3.1.26 \(\int (a x+b x^2)^{5/2} \, dx\) [26]

Optimal. Leaf size=118 \[ \frac {5 a^4 (a+2 b x) \sqrt {a x+b x^2}}{512 b^3}-\frac {5 a^2 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{192 b^2}+\frac {(a+2 b x) \left (a x+b x^2\right )^{5/2}}{12 b}-\frac {5 a^6 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{512 b^{7/2}} \]

[Out]

-5/192*a^2*(2*b*x+a)*(b*x^2+a*x)^(3/2)/b^2+1/12*(2*b*x+a)*(b*x^2+a*x)^(5/2)/b-5/512*a^6*arctanh(x*b^(1/2)/(b*x
^2+a*x)^(1/2))/b^(7/2)+5/512*a^4*(2*b*x+a)*(b*x^2+a*x)^(1/2)/b^3

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {626, 634, 212} \begin {gather*} -\frac {5 a^6 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{512 b^{7/2}}+\frac {5 a^4 (a+2 b x) \sqrt {a x+b x^2}}{512 b^3}-\frac {5 a^2 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{192 b^2}+\frac {(a+2 b x) \left (a x+b x^2\right )^{5/2}}{12 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(5/2),x]

[Out]

(5*a^4*(a + 2*b*x)*Sqrt[a*x + b*x^2])/(512*b^3) - (5*a^2*(a + 2*b*x)*(a*x + b*x^2)^(3/2))/(192*b^2) + ((a + 2*
b*x)*(a*x + b*x^2)^(5/2))/(12*b) - (5*a^6*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(512*b^(7/2))

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 626

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1
))), x] - Dist[p*((b^2 - 4*a*c)/(2*c*(2*p + 1))), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 634

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rubi steps

\begin {align*} \int \left (a x+b x^2\right )^{5/2} \, dx &=\frac {(a+2 b x) \left (a x+b x^2\right )^{5/2}}{12 b}-\frac {\left (5 a^2\right ) \int \left (a x+b x^2\right )^{3/2} \, dx}{24 b}\\ &=-\frac {5 a^2 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{192 b^2}+\frac {(a+2 b x) \left (a x+b x^2\right )^{5/2}}{12 b}+\frac {\left (5 a^4\right ) \int \sqrt {a x+b x^2} \, dx}{128 b^2}\\ &=\frac {5 a^4 (a+2 b x) \sqrt {a x+b x^2}}{512 b^3}-\frac {5 a^2 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{192 b^2}+\frac {(a+2 b x) \left (a x+b x^2\right )^{5/2}}{12 b}-\frac {\left (5 a^6\right ) \int \frac {1}{\sqrt {a x+b x^2}} \, dx}{1024 b^3}\\ &=\frac {5 a^4 (a+2 b x) \sqrt {a x+b x^2}}{512 b^3}-\frac {5 a^2 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{192 b^2}+\frac {(a+2 b x) \left (a x+b x^2\right )^{5/2}}{12 b}-\frac {\left (5 a^6\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right )}{512 b^3}\\ &=\frac {5 a^4 (a+2 b x) \sqrt {a x+b x^2}}{512 b^3}-\frac {5 a^2 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{192 b^2}+\frac {(a+2 b x) \left (a x+b x^2\right )^{5/2}}{12 b}-\frac {5 a^6 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{512 b^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.13, size = 121, normalized size = 1.03 \begin {gather*} \frac {\sqrt {x (a+b x)} \left (\sqrt {b} \left (15 a^5-10 a^4 b x+8 a^3 b^2 x^2+432 a^2 b^3 x^3+640 a b^4 x^4+256 b^5 x^5\right )+\frac {15 a^6 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{\sqrt {x} \sqrt {a+b x}}\right )}{1536 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(5/2),x]

[Out]

(Sqrt[x*(a + b*x)]*(Sqrt[b]*(15*a^5 - 10*a^4*b*x + 8*a^3*b^2*x^2 + 432*a^2*b^3*x^3 + 640*a*b^4*x^4 + 256*b^5*x
^5) + (15*a^6*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/(Sqrt[x]*Sqrt[a + b*x])))/(1536*b^(7/2))

________________________________________________________________________________________

Maple [A]
time = 0.51, size = 118, normalized size = 1.00

method	result	size
risch	\(\frac {\left (256 b^{5} x^{5}+640 a \,b^{4} x^{4}+432 a^{2} b^{3} x^{3}+8 a^{3} x^{2} b^{2}-10 a^{4} b x +15 a^{5}\right ) x \left (b x +a \right )}{1536 b^{3} \sqrt {x \left (b x +a \right )}}-\frac {5 a^{6} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{1024 b^{\frac {7}{2}}}\)	\(106\)
default	\(\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{12 b}-\frac {5 a^{2} \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )}{24 b}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*b*x+a)*(b*x^2+a*x)^(5/2)/b-5/24*a^2/b*(1/8*(2*b*x+a)/b*(b*x^2+a*x)^(3/2)-3/16*a^2/b*(1/4*(2*b*x+a)/b*(
b*x^2+a*x)^(1/2)-1/8*a^2/b^(3/2)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))))

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 141, normalized size = 1.19 \begin {gather*} \frac {1}{6} \, {\left (b x^{2} + a x\right )}^{\frac {5}{2}} x + \frac {5 \, \sqrt {b x^{2} + a x} a^{4} x}{256 \, b^{2}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2} x}{96 \, b} - \frac {5 \, a^{6} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{1024 \, b^{\frac {7}{2}}} + \frac {5 \, \sqrt {b x^{2} + a x} a^{5}}{512 \, b^{3}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{3}}{192 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} a}{12 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(b*x^2 + a*x)^(5/2)*x + 5/256*sqrt(b*x^2 + a*x)*a^4*x/b^2 - 5/96*(b*x^2 + a*x)^(3/2)*a^2*x/b - 5/1024*a^6*
log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2) + 5/512*sqrt(b*x^2 + a*x)*a^5/b^3 - 5/192*(b*x^2 + a*x)^(
3/2)*a^3/b^2 + 1/12*(b*x^2 + a*x)^(5/2)*a/b

________________________________________________________________________________________

Fricas [A]
time = 1.98, size = 213, normalized size = 1.81 \begin {gather*} \left [\frac {15 \, a^{6} \sqrt {b} \log \left (2 \, b x + a - 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (256 \, b^{6} x^{5} + 640 \, a b^{5} x^{4} + 432 \, a^{2} b^{4} x^{3} + 8 \, a^{3} b^{3} x^{2} - 10 \, a^{4} b^{2} x + 15 \, a^{5} b\right )} \sqrt {b x^{2} + a x}}{3072 \, b^{4}}, \frac {15 \, a^{6} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) + {\left (256 \, b^{6} x^{5} + 640 \, a b^{5} x^{4} + 432 \, a^{2} b^{4} x^{3} + 8 \, a^{3} b^{3} x^{2} - 10 \, a^{4} b^{2} x + 15 \, a^{5} b\right )} \sqrt {b x^{2} + a x}}{1536 \, b^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3072*(15*a^6*sqrt(b)*log(2*b*x + a - 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(256*b^6*x^5 + 640*a*b^5*x^4 + 432*a^
2*b^4*x^3 + 8*a^3*b^3*x^2 - 10*a^4*b^2*x + 15*a^5*b)*sqrt(b*x^2 + a*x))/b^4, 1/1536*(15*a^6*sqrt(-b)*arctan(sq
rt(b*x^2 + a*x)*sqrt(-b)/(b*x)) + (256*b^6*x^5 + 640*a*b^5*x^4 + 432*a^2*b^4*x^3 + 8*a^3*b^3*x^2 - 10*a^4*b^2*
x + 15*a^5*b)*sqrt(b*x^2 + a*x))/b^4]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a x + b x^{2}\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*x + b*x**2)**(5/2), x)

________________________________________________________________________________________

Giac [A]
time = 0.76, size = 107, normalized size = 0.91 \begin {gather*} \frac {5 \, a^{6} \log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right )}{1024 \, b^{\frac {7}{2}}} + \frac {1}{1536} \, \sqrt {b x^{2} + a x} {\left (\frac {15 \, a^{5}}{b^{3}} - 2 \, {\left (\frac {5 \, a^{4}}{b^{2}} - 4 \, {\left (\frac {a^{3}}{b} + 2 \, {\left (27 \, a^{2} + 8 \, {\left (2 \, b^{2} x + 5 \, a b\right )} x\right )} x\right )} x\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/1024*a^6*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a))/b^(7/2) + 1/1536*sqrt(b*x^2 + a*x)*(15*a^5
/b^3 - 2*(5*a^4/b^2 - 4*(a^3/b + 2*(27*a^2 + 8*(2*b^2*x + 5*a*b)*x)*x)*x)*x)

________________________________________________________________________________________

Mupad [B]
time = 0.56, size = 119, normalized size = 1.01 \begin {gather*} \frac {{\left (b\,x^2+a\,x\right )}^{5/2}\,\left (\frac {a}{2}+b\,x\right )}{6\,b}-\frac {5\,a^2\,\left (\frac {{\left (b\,x^2+a\,x\right )}^{3/2}\,\left (\frac {a}{2}+b\,x\right )}{4\,b}-\frac {3\,a^2\,\left (\sqrt {b\,x^2+a\,x}\,\left (\frac {x}{2}+\frac {a}{4\,b}\right )-\frac {a^2\,\ln \left (\frac {\frac {a}{2}+b\,x}{\sqrt {b}}+\sqrt {b\,x^2+a\,x}\right )}{8\,b^{3/2}}\right )}{16\,b}\right )}{24\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^2)^(5/2),x)

[Out]

((a*x + b*x^2)^(5/2)*(a/2 + b*x))/(6*b) - (5*a^2*(((a*x + b*x^2)^(3/2)*(a/2 + b*x))/(4*b) - (3*a^2*((a*x + b*x
^2)^(1/2)*(x/2 + a/(4*b)) - (a^2*log((a/2 + b*x)/b^(1/2) + (a*x + b*x^2)^(1/2)))/(8*b^(3/2))))/(16*b)))/(24*b)

________________________________________________________________________________________