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

Optimal. Leaf size=95 \[ \frac{35 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{9/2}}-\frac{7 x^{5/2}}{4 b^2 (a+b x)}-\frac{35 a \sqrt{x}}{4 b^4}-\frac{x^{7/2}}{2 b (a+b x)^2}+\frac{35 x^{3/2}}{12 b^3} \]

[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)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*b^(9/2))

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Rubi [A]  time = 0.0326997, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {47, 50, 63, 205} \[ \frac{35 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{9/2}}-\frac{7 x^{5/2}}{4 b^2 (a+b x)}-\frac{35 a \sqrt{x}}{4 b^4}-\frac{x^{7/2}}{2 b (a+b x)^2}+\frac{35 x^{3/2}}{12 b^3} \]

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)*ArcTan[(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 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{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} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0041032, size = 27, normalized size = 0.28 \[ \frac{2 x^{9/2} \, _2F_1\left (3,\frac{9}{2};\frac{11}{2};-\frac{b x}{a}\right )}{9 a^3} \]

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 = 0.012, size = 79, normalized size = 0.8 \begin{align*}{\frac{2}{3\,{b}^{3}}{x}^{{\frac{3}{2}}}}-6\,{\frac{a\sqrt{x}}{{b}^{4}}}-{\frac{13\,{a}^{2}}{4\,{b}^{3} \left ( bx+a \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{11\,{a}^{3}}{4\,{b}^{4} \left ( bx+a \right ) ^{2}}\sqrt{x}}+{\frac{35\,{a}^{2}}{4\,{b}^{4}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*x^(3/2)/b^3-6*a*x^(1/2)/b^4-13/4/b^3*a^2/(b*x+a)^2*x^(3/2)-11/4/b^4*a^3/(b*x+a)^2*x^(1/2)+35/4/b^4*a^2/(a*
b)^(1/2)*arctan(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 = 1.32586, size = 509, normalized size = 5.36 \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 = 1.25841, size = 104, normalized size = 1.09 \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