3.6.22 \(\int x^{3/2} (a+b x)^{3/2} \, dx\) [522]

Optimal. Leaf size=119 \[ -\frac {3 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^2}+\frac {a^2 x^{3/2} \sqrt {a+b x}}{32 b}+\frac {1}{8} a x^{5/2} \sqrt {a+b x}+\frac {1}{4} x^{5/2} (a+b x)^{3/2}+\frac {3 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{5/2}} \]

[Out]

1/4*x^(5/2)*(b*x+a)^(3/2)+3/64*a^4*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(5/2)+1/32*a^2*x^(3/2)*(b*x+a)^(1/
2)/b+1/8*a*x^(5/2)*(b*x+a)^(1/2)-3/64*a^3*x^(1/2)*(b*x+a)^(1/2)/b^2

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Rubi [A]
time = 0.03, antiderivative size = 119, 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 = {52, 65, 223, 212} \begin {gather*} \frac {3 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{5/2}}-\frac {3 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^2}+\frac {a^2 x^{3/2} \sqrt {a+b x}}{32 b}+\frac {1}{8} a x^{5/2} \sqrt {a+b x}+\frac {1}{4} x^{5/2} (a+b x)^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(a + b*x)^(3/2),x]

[Out]

(-3*a^3*Sqrt[x]*Sqrt[a + b*x])/(64*b^2) + (a^2*x^(3/2)*Sqrt[a + b*x])/(32*b) + (a*x^(5/2)*Sqrt[a + b*x])/8 + (
x^(5/2)*(a + b*x)^(3/2))/4 + (3*a^4*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(64*b^(5/2))

Rule 52

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 65

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 212

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

Rule 223

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^{3/2} (a+b x)^{3/2} \, dx &=\frac {1}{4} x^{5/2} (a+b x)^{3/2}+\frac {1}{8} (3 a) \int x^{3/2} \sqrt {a+b x} \, dx\\ &=\frac {1}{8} a x^{5/2} \sqrt {a+b x}+\frac {1}{4} x^{5/2} (a+b x)^{3/2}+\frac {1}{16} a^2 \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx\\ &=\frac {a^2 x^{3/2} \sqrt {a+b x}}{32 b}+\frac {1}{8} a x^{5/2} \sqrt {a+b x}+\frac {1}{4} x^{5/2} (a+b x)^{3/2}-\frac {\left (3 a^3\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{64 b}\\ &=-\frac {3 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^2}+\frac {a^2 x^{3/2} \sqrt {a+b x}}{32 b}+\frac {1}{8} a x^{5/2} \sqrt {a+b x}+\frac {1}{4} x^{5/2} (a+b x)^{3/2}+\frac {\left (3 a^4\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{128 b^2}\\ &=-\frac {3 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^2}+\frac {a^2 x^{3/2} \sqrt {a+b x}}{32 b}+\frac {1}{8} a x^{5/2} \sqrt {a+b x}+\frac {1}{4} x^{5/2} (a+b x)^{3/2}+\frac {\left (3 a^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{64 b^2}\\ &=-\frac {3 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^2}+\frac {a^2 x^{3/2} \sqrt {a+b x}}{32 b}+\frac {1}{8} a x^{5/2} \sqrt {a+b x}+\frac {1}{4} x^{5/2} (a+b x)^{3/2}+\frac {\left (3 a^4\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^2}\\ &=-\frac {3 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^2}+\frac {a^2 x^{3/2} \sqrt {a+b x}}{32 b}+\frac {1}{8} a x^{5/2} \sqrt {a+b x}+\frac {1}{4} x^{5/2} (a+b x)^{3/2}+\frac {3 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 88, normalized size = 0.74 \begin {gather*} \frac {\sqrt {b} \sqrt {x} \sqrt {a+b x} \left (-3 a^3+2 a^2 b x+24 a b^2 x^2+16 b^3 x^3\right )-3 a^4 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{64 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*(a + b*x)^(3/2),x]

[Out]

(Sqrt[b]*Sqrt[x]*Sqrt[a + b*x]*(-3*a^3 + 2*a^2*b*x + 24*a*b^2*x^2 + 16*b^3*x^3) - 3*a^4*Log[-(Sqrt[b]*Sqrt[x])
+ Sqrt[a + b*x]])/(64*b^(5/2))

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Maple [A]
time = 0.10, size = 122, normalized size = 1.03

method	result	size
risch	\(-\frac {\left (-16 b^{3} x^{3}-24 a \,b^{2} x^{2}-2 a^{2} b x +3 a^{3}\right ) \sqrt {x}\, \sqrt {b x +a}}{64 b^{2}}+\frac {3 a^{4} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {x^{2} b +a x}\right ) \sqrt {x \left (b x +a \right )}}{128 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(98\)
default	\(\frac {x^{\frac {3}{2}} \left (b x +a \right )^{\frac {5}{2}}}{4 b}-\frac {3 a \left (\frac {\sqrt {x}\, \left (b x +a \right )^{\frac {5}{2}}}{3 b}-\frac {a \left (\frac {\left (b x +a \right )^{\frac {3}{2}} \sqrt {x}}{2}+\frac {3 a \left (\sqrt {x}\, \sqrt {b x +a}+\frac {a \sqrt {x \left (b x +a \right )}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {x^{2} b +a x}\right )}{2 \sqrt {b x +a}\, \sqrt {x}\, \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\)	\(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/4/b*x^(3/2)*(b*x+a)^(5/2)-3/8*a/b*(1/3/b*x^(1/2)*(b*x+a)^(5/2)-1/6*a/b*(1/2*(b*x+a)^(3/2)*x^(1/2)+3/4*a*(x^(
1/2)*(b*x+a)^(1/2)+1/2*a*(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))/b^(
1/2))))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (85) = 170\).
time = 0.52, size = 178, normalized size = 1.50 \begin {gather*} -\frac {3 \, 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 {5}{2}}} - \frac {\frac {3 \, \sqrt {b x + a} a^{4} b^{3}}{\sqrt {x}} - \frac {11 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} b^{2}}{x^{\frac {3}{2}}} - \frac {11 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} b}{x^{\frac {5}{2}}} + \frac {3 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{4}}{x^{\frac {7}{2}}}}{64 \, {\left (b^{6} - \frac {4 \, {\left (b x + a\right )} b^{5}}{x} + \frac {6 \, {\left (b x + a\right )}^{2} b^{4}}{x^{2}} - \frac {4 \, {\left (b x + a\right )}^{3} b^{3}}{x^{3}} + \frac {{\left (b x + a\right )}^{4} b^{2}}{x^{4}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/128*a^4*log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt(b*x + a)/sqrt(x)))/b^(5/2) - 1/64*(3*sqrt(b*
x + a)*a^4*b^3/sqrt(x) - 11*(b*x + a)^(3/2)*a^4*b^2/x^(3/2) - 11*(b*x + a)^(5/2)*a^4*b/x^(5/2) + 3*(b*x + a)^(
7/2)*a^4/x^(7/2))/(b^6 - 4*(b*x + a)*b^5/x + 6*(b*x + a)^2*b^4/x^2 - 4*(b*x + a)^3*b^3/x^3 + (b*x + a)^4*b^2/x
^4)

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Fricas [A]
time = 0.46, size = 163, normalized size = 1.37 \begin {gather*} \left [\frac {3 \, a^{4} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (16 \, b^{4} x^{3} + 24 \, a b^{3} x^{2} + 2 \, a^{2} b^{2} x - 3 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{128 \, b^{3}}, -\frac {3 \, a^{4} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (16 \, b^{4} x^{3} + 24 \, a b^{3} x^{2} + 2 \, a^{2} b^{2} x - 3 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{64 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/128*(3*a^4*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(16*b^4*x^3 + 24*a*b^3*x^2 + 2*a^2*
b^2*x - 3*a^3*b)*sqrt(b*x + a)*sqrt(x))/b^3, -1/64*(3*a^4*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x)))
- (16*b^4*x^3 + 24*a*b^3*x^2 + 2*a^2*b^2*x - 3*a^3*b)*sqrt(b*x + a)*sqrt(x))/b^3]

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Sympy [A]
time = 13.32, size = 153, normalized size = 1.29 \begin {gather*} - \frac {3 a^{\frac {7}{2}} \sqrt {x}}{64 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {a^{\frac {5}{2}} x^{\frac {3}{2}}}{64 b \sqrt {1 + \frac {b x}{a}}} + \frac {13 a^{\frac {3}{2}} x^{\frac {5}{2}}}{32 \sqrt {1 + \frac {b x}{a}}} + \frac {5 \sqrt {a} b x^{\frac {7}{2}}}{8 \sqrt {1 + \frac {b x}{a}}} + \frac {3 a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{64 b^{\frac {5}{2}}} + \frac {b^{2} 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**(3/2)*(b*x+a)**(3/2),x)

[Out]

-3*a**(7/2)*sqrt(x)/(64*b**2*sqrt(1 + b*x/a)) - a**(5/2)*x**(3/2)/(64*b*sqrt(1 + b*x/a)) + 13*a**(3/2)*x**(5/2
)/(32*sqrt(1 + b*x/a)) + 5*sqrt(a)*b*x**(7/2)/(8*sqrt(1 + b*x/a)) + 3*a**4*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(64*
b**(5/2)) + b**2*x**(9/2)/(4*sqrt(a)*sqrt(1 + b*x/a))

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Giac [F(-1)] Timed out
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^(3/2)*(b*x+a)^(3/2),x, algorithm="giac")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(a + b*x)^(3/2),x)

[Out]

int(x^(3/2)*(a + b*x)^(3/2), x)

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