3.546 \(\int x^{3/2} (a+b x)^{5/2} \, dx\)

Optimal. Leaf size=140 \[ \frac {3 a^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{5/2}}-\frac {3 a^4 \sqrt {x} \sqrt {a+b x}}{128 b^2}+\frac {a^3 x^{3/2} \sqrt {a+b x}}{64 b}+\frac {1}{16} a^2 x^{5/2} \sqrt {a+b x}+\frac {1}{8} a x^{5/2} (a+b x)^{3/2}+\frac {1}{5} x^{5/2} (a+b x)^{5/2} \]

[Out]

1/8*a*x^(5/2)*(b*x+a)^(3/2)+1/5*x^(5/2)*(b*x+a)^(5/2)+3/128*a^5*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(5/2)
+1/64*a^3*x^(3/2)*(b*x+a)^(1/2)/b+1/16*a^2*x^(5/2)*(b*x+a)^(1/2)-3/128*a^4*x^(1/2)*(b*x+a)^(1/2)/b^2

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Rubi [A]  time = 0.05, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {50, 63, 217, 206} \[ -\frac {3 a^4 \sqrt {x} \sqrt {a+b x}}{128 b^2}+\frac {3 a^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{5/2}}+\frac {a^3 x^{3/2} \sqrt {a+b x}}{64 b}+\frac {1}{16} a^2 x^{5/2} \sqrt {a+b x}+\frac {1}{8} a x^{5/2} (a+b x)^{3/2}+\frac {1}{5} x^{5/2} (a+b x)^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(a + b*x)^(5/2),x]

[Out]

(-3*a^4*Sqrt[x]*Sqrt[a + b*x])/(128*b^2) + (a^3*x^(3/2)*Sqrt[a + b*x])/(64*b) + (a^2*x^(5/2)*Sqrt[a + b*x])/16
+ (a*x^(5/2)*(a + b*x)^(3/2))/8 + (x^(5/2)*(a + b*x)^(5/2))/5 + (3*a^5*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x
]])/(128*b^(5/2))

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^{3/2} (a+b x)^{5/2} \, dx &=\frac {1}{5} x^{5/2} (a+b x)^{5/2}+\frac {1}{2} a \int x^{3/2} (a+b x)^{3/2} \, dx\\ &=\frac {1}{8} a x^{5/2} (a+b x)^{3/2}+\frac {1}{5} x^{5/2} (a+b x)^{5/2}+\frac {1}{16} \left (3 a^2\right ) \int x^{3/2} \sqrt {a+b x} \, dx\\ &=\frac {1}{16} a^2 x^{5/2} \sqrt {a+b x}+\frac {1}{8} a x^{5/2} (a+b x)^{3/2}+\frac {1}{5} x^{5/2} (a+b x)^{5/2}+\frac {1}{32} a^3 \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx\\ &=\frac {a^3 x^{3/2} \sqrt {a+b x}}{64 b}+\frac {1}{16} a^2 x^{5/2} \sqrt {a+b x}+\frac {1}{8} a x^{5/2} (a+b x)^{3/2}+\frac {1}{5} x^{5/2} (a+b x)^{5/2}-\frac {\left (3 a^4\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{128 b}\\ &=-\frac {3 a^4 \sqrt {x} \sqrt {a+b x}}{128 b^2}+\frac {a^3 x^{3/2} \sqrt {a+b x}}{64 b}+\frac {1}{16} a^2 x^{5/2} \sqrt {a+b x}+\frac {1}{8} a x^{5/2} (a+b x)^{3/2}+\frac {1}{5} x^{5/2} (a+b x)^{5/2}+\frac {\left (3 a^5\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{256 b^2}\\ &=-\frac {3 a^4 \sqrt {x} \sqrt {a+b x}}{128 b^2}+\frac {a^3 x^{3/2} \sqrt {a+b x}}{64 b}+\frac {1}{16} a^2 x^{5/2} \sqrt {a+b x}+\frac {1}{8} a x^{5/2} (a+b x)^{3/2}+\frac {1}{5} x^{5/2} (a+b x)^{5/2}+\frac {\left (3 a^5\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{128 b^2}\\ &=-\frac {3 a^4 \sqrt {x} \sqrt {a+b x}}{128 b^2}+\frac {a^3 x^{3/2} \sqrt {a+b x}}{64 b}+\frac {1}{16} a^2 x^{5/2} \sqrt {a+b x}+\frac {1}{8} a x^{5/2} (a+b x)^{3/2}+\frac {1}{5} x^{5/2} (a+b x)^{5/2}+\frac {\left (3 a^5\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^2}\\ &=-\frac {3 a^4 \sqrt {x} \sqrt {a+b x}}{128 b^2}+\frac {a^3 x^{3/2} \sqrt {a+b x}}{64 b}+\frac {1}{16} a^2 x^{5/2} \sqrt {a+b x}+\frac {1}{8} a x^{5/2} (a+b x)^{3/2}+\frac {1}{5} x^{5/2} (a+b x)^{5/2}+\frac {3 a^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 107, normalized size = 0.76 \[ \frac {\sqrt {a+b x} \left (\frac {15 a^{9/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {\frac {b x}{a}+1}}+\sqrt {b} \sqrt {x} \left (-15 a^4+10 a^3 b x+248 a^2 b^2 x^2+336 a b^3 x^3+128 b^4 x^4\right )\right )}{640 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*(a + b*x)^(5/2),x]

[Out]

(Sqrt[a + b*x]*(Sqrt[b]*Sqrt[x]*(-15*a^4 + 10*a^3*b*x + 248*a^2*b^2*x^2 + 336*a*b^3*x^3 + 128*b^4*x^4) + (15*a
^(9/2)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/Sqrt[1 + (b*x)/a]))/(640*b^(5/2))

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fricas [A]  time = 0.44, size = 185, normalized size = 1.32 \[ \left [\frac {15 \, a^{5} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (128 \, b^{5} x^{4} + 336 \, a b^{4} x^{3} + 248 \, a^{2} b^{3} x^{2} + 10 \, a^{3} b^{2} x - 15 \, a^{4} b\right )} \sqrt {b x + a} \sqrt {x}}{1280 \, b^{3}}, -\frac {15 \, a^{5} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (128 \, b^{5} x^{4} + 336 \, a b^{4} x^{3} + 248 \, a^{2} b^{3} x^{2} + 10 \, a^{3} b^{2} x - 15 \, a^{4} b\right )} \sqrt {b x + a} \sqrt {x}}{640 \, b^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[1/1280*(15*a^5*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(128*b^5*x^4 + 336*a*b^4*x^3 + 24
8*a^2*b^3*x^2 + 10*a^3*b^2*x - 15*a^4*b)*sqrt(b*x + a)*sqrt(x))/b^3, -1/640*(15*a^5*sqrt(-b)*arctan(sqrt(b*x +
a)*sqrt(-b)/(b*sqrt(x))) - (128*b^5*x^4 + 336*a*b^4*x^3 + 248*a^2*b^3*x^2 + 10*a^3*b^2*x - 15*a^4*b)*sqrt(b*x
+ a)*sqrt(x))/b^3]

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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^(3/2)*(b*x+a)^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 138, normalized size = 0.99 \[ \frac {3 \sqrt {\left (b x +a \right ) x}\, a^{5} \ln \left (\frac {b x +\frac {a}{2}}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{256 \sqrt {b x +a}\, b^{\frac {5}{2}} \sqrt {x}}+\frac {3 \sqrt {b x +a}\, a^{4} \sqrt {x}}{128 b^{2}}+\frac {\left (b x +a \right )^{\frac {3}{2}} a^{3} \sqrt {x}}{64 b^{2}}+\frac {\left (b x +a \right )^{\frac {5}{2}} a^{2} \sqrt {x}}{80 b^{2}}+\frac {\left (b x +a \right )^{\frac {7}{2}} x^{\frac {3}{2}}}{5 b}-\frac {3 \left (b x +a \right )^{\frac {7}{2}} a \sqrt {x}}{40 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(b*x+a)^(5/2),x)

[Out]

1/5/b*x^(3/2)*(b*x+a)^(7/2)-3/40*a/b^2*x^(1/2)*(b*x+a)^(7/2)+1/80*a^2/b^2*(b*x+a)^(5/2)*x^(1/2)+1/64*a^3/b^2*(
b*x+a)^(3/2)*x^(1/2)+3/128*a^4*x^(1/2)*(b*x+a)^(1/2)/b^2+3/256*a^5/b^(5/2)*((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 = 2.98, size = 212, normalized size = 1.51 \[ -\frac {3 \, a^{5} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{256 \, b^{\frac {5}{2}}} - \frac {\frac {15 \, \sqrt {b x + a} a^{5} b^{4}}{\sqrt {x}} - \frac {70 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{5} b^{3}}{x^{\frac {3}{2}}} + \frac {128 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{5} b^{2}}{x^{\frac {5}{2}}} + \frac {70 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{5} b}{x^{\frac {7}{2}}} - \frac {15 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{5}}{x^{\frac {9}{2}}}}{640 \, {\left (b^{7} - \frac {5 \, {\left (b x + a\right )} b^{6}}{x} + \frac {10 \, {\left (b x + a\right )}^{2} b^{5}}{x^{2}} - \frac {10 \, {\left (b x + a\right )}^{3} b^{4}}{x^{3}} + \frac {5 \, {\left (b x + a\right )}^{4} b^{3}}{x^{4}} - \frac {{\left (b x + a\right )}^{5} b^{2}}{x^{5}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

-3/256*a^5*log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt(b*x + a)/sqrt(x)))/b^(5/2) - 1/640*(15*sqrt(
b*x + a)*a^5*b^4/sqrt(x) - 70*(b*x + a)^(3/2)*a^5*b^3/x^(3/2) + 128*(b*x + a)^(5/2)*a^5*b^2/x^(5/2) + 70*(b*x
+ a)^(7/2)*a^5*b/x^(7/2) - 15*(b*x + a)^(9/2)*a^5/x^(9/2))/(b^7 - 5*(b*x + a)*b^6/x + 10*(b*x + a)^2*b^5/x^2 -
10*(b*x + a)^3*b^4/x^3 + 5*(b*x + a)^4*b^3/x^4 - (b*x + a)^5*b^2/x^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{3/2}\,{\left (a+b\,x\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(a + b*x)^(5/2),x)

[Out]

int(x^(3/2)*(a + b*x)^(5/2), x)

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sympy [A]  time = 16.41, size = 180, normalized size = 1.29 \[ - \frac {3 a^{\frac {9}{2}} \sqrt {x}}{128 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {a^{\frac {7}{2}} x^{\frac {3}{2}}}{128 b \sqrt {1 + \frac {b x}{a}}} + \frac {129 a^{\frac {5}{2}} x^{\frac {5}{2}}}{320 \sqrt {1 + \frac {b x}{a}}} + \frac {73 a^{\frac {3}{2}} b x^{\frac {7}{2}}}{80 \sqrt {1 + \frac {b x}{a}}} + \frac {29 \sqrt {a} b^{2} x^{\frac {9}{2}}}{40 \sqrt {1 + \frac {b x}{a}}} + \frac {3 a^{5} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{128 b^{\frac {5}{2}}} + \frac {b^{3} x^{\frac {11}{2}}}{5 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(b*x+a)**(5/2),x)

[Out]

-3*a**(9/2)*sqrt(x)/(128*b**2*sqrt(1 + b*x/a)) - a**(7/2)*x**(3/2)/(128*b*sqrt(1 + b*x/a)) + 129*a**(5/2)*x**(
5/2)/(320*sqrt(1 + b*x/a)) + 73*a**(3/2)*b*x**(7/2)/(80*sqrt(1 + b*x/a)) + 29*sqrt(a)*b**2*x**(9/2)/(40*sqrt(1
+ b*x/a)) + 3*a**5*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(128*b**(5/2)) + b**3*x**(11/2)/(5*sqrt(a)*sqrt(1 + b*x/a))

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