3.545 \(\int x^{5/2} (a+b x)^{5/2} \, dx\)

Optimal. Leaf size=164 \[ -\frac{5 a^4 x^{3/2} \sqrt{a+b x}}{768 b^2}+\frac{5 a^5 \sqrt{x} \sqrt{a+b x}}{512 b^3}-\frac{5 a^6 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{512 b^{7/2}}+\frac{a^3 x^{5/2} \sqrt{a+b x}}{192 b}+\frac{1}{32} a^2 x^{7/2} \sqrt{a+b x}+\frac{1}{12} a x^{7/2} (a+b x)^{3/2}+\frac{1}{6} x^{7/2} (a+b x)^{5/2} \]

[Out]

(5*a^5*Sqrt[x]*Sqrt[a + b*x])/(512*b^3) - (5*a^4*x^(3/2)*Sqrt[a + b*x])/(768*b^2) + (a^3*x^(5/2)*Sqrt[a + b*x]
)/(192*b) + (a^2*x^(7/2)*Sqrt[a + b*x])/32 + (a*x^(7/2)*(a + b*x)^(3/2))/12 + (x^(7/2)*(a + b*x)^(5/2))/6 - (5
*a^6*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(512*b^(7/2))

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Rubi [A]  time = 0.0632622, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {50, 63, 217, 206} \[ -\frac{5 a^4 x^{3/2} \sqrt{a+b x}}{768 b^2}+\frac{5 a^5 \sqrt{x} \sqrt{a+b x}}{512 b^3}-\frac{5 a^6 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{512 b^{7/2}}+\frac{a^3 x^{5/2} \sqrt{a+b x}}{192 b}+\frac{1}{32} a^2 x^{7/2} \sqrt{a+b x}+\frac{1}{12} a x^{7/2} (a+b x)^{3/2}+\frac{1}{6} x^{7/2} (a+b x)^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(a + b*x)^(5/2),x]

[Out]

(5*a^5*Sqrt[x]*Sqrt[a + b*x])/(512*b^3) - (5*a^4*x^(3/2)*Sqrt[a + b*x])/(768*b^2) + (a^3*x^(5/2)*Sqrt[a + b*x]
)/(192*b) + (a^2*x^(7/2)*Sqrt[a + b*x])/32 + (a*x^(7/2)*(a + b*x)^(3/2))/12 + (x^(7/2)*(a + b*x)^(5/2))/6 - (5
*a^6*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(512*b^(7/2))

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int x^{5/2} (a+b x)^{5/2} \, dx &=\frac{1}{6} x^{7/2} (a+b x)^{5/2}+\frac{1}{12} (5 a) \int x^{5/2} (a+b x)^{3/2} \, dx\\ &=\frac{1}{12} a x^{7/2} (a+b x)^{3/2}+\frac{1}{6} x^{7/2} (a+b x)^{5/2}+\frac{1}{8} a^2 \int x^{5/2} \sqrt{a+b x} \, dx\\ &=\frac{1}{32} a^2 x^{7/2} \sqrt{a+b x}+\frac{1}{12} a x^{7/2} (a+b x)^{3/2}+\frac{1}{6} x^{7/2} (a+b x)^{5/2}+\frac{1}{64} a^3 \int \frac{x^{5/2}}{\sqrt{a+b x}} \, dx\\ &=\frac{a^3 x^{5/2} \sqrt{a+b x}}{192 b}+\frac{1}{32} a^2 x^{7/2} \sqrt{a+b x}+\frac{1}{12} a x^{7/2} (a+b x)^{3/2}+\frac{1}{6} x^{7/2} (a+b x)^{5/2}-\frac{\left (5 a^4\right ) \int \frac{x^{3/2}}{\sqrt{a+b x}} \, dx}{384 b}\\ &=-\frac{5 a^4 x^{3/2} \sqrt{a+b x}}{768 b^2}+\frac{a^3 x^{5/2} \sqrt{a+b x}}{192 b}+\frac{1}{32} a^2 x^{7/2} \sqrt{a+b x}+\frac{1}{12} a x^{7/2} (a+b x)^{3/2}+\frac{1}{6} x^{7/2} (a+b x)^{5/2}+\frac{\left (5 a^5\right ) \int \frac{\sqrt{x}}{\sqrt{a+b x}} \, dx}{512 b^2}\\ &=\frac{5 a^5 \sqrt{x} \sqrt{a+b x}}{512 b^3}-\frac{5 a^4 x^{3/2} \sqrt{a+b x}}{768 b^2}+\frac{a^3 x^{5/2} \sqrt{a+b x}}{192 b}+\frac{1}{32} a^2 x^{7/2} \sqrt{a+b x}+\frac{1}{12} a x^{7/2} (a+b x)^{3/2}+\frac{1}{6} x^{7/2} (a+b x)^{5/2}-\frac{\left (5 a^6\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx}{1024 b^3}\\ &=\frac{5 a^5 \sqrt{x} \sqrt{a+b x}}{512 b^3}-\frac{5 a^4 x^{3/2} \sqrt{a+b x}}{768 b^2}+\frac{a^3 x^{5/2} \sqrt{a+b x}}{192 b}+\frac{1}{32} a^2 x^{7/2} \sqrt{a+b x}+\frac{1}{12} a x^{7/2} (a+b x)^{3/2}+\frac{1}{6} x^{7/2} (a+b x)^{5/2}-\frac{\left (5 a^6\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )}{512 b^3}\\ &=\frac{5 a^5 \sqrt{x} \sqrt{a+b x}}{512 b^3}-\frac{5 a^4 x^{3/2} \sqrt{a+b x}}{768 b^2}+\frac{a^3 x^{5/2} \sqrt{a+b x}}{192 b}+\frac{1}{32} a^2 x^{7/2} \sqrt{a+b x}+\frac{1}{12} a x^{7/2} (a+b x)^{3/2}+\frac{1}{6} x^{7/2} (a+b x)^{5/2}-\frac{\left (5 a^6\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a+b x}}\right )}{512 b^3}\\ &=\frac{5 a^5 \sqrt{x} \sqrt{a+b x}}{512 b^3}-\frac{5 a^4 x^{3/2} \sqrt{a+b x}}{768 b^2}+\frac{a^3 x^{5/2} \sqrt{a+b x}}{192 b}+\frac{1}{32} a^2 x^{7/2} \sqrt{a+b x}+\frac{1}{12} a x^{7/2} (a+b x)^{3/2}+\frac{1}{6} x^{7/2} (a+b x)^{5/2}-\frac{5 a^6 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{512 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.233578, size = 118, normalized size = 0.72 \[ \frac{\sqrt{a+b x} \left (\sqrt{b} \sqrt{x} \left (8 a^3 b^2 x^2+432 a^2 b^3 x^3-10 a^4 b x+15 a^5+640 a b^4 x^4+256 b^5 x^5\right )-\frac{15 a^{11/2} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{\frac{b x}{a}+1}}\right )}{1536 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(a + b*x)^(5/2),x]

[Out]

(Sqrt[a + b*x]*(Sqrt[b]*Sqrt[x]*(15*a^5 - 10*a^4*b*x + 8*a^3*b^2*x^2 + 432*a^2*b^3*x^3 + 640*a*b^4*x^4 + 256*b
^5*x^5) - (15*a^(11/2)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/Sqrt[1 + (b*x)/a]))/(1536*b^(7/2))

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Maple [A]  time = 0.003, size = 156, normalized size = 1. \begin{align*}{\frac{1}{6\,b}{x}^{{\frac{5}{2}}} \left ( bx+a \right ) ^{{\frac{7}{2}}}}-{\frac{a}{12\,{b}^{2}}{x}^{{\frac{3}{2}}} \left ( bx+a \right ) ^{{\frac{7}{2}}}}+{\frac{{a}^{2}}{32\,{b}^{3}}\sqrt{x} \left ( bx+a \right ) ^{{\frac{7}{2}}}}-{\frac{{a}^{3}}{192\,{b}^{3}} \left ( bx+a \right ) ^{{\frac{5}{2}}}\sqrt{x}}-{\frac{5\,{a}^{4}}{768\,{b}^{3}} \left ( bx+a \right ) ^{{\frac{3}{2}}}\sqrt{x}}-{\frac{5\,{a}^{5}}{512\,{b}^{3}}\sqrt{x}\sqrt{bx+a}}-{\frac{5\,{a}^{6}}{1024}\sqrt{x \left ( bx+a \right ) }\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ){b}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(b*x+a)^(5/2),x)

[Out]

1/6/b*x^(5/2)*(b*x+a)^(7/2)-1/12/b^2*a*x^(3/2)*(b*x+a)^(7/2)+1/32/b^3*a^2*x^(1/2)*(b*x+a)^(7/2)-1/192/b^3*a^3*
(b*x+a)^(5/2)*x^(1/2)-5/768/b^3*a^4*(b*x+a)^(3/2)*x^(1/2)-5/512*a^5*x^(1/2)*(b*x+a)^(1/2)/b^3-5/1024/b^(7/2)*a
^6*(x*(b*x+a))^(1/2)/(b*x+a)^(1/2)/x^(1/2)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(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^(5/2)*(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.9423, size = 520, normalized size = 3.17 \begin{align*} \left [\frac{15 \, a^{6} \sqrt{b} \log \left (2 \, b x - 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (256 \, b^{6} x^{5} + 640 \, a b^{5} x^{4} + 432 \, a^{2} b^{4} x^{3} + 8 \, a^{3} b^{3} x^{2} - 10 \, a^{4} b^{2} x + 15 \, a^{5} b\right )} \sqrt{b x + a} \sqrt{x}}{3072 \, b^{4}}, \frac{15 \, a^{6} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (256 \, b^{6} x^{5} + 640 \, a b^{5} x^{4} + 432 \, a^{2} b^{4} x^{3} + 8 \, a^{3} b^{3} x^{2} - 10 \, a^{4} b^{2} x + 15 \, a^{5} b\right )} \sqrt{b x + a} \sqrt{x}}{1536 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3072*(15*a^6*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(256*b^6*x^5 + 640*a*b^5*x^4 + 43
2*a^2*b^4*x^3 + 8*a^3*b^3*x^2 - 10*a^4*b^2*x + 15*a^5*b)*sqrt(b*x + a)*sqrt(x))/b^4, 1/1536*(15*a^6*sqrt(-b)*a
rctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (256*b^6*x^5 + 640*a*b^5*x^4 + 432*a^2*b^4*x^3 + 8*a^3*b^3*x^2 - 1
0*a^4*b^2*x + 15*a^5*b)*sqrt(b*x + a)*sqrt(x))/b^4]

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Sympy [A]  time = 102.469, size = 207, normalized size = 1.26 \begin{align*} \frac{5 a^{\frac{11}{2}} \sqrt{x}}{512 b^{3} \sqrt{1 + \frac{b x}{a}}} + \frac{5 a^{\frac{9}{2}} x^{\frac{3}{2}}}{1536 b^{2} \sqrt{1 + \frac{b x}{a}}} - \frac{a^{\frac{7}{2}} x^{\frac{5}{2}}}{768 b \sqrt{1 + \frac{b x}{a}}} + \frac{55 a^{\frac{5}{2}} x^{\frac{7}{2}}}{192 \sqrt{1 + \frac{b x}{a}}} + \frac{67 a^{\frac{3}{2}} b x^{\frac{9}{2}}}{96 \sqrt{1 + \frac{b x}{a}}} + \frac{7 \sqrt{a} b^{2} x^{\frac{11}{2}}}{12 \sqrt{1 + \frac{b x}{a}}} - \frac{5 a^{6} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{512 b^{\frac{7}{2}}} + \frac{b^{3} x^{\frac{13}{2}}}{6 \sqrt{a} \sqrt{1 + \frac{b x}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(b*x+a)**(5/2),x)

[Out]

5*a**(11/2)*sqrt(x)/(512*b**3*sqrt(1 + b*x/a)) + 5*a**(9/2)*x**(3/2)/(1536*b**2*sqrt(1 + b*x/a)) - a**(7/2)*x*
*(5/2)/(768*b*sqrt(1 + b*x/a)) + 55*a**(5/2)*x**(7/2)/(192*sqrt(1 + b*x/a)) + 67*a**(3/2)*b*x**(9/2)/(96*sqrt(
1 + b*x/a)) + 7*sqrt(a)*b**2*x**(11/2)/(12*sqrt(1 + b*x/a)) - 5*a**6*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(512*b**(7
/2)) + b**3*x**(13/2)/(6*sqrt(a)*sqrt(1 + b*x/a))

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Giac [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^(5/2)*(b*x+a)^(5/2),x, algorithm="giac")

[Out]

Timed out