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

Optimal. Leaf size=122 \[ -\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{7/2}}+\frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x} \]

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Rubi [A]  time = 0.04, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {50, 63, 217, 206} \begin {gather*} -\frac {5 a^2 x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^3}-\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{7/2}}+\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*Sqrt[a + b*x],x]

[Out]

(5*a^3*Sqrt[x]*Sqrt[a + b*x])/(64*b^3) - (5*a^2*x^(3/2)*Sqrt[a + b*x])/(96*b^2) + (a*x^(5/2)*Sqrt[a + b*x])/(2
4*b) + (x^(7/2)*Sqrt[a + b*x])/4 - (5*a^4*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(64*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} \sqrt {a+b x} \, dx &=\frac {1}{4} x^{7/2} \sqrt {a+b x}+\frac {1}{8} a \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx\\ &=\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x}-\frac {\left (5 a^2\right ) \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{48 b}\\ &=-\frac {5 a^2 x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x}+\frac {\left (5 a^3\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{64 b^2}\\ &=\frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x}-\frac {\left (5 a^4\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{128 b^3}\\ &=\frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x}-\frac {\left (5 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{64 b^3}\\ &=\frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x}-\frac {\left (5 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^3}\\ &=\frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x}-\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 96, normalized size = 0.79 \begin {gather*} \frac {\sqrt {a+b x} \left (\sqrt {b} \sqrt {x} \left (15 a^3-10 a^2 b x+8 a b^2 x^2+48 b^3 x^3\right )-\frac {15 a^{7/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {\frac {b x}{a}+1}}\right )}{192 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*Sqrt[a + b*x],x]

[Out]

(Sqrt[a + b*x]*(Sqrt[b]*Sqrt[x]*(15*a^3 - 10*a^2*b*x + 8*a*b^2*x^2 + 48*b^3*x^3) - (15*a^(7/2)*ArcSinh[(Sqrt[b
]*Sqrt[x])/Sqrt[a]])/Sqrt[1 + (b*x)/a]))/(192*b^(7/2))

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IntegrateAlgebraic [A]  time = 0.14, size = 95, normalized size = 0.78 \begin {gather*} \frac {5 a^4 \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{64 b^{7/2}}+\frac {\sqrt {a+b x} \left (15 a^3 \sqrt {x}-10 a^2 b x^{3/2}+8 a b^2 x^{5/2}+48 b^3 x^{7/2}\right )}{192 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)*Sqrt[a + b*x],x]

[Out]

(Sqrt[a + b*x]*(15*a^3*Sqrt[x] - 10*a^2*b*x^(3/2) + 8*a*b^2*x^(5/2) + 48*b^3*x^(7/2)))/(192*b^3) + (5*a^4*Log[
-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/(64*b^(7/2))

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fricas [A]  time = 0.95, size = 162, normalized size = 1.33 \begin {gather*} \left [\frac {15 \, a^{4} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (48 \, b^{4} x^{3} + 8 \, a b^{3} x^{2} - 10 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{384 \, b^{4}}, \frac {15 \, a^{4} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (48 \, b^{4} x^{3} + 8 \, a b^{3} x^{2} - 10 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{192 \, b^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/384*(15*a^4*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(48*b^4*x^3 + 8*a*b^3*x^2 - 10*a^2
*b^2*x + 15*a^3*b)*sqrt(b*x + a)*sqrt(x))/b^4, 1/192*(15*a^4*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x)
)) + (48*b^4*x^3 + 8*a*b^3*x^2 - 10*a^2*b^2*x + 15*a^3*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 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.01, size = 120, normalized size = 0.98 \begin {gather*} \frac {\left (b x +a \right )^{\frac {3}{2}} x^{\frac {5}{2}}}{4 b}-\frac {5 \sqrt {\left (b x +a \right ) x}\, a^{4} \ln \left (\frac {b x +\frac {a}{2}}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{128 \sqrt {b x +a}\, b^{\frac {7}{2}} \sqrt {x}}-\frac {5 \sqrt {b x +a}\, a^{3} \sqrt {x}}{64 b^{3}}-\frac {5 \left (b x +a \right )^{\frac {3}{2}} a \,x^{\frac {3}{2}}}{24 b^{2}}+\frac {5 \left (b x +a \right )^{\frac {3}{2}} a^{2} \sqrt {x}}{32 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/b*x^(5/2)*(b*x+a)^(3/2)-5/24*a/b^2*x^(3/2)*(b*x+a)^(3/2)+5/32*a^2/b^3*x^(1/2)*(b*x+a)^(3/2)-5/64*a^3*x^(1/
2)*(b*x+a)^(1/2)/b^3-5/128*a^4/b^(7/2)*(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 [B]  time = 2.95, size = 178, normalized size = 1.46 \begin {gather*} \frac {5 \, a^{4} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{128 \, b^{\frac {7}{2}}} + \frac {\frac {15 \, \sqrt {b x + a} a^{4} b^{3}}{\sqrt {x}} + \frac {73 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} b^{2}}{x^{\frac {3}{2}}} - \frac {55 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} b}{x^{\frac {5}{2}}} + \frac {15 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{4}}{x^{\frac {7}{2}}}}{192 \, {\left (b^{7} - \frac {4 \, {\left (b x + a\right )} b^{6}}{x} + \frac {6 \, {\left (b x + a\right )}^{2} b^{5}}{x^{2}} - \frac {4 \, {\left (b x + a\right )}^{3} b^{4}}{x^{3}} + \frac {{\left (b x + a\right )}^{4} b^{3}}{x^{4}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/128*a^4*log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt(b*x + a)/sqrt(x)))/b^(7/2) + 1/192*(15*sqrt(b
*x + a)*a^4*b^3/sqrt(x) + 73*(b*x + a)^(3/2)*a^4*b^2/x^(3/2) - 55*(b*x + a)^(5/2)*a^4*b/x^(5/2) + 15*(b*x + a)
^(7/2)*a^4/x^(7/2))/(b^7 - 4*(b*x + a)*b^6/x + 6*(b*x + a)^2*b^5/x^2 - 4*(b*x + a)^3*b^4/x^3 + (b*x + a)^4*b^3
/x^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{5/2}\,\sqrt {a+b\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(a + b*x)^(1/2),x)

[Out]

int(x^(5/2)*(a + b*x)^(1/2), x)

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sympy [A]  time = 11.70, size = 153, normalized size = 1.25 \begin {gather*} \frac {5 a^{\frac {7}{2}} \sqrt {x}}{64 b^{3} \sqrt {1 + \frac {b x}{a}}} + \frac {5 a^{\frac {5}{2}} x^{\frac {3}{2}}}{192 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {a^{\frac {3}{2}} x^{\frac {5}{2}}}{96 b \sqrt {1 + \frac {b x}{a}}} + \frac {7 \sqrt {a} x^{\frac {7}{2}}}{24 \sqrt {1 + \frac {b x}{a}}} - \frac {5 a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{64 b^{\frac {7}{2}}} + \frac {b x^{\frac {9}{2}}}{4 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*a**(7/2)*sqrt(x)/(64*b**3*sqrt(1 + b*x/a)) + 5*a**(5/2)*x**(3/2)/(192*b**2*sqrt(1 + b*x/a)) - a**(3/2)*x**(5
/2)/(96*b*sqrt(1 + b*x/a)) + 7*sqrt(a)*x**(7/2)/(24*sqrt(1 + b*x/a)) - 5*a**4*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(
64*b**(7/2)) + b*x**(9/2)/(4*sqrt(a)*sqrt(1 + b*x/a))

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