\(\int \frac {x^{7/2}}{\sqrt {a x^2+b x^3}} \, dx\) [269]

Optimal result

Integrand size = 21, antiderivative size = 125 \[ \int \frac {x^{7/2}}{\sqrt {a x^2+b x^3}} \, dx=\frac {5 a^2 \sqrt {a x^2+b x^3}}{8 b^3 \sqrt {x}}-\frac {5 a \sqrt {x} \sqrt {a x^2+b x^3}}{12 b^2}+\frac {x^{3/2} \sqrt {a x^2+b x^3}}{3 b}-\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{8 b^{7/2}} \]

[Out]

-5/8*a^3*arctanh(x^(3/2)*b^(1/2)/(b*x^3+a*x^2)^(1/2))/b^(7/2)+1/3*x^(3/2)*(b*x^3+a*x^2)^(1/2)/b+5/8*a^2*(b*x^3
+a*x^2)^(1/2)/b^3/x^(1/2)-5/12*a*x^(1/2)*(b*x^3+a*x^2)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2049, 2054, 212} \[ \int \frac {x^{7/2}}{\sqrt {a x^2+b x^3}} \, dx=-\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{8 b^{7/2}}+\frac {5 a^2 \sqrt {a x^2+b x^3}}{8 b^3 \sqrt {x}}-\frac {5 a \sqrt {x} \sqrt {a x^2+b x^3}}{12 b^2}+\frac {x^{3/2} \sqrt {a x^2+b x^3}}{3 b} \]

[In]

Int[x^(7/2)/Sqrt[a*x^2 + b*x^3],x]

[Out]

(5*a^2*Sqrt[a*x^2 + b*x^3])/(8*b^3*Sqrt[x]) - (5*a*Sqrt[x]*Sqrt[a*x^2 + b*x^3])/(12*b^2) + (x^(3/2)*Sqrt[a*x^2
+ b*x^3])/(3*b) - (5*a^3*ArcTanh[(Sqrt[b]*x^(3/2))/Sqrt[a*x^2 + b*x^3]])/(8*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 2049

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n +
1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1))
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2054

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rubi steps \begin{align*} \text {integral}& = \frac {x^{3/2} \sqrt {a x^2+b x^3}}{3 b}-\frac {(5 a) \int \frac {x^{5/2}}{\sqrt {a x^2+b x^3}} \, dx}{6 b} \\ & = -\frac {5 a \sqrt {x} \sqrt {a x^2+b x^3}}{12 b^2}+\frac {x^{3/2} \sqrt {a x^2+b x^3}}{3 b}+\frac {\left (5 a^2\right ) \int \frac {x^{3/2}}{\sqrt {a x^2+b x^3}} \, dx}{8 b^2} \\ & = \frac {5 a^2 \sqrt {a x^2+b x^3}}{8 b^3 \sqrt {x}}-\frac {5 a \sqrt {x} \sqrt {a x^2+b x^3}}{12 b^2}+\frac {x^{3/2} \sqrt {a x^2+b x^3}}{3 b}-\frac {\left (5 a^3\right ) \int \frac {\sqrt {x}}{\sqrt {a x^2+b x^3}} \, dx}{16 b^3} \\ & = \frac {5 a^2 \sqrt {a x^2+b x^3}}{8 b^3 \sqrt {x}}-\frac {5 a \sqrt {x} \sqrt {a x^2+b x^3}}{12 b^2}+\frac {x^{3/2} \sqrt {a x^2+b x^3}}{3 b}-\frac {\left (5 a^3\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{8 b^3} \\ & = \frac {5 a^2 \sqrt {a x^2+b x^3}}{8 b^3 \sqrt {x}}-\frac {5 a \sqrt {x} \sqrt {a x^2+b x^3}}{12 b^2}+\frac {x^{3/2} \sqrt {a x^2+b x^3}}{3 b}-\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{8 b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.88 \[ \int \frac {x^{7/2}}{\sqrt {a x^2+b x^3}} \, dx=\frac {\sqrt {b} x^{3/2} \left (15 a^3+5 a^2 b x-2 a b^2 x^2+8 b^3 x^3\right )+30 a^3 x \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{24 b^{7/2} \sqrt {x^2 (a+b x)}} \]

[In]

Integrate[x^(7/2)/Sqrt[a*x^2 + b*x^3],x]

[Out]

(Sqrt[b]*x^(3/2)*(15*a^3 + 5*a^2*b*x - 2*a*b^2*x^2 + 8*b^3*x^3) + 30*a^3*x*Sqrt[a + b*x]*ArcTanh[(Sqrt[b]*Sqrt
[x])/(Sqrt[a] - Sqrt[a + b*x])])/(24*b^(7/2)*Sqrt[x^2*(a + b*x)])

Maple [A] (verified)

Time = 1.94 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.80

method	result	size
risch	\(\frac {\left (8 b^{2} x^{2}-10 a b x +15 a^{2}\right ) x^{\frac {3}{2}} \left (b x +a \right )}{24 b^{3} \sqrt {x^{2} \left (b x +a \right )}}-\frac {5 a^{3} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x}\, \sqrt {x \left (b x +a \right )}}{16 b^{\frac {7}{2}} \sqrt {x^{2} \left (b x +a \right )}}\)	\(100\)
default	\(\frac {\sqrt {x}\, \left (16 b^{\frac {9}{2}} x^{4}-4 b^{\frac {7}{2}} a \,x^{3}+10 b^{\frac {5}{2}} a^{2} x^{2}+30 a^{3} b^{\frac {3}{2}} x -15 \sqrt {x \left (b x +a \right )}\, \ln \left (\frac {2 \sqrt {b \,x^{2}+a x}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3} b \right )}{48 \sqrt {b \,x^{3}+a \,x^{2}}\, b^{\frac {9}{2}}}\)	\(103\)

[In]

int(x^(7/2)/(b*x^3+a*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/24*(8*b^2*x^2-10*a*b*x+15*a^2)*x^(3/2)*(b*x+a)/b^3/(x^2*(b*x+a))^(1/2)-5/16*a^3/b^(7/2)*ln((1/2*a+b*x)/b^(1/
2)+(b*x^2+a*x)^(1/2))/(x^2*(b*x+a))^(1/2)*x^(1/2)*(x*(b*x+a))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.44 \[ \int \frac {x^{7/2}}{\sqrt {a x^2+b x^3}} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} x \log \left (\frac {2 \, b x^{2} + a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {b} \sqrt {x}}{x}\right ) + 2 \, {\left (8 \, b^{3} x^{2} - 10 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {x}}{48 \, b^{4} x}, \frac {15 \, a^{3} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-b}}{b x^{\frac {3}{2}}}\right ) + {\left (8 \, b^{3} x^{2} - 10 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {x}}{24 \, b^{4} x}\right ] \]

[In]

integrate(x^(7/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(15*a^3*sqrt(b)*x*log((2*b*x^2 + a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(b)*sqrt(x))/x) + 2*(8*b^3*x^2 - 10*a*b
^2*x + 15*a^2*b)*sqrt(b*x^3 + a*x^2)*sqrt(x))/(b^4*x), 1/24*(15*a^3*sqrt(-b)*x*arctan(sqrt(b*x^3 + a*x^2)*sqrt
(-b)/(b*x^(3/2))) + (8*b^3*x^2 - 10*a*b^2*x + 15*a^2*b)*sqrt(b*x^3 + a*x^2)*sqrt(x))/(b^4*x)]

Sympy [F]

\[ \int \frac {x^{7/2}}{\sqrt {a x^2+b x^3}} \, dx=\int \frac {x^{\frac {7}{2}}}{\sqrt {x^{2} \left (a + b x\right )}}\, dx \]

[In]

integrate(x**(7/2)/(b*x**3+a*x**2)**(1/2),x)

[Out]

Integral(x**(7/2)/sqrt(x**2*(a + b*x)), x)

Maxima [F]

\[ \int \frac {x^{7/2}}{\sqrt {a x^2+b x^3}} \, dx=\int { \frac {x^{\frac {7}{2}}}{\sqrt {b x^{3} + a x^{2}}} \,d x } \]

[In]

integrate(x^(7/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^(7/2)/sqrt(b*x^3 + a*x^2), x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.66 \[ \int \frac {x^{7/2}}{\sqrt {a x^2+b x^3}} \, dx=-\frac {5 \, a^{3} \log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, b^{\frac {7}{2}}} + \frac {\sqrt {b x + a} {\left (2 \, x {\left (\frac {4 \, x}{b} - \frac {5 \, a}{b^{2}}\right )} + \frac {15 \, a^{2}}{b^{3}}\right )} \sqrt {x} + \frac {15 \, a^{3} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {7}{2}}}}{24 \, \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^(7/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="giac")

[Out]

-5/16*a^3*log(abs(a))*sgn(x)/b^(7/2) + 1/24*(sqrt(b*x + a)*(2*x*(4*x/b - 5*a/b^2) + 15*a^2/b^3)*sqrt(x) + 15*a
^3*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/b^(7/2))/sgn(x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{7/2}}{\sqrt {a x^2+b x^3}} \, dx=\int \frac {x^{7/2}}{\sqrt {b\,x^3+a\,x^2}} \,d x \]

[In]

int(x^(7/2)/(a*x^2 + b*x^3)^(1/2),x)

[Out]

int(x^(7/2)/(a*x^2 + b*x^3)^(1/2), x)