3.6.23 \(\int \sqrt {x} (a+b x)^{3/2} \, dx\)

Optimal. Leaf size=95 \[ -\frac {a^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{3/2}}+\frac {a^2 \sqrt {x} \sqrt {a+b x}}{8 b}+\frac {1}{4} a x^{3/2} \sqrt {a+b x}+\frac {1}{3} x^{3/2} (a+b x)^{3/2} \]

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Rubi [A]  time = 0.03, antiderivative size = 95, normalized size of antiderivative = 1.00, 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 = {50, 63, 217, 206} \begin {gather*} -\frac {a^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{3/2}}+\frac {a^2 \sqrt {x} \sqrt {a+b x}}{8 b}+\frac {1}{4} a x^{3/2} \sqrt {a+b x}+\frac {1}{3} x^{3/2} (a+b x)^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(a + b*x)^(3/2),x]

[Out]

(a^2*Sqrt[x]*Sqrt[a + b*x])/(8*b) + (a*x^(3/2)*Sqrt[a + b*x])/4 + (x^(3/2)*(a + b*x)^(3/2))/3 - (a^3*ArcTanh[(
Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(8*b^(3/2))

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 \sqrt {x} (a+b x)^{3/2} \, dx &=\frac {1}{3} x^{3/2} (a+b x)^{3/2}+\frac {1}{2} a \int \sqrt {x} \sqrt {a+b x} \, dx\\ &=\frac {1}{4} a x^{3/2} \sqrt {a+b x}+\frac {1}{3} x^{3/2} (a+b x)^{3/2}+\frac {1}{8} a^2 \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx\\ &=\frac {a^2 \sqrt {x} \sqrt {a+b x}}{8 b}+\frac {1}{4} a x^{3/2} \sqrt {a+b x}+\frac {1}{3} x^{3/2} (a+b x)^{3/2}-\frac {a^3 \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{16 b}\\ &=\frac {a^2 \sqrt {x} \sqrt {a+b x}}{8 b}+\frac {1}{4} a x^{3/2} \sqrt {a+b x}+\frac {1}{3} x^{3/2} (a+b x)^{3/2}-\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{8 b}\\ &=\frac {a^2 \sqrt {x} \sqrt {a+b x}}{8 b}+\frac {1}{4} a x^{3/2} \sqrt {a+b x}+\frac {1}{3} x^{3/2} (a+b x)^{3/2}-\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{8 b}\\ &=\frac {a^2 \sqrt {x} \sqrt {a+b x}}{8 b}+\frac {1}{4} a x^{3/2} \sqrt {a+b x}+\frac {1}{3} x^{3/2} (a+b x)^{3/2}-\frac {a^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 85, normalized size = 0.89 \begin {gather*} \frac {\sqrt {a+b x} \left (\sqrt {b} \sqrt {x} \left (3 a^2+14 a b x+8 b^2 x^2\right )-\frac {3 a^{5/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {\frac {b x}{a}+1}}\right )}{24 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*(a + b*x)^(3/2),x]

[Out]

(Sqrt[a + b*x]*(Sqrt[b]*Sqrt[x]*(3*a^2 + 14*a*b*x + 8*b^2*x^2) - (3*a^(5/2)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]]
)/Sqrt[1 + (b*x)/a]))/(24*b^(3/2))

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IntegrateAlgebraic [A]  time = 0.11, size = 82, normalized size = 0.86 \begin {gather*} \frac {a^3 \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{8 b^{3/2}}+\frac {\sqrt {a+b x} \left (3 a^2 \sqrt {x}+14 a b x^{3/2}+8 b^2 x^{5/2}\right )}{24 b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[x]*(a + b*x)^(3/2),x]

[Out]

(Sqrt[a + b*x]*(3*a^2*Sqrt[x] + 14*a*b*x^(3/2) + 8*b^2*x^(5/2)))/(24*b) + (a^3*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a
+ b*x]])/(8*b^(3/2))

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fricas [A]  time = 0.72, size = 140, normalized size = 1.47 \begin {gather*} \left [\frac {3 \, a^{3} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (8 \, b^{3} x^{2} + 14 \, a b^{2} x + 3 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b^{2}}, \frac {3 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (8 \, b^{3} x^{2} + 14 \, a b^{2} x + 3 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*x^(1/2),x, algorithm="fricas")

[Out]

[1/48*(3*a^3*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(8*b^3*x^2 + 14*a*b^2*x + 3*a^2*b)*s
qrt(b*x + a)*sqrt(x))/b^2, 1/24*(3*a^3*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (8*b^3*x^2 + 14*a
*b^2*x + 3*a^2*b)*sqrt(b*x + a)*sqrt(x))/b^2]

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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((b*x+a)^(3/2)*x^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 96, normalized size = 1.01 \begin {gather*} \frac {\sqrt {b x +a}\, a \,x^{\frac {3}{2}}}{4}-\frac {\sqrt {\left (b x +a \right ) x}\, a^{3} \ln \left (\frac {b x +\frac {a}{2}}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{16 \sqrt {b x +a}\, b^{\frac {3}{2}} \sqrt {x}}+\frac {\sqrt {b x +a}\, a^{2} \sqrt {x}}{8 b}+\frac {\left (b x +a \right )^{\frac {3}{2}} x^{\frac {3}{2}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*x^(1/2),x)

[Out]

1/3*x^(3/2)*(b*x+a)^(3/2)+1/4*a*x^(3/2)*(b*x+a)^(1/2)+1/8*a^2*x^(1/2)*(b*x+a)^(1/2)/b-1/16*a^3/b^(3/2)*((b*x+a
)*x)^(1/2)/x^(1/2)/(b*x+a)^(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 = 3.02, size = 144, normalized size = 1.52 \begin {gather*} \frac {a^{3} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{16 \, b^{\frac {3}{2}}} + \frac {\frac {3 \, \sqrt {b x + a} a^{3} b^{2}}{\sqrt {x}} - \frac {8 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b}{x^{\frac {3}{2}}} - \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3}}{x^{\frac {5}{2}}}}{24 \, {\left (b^{4} - \frac {3 \, {\left (b x + a\right )} b^{3}}{x} + \frac {3 \, {\left (b x + a\right )}^{2} b^{2}}{x^{2}} - \frac {{\left (b x + a\right )}^{3} b}{x^{3}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*x^(1/2),x, algorithm="maxima")

[Out]

1/16*a^3*log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt(b*x + a)/sqrt(x)))/b^(3/2) + 1/24*(3*sqrt(b*x
+ a)*a^3*b^2/sqrt(x) - 8*(b*x + a)^(3/2)*a^3*b/x^(3/2) - 3*(b*x + a)^(5/2)*a^3/x^(5/2))/(b^4 - 3*(b*x + a)*b^3
/x + 3*(b*x + a)^2*b^2/x^2 - (b*x + a)^3*b/x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {x}\,{\left (a+b\,x\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(a + b*x)^(3/2),x)

[Out]

int(x^(1/2)*(a + b*x)^(3/2), x)

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sympy [A]  time = 5.59, size = 124, normalized size = 1.31 \begin {gather*} \frac {a^{\frac {5}{2}} \sqrt {x}}{8 b \sqrt {1 + \frac {b x}{a}}} + \frac {17 a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 \sqrt {1 + \frac {b x}{a}}} + \frac {11 \sqrt {a} b x^{\frac {5}{2}}}{12 \sqrt {1 + \frac {b x}{a}}} - \frac {a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 b^{\frac {3}{2}}} + \frac {b^{2} x^{\frac {7}{2}}}{3 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)*x**(1/2),x)

[Out]

a**(5/2)*sqrt(x)/(8*b*sqrt(1 + b*x/a)) + 17*a**(3/2)*x**(3/2)/(24*sqrt(1 + b*x/a)) + 11*sqrt(a)*b*x**(5/2)/(12
*sqrt(1 + b*x/a)) - a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(3/2)) + b**2*x**(7/2)/(3*sqrt(a)*sqrt(1 + b*x/a
))

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