^2}+\frac{7 x^{5/2}}{4 b^2 (a-b x)}+\frac{\left (35 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{-a+b x^2} \, dx,x,\sqrt{x}\right )}{4 b^4}\\ &=\frac{35 a \sqrt{x}}{4 b^4}+\frac{35 x^{3/2}}{12 b^3}-\frac{x^{7/2}}{2 b (a-b x)^2}+\frac{7 x^{5/2}}{4 b^2 (a-b x)}-\frac{35 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0052807, size = 26, normalized size = 0.27 \[ -\frac{2 x^{9/2} \, _2F_1\left (3,\frac{9}{2};\frac{11}{2};\frac{b x}{a}\right )}{9 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/(-a + b*x)^3,x]

[Out]

(-2*x^(9/2)*Hypergeometric2F1[3, 9/2, 11/2, (b*x)/a])/(9*a^3)

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Maple [A]  time = 0.011, size = 70, normalized size = 0.7 \begin{align*} 2\,{\frac{1/3\,b{x}^{3/2}+3\,a\sqrt{x}}{{b}^{4}}}+2\,{\frac{{a}^{2}}{{b}^{4}} \left ({\frac{1}{ \left ( bx-a \right ) ^{2}} \left ( -{\frac{13\,b{x}^{3/2}}{8}}+{\frac{11\,a\sqrt{x}}{8}} \right ) }-{\frac{35}{8\,\sqrt{ab}}{\it Artanh} \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/b^4*(1/3*b*x^(3/2)+3*a*x^(1/2))+2/b^4*a^2*((-13/8*b*x^(3/2)+11/8*a*x^(1/2))/(b*x-a)^2-35/8/(a*b)^(1/2)*arcta
nh(b*x^(1/2)/(a*b)^(1/2)))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.5717, size = 509, normalized size = 5.25 \begin{align*} \left [\frac{105 \,{\left (a b^{2} x^{2} - 2 \, a^{2} b x + a^{3}\right )} \sqrt{\frac{a}{b}} \log \left (\frac{b x - 2 \, b \sqrt{x} \sqrt{\frac{a}{b}} + a}{b x - a}\right ) + 2 \,{\left (8 \, b^{3} x^{3} + 56 \, a b^{2} x^{2} - 175 \, a^{2} b x + 105 \, a^{3}\right )} \sqrt{x}}{24 \,{\left (b^{6} x^{2} - 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac{105 \,{\left (a b^{2} x^{2} - 2 \, a^{2} b x + a^{3}\right )} \sqrt{-\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{-\frac{a}{b}}}{a}\right ) +{\left (8 \, b^{3} x^{3} + 56 \, a b^{2} x^{2} - 175 \, a^{2} b x + 105 \, a^{3}\right )} \sqrt{x}}{12 \,{\left (b^{6} x^{2} - 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(105*(a*b^2*x^2 - 2*a^2*b*x + a^3)*sqrt(a/b)*log((b*x - 2*b*sqrt(x)*sqrt(a/b) + a)/(b*x - a)) + 2*(8*b^3
*x^3 + 56*a*b^2*x^2 - 175*a^2*b*x + 105*a^3)*sqrt(x))/(b^6*x^2 - 2*a*b^5*x + a^2*b^4), 1/12*(105*(a*b^2*x^2 -
2*a^2*b*x + a^3)*sqrt(-a/b)*arctan(b*sqrt(x)*sqrt(-a/b)/a) + (8*b^3*x^3 + 56*a*b^2*x^2 - 175*a^2*b*x + 105*a^3
)*sqrt(x))/(b^6*x^2 - 2*a*b^5*x + a^2*b^4)]

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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(x**(7/2)/(b*x-a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.25906, size = 109, normalized size = 1.12 \begin{align*} \frac{35 \, a^{2} \arctan \left (\frac{b \sqrt{x}}{\sqrt{-a b}}\right )}{4 \, \sqrt{-a b} b^{4}} - \frac{13 \, a^{2} b x^{\frac{3}{2}} - 11 \, a^{3} \sqrt{x}}{4 \,{\left (b x - a\right )}^{2} b^{4}} + \frac{2 \,{\left (b^{6} x^{\frac{3}{2}} + 9 \, a b^{5} \sqrt{x}\right )}}{3 \, b^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/4*a^2*arctan(b*sqrt(x)/sqrt(-a*b))/(sqrt(-a*b)*b^4) - 1/4*(13*a^2*b*x^(3/2) - 11*a^3*sqrt(x))/((b*x - a)^2*
b^4) + 2/3*(b^6*x^(3/2) + 9*a*b^5*sqrt(x))/b^9