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

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

[Out]

1/12*a*x^(7/2)*(b*x+a)^(3/2)+1/6*x^(7/2)*(b*x+a)^(5/2)-5/512*a^6*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(7/2
)-5/768*a^4*x^(3/2)*(b*x+a)^(1/2)/b^2+1/192*a^3*x^(5/2)*(b*x+a)^(1/2)/b+1/32*a^2*x^(7/2)*(b*x+a)^(1/2)+5/512*a
^5*x^(1/2)*(b*x+a)^(1/2)/b^3

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Rubi [A]  time = 0.06, antiderivative size = 164, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.24, size = 118, normalized size = 0.72 \[ \frac {\sqrt {a+b x} \left (\sqrt {b} \sqrt {x} \left (15 a^5-10 a^4 b x+8 a^3 b^2 x^2+432 a^2 b^3 x^3+640 a b^4 x^4+256 b^5 x^5\right )-\frac {15 a^{11/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {\frac {b x}{a}+1}}\right )}{1536 b^{7/2}} \]

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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fricas [A]  time = 0.49, size = 206, normalized size = 1.26 \[ \left [\frac {15 \, a^{6} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (256 \, b^{6} x^{5} + 640 \, a b^{5} x^{4} + 432 \, a^{2} b^{4} x^{3} + 8 \, a^{3} b^{3} x^{2} - 10 \, a^{4} b^{2} x + 15 \, a^{5} b\right )} \sqrt {b x + a} \sqrt {x}}{3072 \, b^{4}}, \frac {15 \, a^{6} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (256 \, b^{6} x^{5} + 640 \, a b^{5} x^{4} + 432 \, a^{2} b^{4} x^{3} + 8 \, a^{3} b^{3} x^{2} - 10 \, a^{4} b^{2} x + 15 \, a^{5} b\right )} \sqrt {b x + a} \sqrt {x}}{1536 \, b^{4}}\right ] \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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maple [A]  time = 0.00, size = 156, normalized size = 0.95 \[ -\frac {5 \sqrt {\left (b x +a \right ) x}\, a^{6} \ln \left (\frac {b x +\frac {a}{2}}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{1024 \sqrt {b x +a}\, b^{\frac {7}{2}} \sqrt {x}}-\frac {5 \sqrt {b x +a}\, a^{5} \sqrt {x}}{512 b^{3}}-\frac {5 \left (b x +a \right )^{\frac {3}{2}} a^{4} \sqrt {x}}{768 b^{3}}+\frac {\left (b x +a \right )^{\frac {7}{2}} x^{\frac {5}{2}}}{6 b}-\frac {\left (b x +a \right )^{\frac {5}{2}} a^{3} \sqrt {x}}{192 b^{3}}-\frac {\left (b x +a \right )^{\frac {7}{2}} a \,x^{\frac {3}{2}}}{12 b^{2}}+\frac {\left (b x +a \right )^{\frac {7}{2}} a^{2} \sqrt {x}}{32 b^{3}} \]

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*a/b^2*x^(3/2)*(b*x+a)^(7/2)+1/32*a^2/b^3*x^(1/2)*(b*x+a)^(7/2)-1/192*a^3/b^3*
(b*x+a)^(5/2)*x^(1/2)-5/768*a^4/b^3*(b*x+a)^(3/2)*x^(1/2)-5/512*a^5*x^(1/2)*(b*x+a)^(1/2)/b^3-5/1024*a^6/b^(7/
2)*((b*x+a)*x)^(1/2)/(b*x+a)^(1/2)/x^(1/2)*ln((b*x+1/2*a)/b^(1/2)+(b*x^2+a*x)^(1/2))

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maxima [B]  time = 2.99, size = 244, normalized size = 1.49 \[ \frac {5 \, a^{6} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{1024 \, b^{\frac {7}{2}}} + \frac {\frac {15 \, \sqrt {b x + a} a^{6} b^{5}}{\sqrt {x}} - \frac {85 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{6} b^{4}}{x^{\frac {3}{2}}} + \frac {198 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{6} b^{3}}{x^{\frac {5}{2}}} + \frac {198 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{6} b^{2}}{x^{\frac {7}{2}}} - \frac {85 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{6} b}{x^{\frac {9}{2}}} + \frac {15 \, {\left (b x + a\right )}^{\frac {11}{2}} a^{6}}{x^{\frac {11}{2}}}}{1536 \, {\left (b^{9} - \frac {6 \, {\left (b x + a\right )} b^{8}}{x} + \frac {15 \, {\left (b x + a\right )}^{2} b^{7}}{x^{2}} - \frac {20 \, {\left (b x + a\right )}^{3} b^{6}}{x^{3}} + \frac {15 \, {\left (b x + a\right )}^{4} b^{5}}{x^{4}} - \frac {6 \, {\left (b x + a\right )}^{5} b^{4}}{x^{5}} + \frac {{\left (b x + a\right )}^{6} b^{3}}{x^{6}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/1024*a^6*log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt(b*x + a)/sqrt(x)))/b^(7/2) + 1/1536*(15*sqrt
(b*x + a)*a^6*b^5/sqrt(x) - 85*(b*x + a)^(3/2)*a^6*b^4/x^(3/2) + 198*(b*x + a)^(5/2)*a^6*b^3/x^(5/2) + 198*(b*
x + a)^(7/2)*a^6*b^2/x^(7/2) - 85*(b*x + a)^(9/2)*a^6*b/x^(9/2) + 15*(b*x + a)^(11/2)*a^6/x^(11/2))/(b^9 - 6*(
b*x + a)*b^8/x + 15*(b*x + a)^2*b^7/x^2 - 20*(b*x + a)^3*b^6/x^3 + 15*(b*x + a)^4*b^5/x^4 - 6*(b*x + a)^5*b^4/
x^5 + (b*x + a)^6*b^3/x^6)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{5/2}\,{\left (a+b\,x\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(a + b*x)^(5/2),x)

[Out]

int(x^(5/2)*(a + b*x)^(5/2), x)

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sympy [A]  time = 25.94, size = 207, normalized size = 1.26 \[ \frac {5 a^{\frac {11}{2}} \sqrt {x}}{512 b^{3} \sqrt {1 + \frac {b x}{a}}} + \frac {5 a^{\frac {9}{2}} x^{\frac {3}{2}}}{1536 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {a^{\frac {7}{2}} x^{\frac {5}{2}}}{768 b \sqrt {1 + \frac {b x}{a}}} + \frac {55 a^{\frac {5}{2}} x^{\frac {7}{2}}}{192 \sqrt {1 + \frac {b x}{a}}} + \frac {67 a^{\frac {3}{2}} b x^{\frac {9}{2}}}{96 \sqrt {1 + \frac {b x}{a}}} + \frac {7 \sqrt {a} b^{2} x^{\frac {11}{2}}}{12 \sqrt {1 + \frac {b x}{a}}} - \frac {5 a^{6} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{512 b^{\frac {7}{2}}} + \frac {b^{3} x^{\frac {13}{2}}}{6 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]

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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