3.6.47 \(\int \sqrt {x} (a+b x)^{5/2} \, dx\) [547]

Optimal. Leaf size=116 \[ \frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b}+\frac {5}{32} a^2 x^{3/2} \sqrt {a+b x}+\frac {5}{24} a x^{3/2} (a+b x)^{3/2}+\frac {1}{4} x^{3/2} (a+b x)^{5/2}-\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{3/2}} \]

[Out]

5/24*a*x^(3/2)*(b*x+a)^(3/2)+1/4*x^(3/2)*(b*x+a)^(5/2)-5/64*a^4*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(3/2)
+5/32*a^2*x^(3/2)*(b*x+a)^(1/2)+5/64*a^3*x^(1/2)*(b*x+a)^(1/2)/b

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Rubi [A]
time = 0.03, antiderivative size = 116, 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 {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{3/2}}+\frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b}+\frac {5}{32} a^2 x^{3/2} \sqrt {a+b x}+\frac {5}{24} a x^{3/2} (a+b x)^{3/2}+\frac {1}{4} x^{3/2} (a+b x)^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(a + b*x)^(5/2),x]

[Out]

(5*a^3*Sqrt[x]*Sqrt[a + b*x])/(64*b) + (5*a^2*x^(3/2)*Sqrt[a + b*x])/32 + (5*a*x^(3/2)*(a + b*x)^(3/2))/24 + (
x^(3/2)*(a + b*x)^(5/2))/4 - (5*a^4*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(64*b^(3/2))

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 \sqrt {x} (a+b x)^{5/2} \, dx &=\frac {1}{4} x^{3/2} (a+b x)^{5/2}+\frac {1}{8} (5 a) \int \sqrt {x} (a+b x)^{3/2} \, dx\\ &=\frac {5}{24} a x^{3/2} (a+b x)^{3/2}+\frac {1}{4} x^{3/2} (a+b x)^{5/2}+\frac {1}{16} \left (5 a^2\right ) \int \sqrt {x} \sqrt {a+b x} \, dx\\ &=\frac {5}{32} a^2 x^{3/2} \sqrt {a+b x}+\frac {5}{24} a x^{3/2} (a+b x)^{3/2}+\frac {1}{4} x^{3/2} (a+b x)^{5/2}+\frac {1}{64} \left (5 a^3\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx\\ &=\frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b}+\frac {5}{32} a^2 x^{3/2} \sqrt {a+b x}+\frac {5}{24} a x^{3/2} (a+b x)^{3/2}+\frac {1}{4} x^{3/2} (a+b x)^{5/2}-\frac {\left (5 a^4\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{128 b}\\ &=\frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b}+\frac {5}{32} a^2 x^{3/2} \sqrt {a+b x}+\frac {5}{24} a x^{3/2} (a+b x)^{3/2}+\frac {1}{4} x^{3/2} (a+b x)^{5/2}-\frac {\left (5 a^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{64 b}\\ &=\frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b}+\frac {5}{32} a^2 x^{3/2} \sqrt {a+b x}+\frac {5}{24} a x^{3/2} (a+b x)^{3/2}+\frac {1}{4} x^{3/2} (a+b x)^{5/2}-\frac {\left (5 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}\\ &=\frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b}+\frac {5}{32} a^2 x^{3/2} \sqrt {a+b x}+\frac {5}{24} a x^{3/2} (a+b x)^{3/2}+\frac {1}{4} x^{3/2} (a+b x)^{5/2}-\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 87, normalized size = 0.75 \begin {gather*} \frac {\sqrt {x} \sqrt {a+b x} \left (15 a^3+118 a^2 b x+136 a b^2 x^2+48 b^3 x^3\right )}{192 b}+\frac {5 a^4 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{64 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*(a + b*x)^(5/2),x]

[Out]

(Sqrt[x]*Sqrt[a + b*x]*(15*a^3 + 118*a^2*b*x + 136*a*b^2*x^2 + 48*b^3*x^3))/(192*b) + (5*a^4*Log[-(Sqrt[b]*Sqr
t[x]) + Sqrt[a + b*x]])/(64*b^(3/2))

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Maple [A]
time = 0.12, size = 113, normalized size = 0.97

method	result	size
risch	\(\frac {\left (48 b^{3} x^{3}+136 a \,b^{2} x^{2}+118 a^{2} b x +15 a^{3}\right ) \sqrt {x}\, \sqrt {b x +a}}{192 b}-\frac {5 a^{4} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {x^{2} b +a x}\right ) \sqrt {x \left (b x +a \right )}}{128 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(98\)
default	\(\frac {x^{\frac {3}{2}} \left (b x +a \right )^{\frac {5}{2}}}{4}+\frac {5 a \left (\frac {x^{\frac {3}{2}} \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {a \left (\frac {x^{\frac {3}{2}} \sqrt {b x +a}}{2}+\frac {a \left (\frac {\sqrt {x}\, \sqrt {b x +a}}{b}-\frac {a \sqrt {x \left (b x +a \right )}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {x^{2} b +a x}\right )}{2 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {b x +a}}\right )}{4}\right )}{2}\right )}{8}\)	\(113\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 176 vs. \(2 (82) = 164\).
time = 0.52, size = 176, normalized size = 1.52 \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 {3}{2}}} + \frac {\frac {15 \, \sqrt {b x + a} a^{4} b^{3}}{\sqrt {x}} - \frac {55 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} b^{2}}{x^{\frac {3}{2}}} + \frac {73 \, {\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^{5} - \frac {4 \, {\left (b x + a\right )} b^{4}}{x} + \frac {6 \, {\left (b x + a\right )}^{2} b^{3}}{x^{2}} - \frac {4 \, {\left (b x + a\right )}^{3} b^{2}}{x^{3}} + \frac {{\left (b x + a\right )}^{4} b}{x^{4}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]
time = 0.43, size = 162, normalized size = 1.40 \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} + 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{384 \, b^{2}}, \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} + 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{192 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*x^(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 + 136*a*b^3*x^2 + 118*
a^2*b^2*x + 15*a^3*b)*sqrt(b*x + a)*sqrt(x))/b^2, 1/192*(15*a^4*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt
(x))) + (48*b^4*x^3 + 136*a*b^3*x^2 + 118*a^2*b^2*x + 15*a^3*b)*sqrt(b*x + a)*sqrt(x))/b^2]

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Sympy [A]
time = 8.52, size = 155, normalized size = 1.34 \begin {gather*} \frac {5 a^{\frac {7}{2}} \sqrt {x}}{64 b \sqrt {1 + \frac {b x}{a}}} + \frac {133 a^{\frac {5}{2}} x^{\frac {3}{2}}}{192 \sqrt {1 + \frac {b x}{a}}} + \frac {127 a^{\frac {3}{2}} b x^{\frac {5}{2}}}{96 \sqrt {1 + \frac {b x}{a}}} + \frac {23 \sqrt {a} b^{2} 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 {3}{2}}} + \frac {b^{3} 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((b*x+a)**(5/2)*x**(1/2),x)

[Out]

5*a**(7/2)*sqrt(x)/(64*b*sqrt(1 + b*x/a)) + 133*a**(5/2)*x**(3/2)/(192*sqrt(1 + b*x/a)) + 127*a**(3/2)*b*x**(5
/2)/(96*sqrt(1 + b*x/a)) + 23*sqrt(a)*b**2*x**(7/2)/(24*sqrt(1 + b*x/a)) - 5*a**4*asinh(sqrt(b)*sqrt(x)/sqrt(a
))/(64*b**(3/2)) + b**3*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((b*x+a)^(5/2)*x^(1/2),x, algorithm="giac")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(a + b*x)^(5/2),x)

[Out]

int(x^(1/2)*(a + b*x)^(5/2), x)

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