3.1.75 \(\int \frac {x^{29/2}}{(a x+b x^3)^{9/2}} \, dx\) [75]

Optimal. Leaf size=159 \[ -\frac {x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {3 x^{13/2}}{5 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {3 x^{7/2}}{b^4 \sqrt {a x+b x^3}}+\frac {9 \sqrt {x} \sqrt {a x+b x^3}}{2 b^5}-\frac {9 a \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x+b x^3}}\right )}{2 b^{11/2}} \]

[Out]

-1/7*x^(25/2)/b/(b*x^3+a*x)^(7/2)-9/35*x^(19/2)/b^2/(b*x^3+a*x)^(5/2)-3/5*x^(13/2)/b^3/(b*x^3+a*x)^(3/2)-9/2*a
*arctanh(x^(3/2)*b^(1/2)/(b*x^3+a*x)^(1/2))/b^(11/2)-3*x^(7/2)/b^4/(b*x^3+a*x)^(1/2)+9/2*x^(1/2)*(b*x^3+a*x)^(
1/2)/b^5

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Rubi [A]
time = 0.16, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2047, 2049, 2054, 212} \begin {gather*} -\frac {9 a \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x+b x^3}}\right )}{2 b^{11/2}}+\frac {9 \sqrt {x} \sqrt {a x+b x^3}}{2 b^5}-\frac {3 x^{7/2}}{b^4 \sqrt {a x+b x^3}}-\frac {3 x^{13/2}}{5 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(29/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-1/7*x^(25/2)/(b*(a*x + b*x^3)^(7/2)) - (9*x^(19/2))/(35*b^2*(a*x + b*x^3)^(5/2)) - (3*x^(13/2))/(5*b^3*(a*x +
b*x^3)^(3/2)) - (3*x^(7/2))/(b^4*Sqrt[a*x + b*x^3]) + (9*Sqrt[x]*Sqrt[a*x + b*x^3])/(2*b^5) - (9*a*ArcTanh[(S
qrt[b]*x^(3/2))/Sqrt[a*x + b*x^3]])/(2*b^(11/2))

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 2047

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*(n - j)*(p + 1))), x] - Dist[c^n*((m + j*p - n + j + 1)/(b*(n - j)*(p + 1))), I
nt[(c*x)^(m - n)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (I
ntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1] && GtQ[m + j*p + 1, n - j]

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*} \int \frac {x^{29/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=-\frac {x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac {9 \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 b}\\ &=-\frac {x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}+\frac {9 \int \frac {x^{17/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{5 b^2}\\ &=-\frac {x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {3 x^{13/2}}{5 b^3 \left (a x+b x^3\right )^{3/2}}+\frac {3 \int \frac {x^{11/2}}{\left (a x+b x^3\right )^{3/2}} \, dx}{b^3}\\ &=-\frac {x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {3 x^{13/2}}{5 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {3 x^{7/2}}{b^4 \sqrt {a x+b x^3}}+\frac {9 \int \frac {x^{5/2}}{\sqrt {a x+b x^3}} \, dx}{b^4}\\ &=-\frac {x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {3 x^{13/2}}{5 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {3 x^{7/2}}{b^4 \sqrt {a x+b x^3}}+\frac {9 \sqrt {x} \sqrt {a x+b x^3}}{2 b^5}-\frac {(9 a) \int \frac {\sqrt {x}}{\sqrt {a x+b x^3}} \, dx}{2 b^5}\\ &=-\frac {x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {3 x^{13/2}}{5 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {3 x^{7/2}}{b^4 \sqrt {a x+b x^3}}+\frac {9 \sqrt {x} \sqrt {a x+b x^3}}{2 b^5}-\frac {(9 a) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {a x+b x^3}}\right )}{2 b^5}\\ &=-\frac {x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {3 x^{13/2}}{5 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {3 x^{7/2}}{b^4 \sqrt {a x+b x^3}}+\frac {9 \sqrt {x} \sqrt {a x+b x^3}}{2 b^5}-\frac {9 a \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x+b x^3}}\right )}{2 b^{11/2}}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 122, normalized size = 0.77 \begin {gather*} \frac {\sqrt {x} \left (\sqrt {b} x \left (315 a^4+1050 a^3 b x^2+1218 a^2 b^2 x^4+528 a b^3 x^6+35 b^4 x^8\right )+315 a \left (a+b x^2\right )^{7/2} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )\right )}{70 b^{11/2} \left (a+b x^2\right )^3 \sqrt {x \left (a+b x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(29/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(Sqrt[x]*(Sqrt[b]*x*(315*a^4 + 1050*a^3*b*x^2 + 1218*a^2*b^2*x^4 + 528*a*b^3*x^6 + 35*b^4*x^8) + 315*a*(a + b*
x^2)^(7/2)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]]))/(70*b^(11/2)*(a + b*x^2)^3*Sqrt[x*(a + b*x^2)])

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Maple [A]
time = 0.42, size = 212, normalized size = 1.33

method	result	size
default	\(-\frac {\sqrt {x \left (b \,x^{2}+a \right )}\, \left (-35 x^{9} b^{\frac {9}{2}}+315 \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) a \,b^{3} x^{6} \sqrt {b \,x^{2}+a}-528 b^{\frac {7}{2}} a \,x^{7}+945 \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) a^{2} b^{2} x^{4} \sqrt {b \,x^{2}+a}-1218 b^{\frac {5}{2}} a^{2} x^{5}+945 \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) a^{3} b \,x^{2} \sqrt {b \,x^{2}+a}-1050 b^{\frac {3}{2}} a^{3} x^{3}+315 \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) a^{4} \sqrt {b \,x^{2}+a}-315 \sqrt {b}\, a^{4} x \right )}{70 b^{\frac {11}{2}} \sqrt {x}\, \left (b \,x^{2}+a \right )^{4}}\)	\(212\)
risch	\(\frac {x^{\frac {3}{2}} \left (b \,x^{2}+a \right )}{2 b^{5} \sqrt {x \left (b \,x^{2}+a \right )}}+\frac {\left (-\frac {9 a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {11}{2}}}+\frac {a^{3} \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{112 b^{7} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )^{4}}-\frac {53 a^{2} \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{560 b^{7} \left (x -\frac {\sqrt {-a b}}{b}\right )^{3}}-\frac {571 a^{2} \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{1120 b^{6} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}+\frac {97 a \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{35 b^{6} \left (x -\frac {\sqrt {-a b}}{b}\right )}-\frac {a^{3} \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{112 b^{7} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )^{4}}-\frac {53 a^{2} \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{560 b^{7} \left (x +\frac {\sqrt {-a b}}{b}\right )^{3}}+\frac {571 a^{2} \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{1120 b^{6} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}+\frac {97 a \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{35 b^{6} \left (x +\frac {\sqrt {-a b}}{b}\right )}\right ) \sqrt {b \,x^{2}+a}\, \sqrt {x}}{\sqrt {x \left (b \,x^{2}+a \right )}}\)	\(595\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(29/2)/(b*x^3+a*x)^(9/2),x,method=_RETURNVERBOSE)

[Out]

-1/70*(x*(b*x^2+a))^(1/2)/b^(11/2)*(-35*x^9*b^(9/2)+315*ln(x*b^(1/2)+(b*x^2+a)^(1/2))*a*b^3*x^6*(b*x^2+a)^(1/2
)-528*b^(7/2)*a*x^7+945*ln(x*b^(1/2)+(b*x^2+a)^(1/2))*a^2*b^2*x^4*(b*x^2+a)^(1/2)-1218*b^(5/2)*a^2*x^5+945*ln(
x*b^(1/2)+(b*x^2+a)^(1/2))*a^3*b*x^2*(b*x^2+a)^(1/2)-1050*b^(3/2)*a^3*x^3+315*ln(x*b^(1/2)+(b*x^2+a)^(1/2))*a^
4*(b*x^2+a)^(1/2)-315*b^(1/2)*a^4*x)/x^(1/2)/(b*x^2+a)^4

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(29/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(29/2)/(b*x^3 + a*x)^(9/2), x)

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Fricas [A]
time = 2.84, size = 376, normalized size = 2.36 \begin {gather*} \left [\frac {315 \, {\left (a b^{4} x^{8} + 4 \, a^{2} b^{3} x^{6} + 6 \, a^{3} b^{2} x^{4} + 4 \, a^{4} b x^{2} + a^{5}\right )} \sqrt {b} \log \left (2 \, b x^{2} - 2 \, \sqrt {b x^{3} + a x} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (35 \, b^{5} x^{8} + 528 \, a b^{4} x^{6} + 1218 \, a^{2} b^{3} x^{4} + 1050 \, a^{3} b^{2} x^{2} + 315 \, a^{4} b\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{140 \, {\left (b^{10} x^{8} + 4 \, a b^{9} x^{6} + 6 \, a^{2} b^{8} x^{4} + 4 \, a^{3} b^{7} x^{2} + a^{4} b^{6}\right )}}, \frac {315 \, {\left (a b^{4} x^{8} + 4 \, a^{2} b^{3} x^{6} + 6 \, a^{3} b^{2} x^{4} + 4 \, a^{4} b x^{2} + a^{5}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{3} + a x} \sqrt {-b}}{b x^{\frac {3}{2}}}\right ) + {\left (35 \, b^{5} x^{8} + 528 \, a b^{4} x^{6} + 1218 \, a^{2} b^{3} x^{4} + 1050 \, a^{3} b^{2} x^{2} + 315 \, a^{4} b\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{70 \, {\left (b^{10} x^{8} + 4 \, a b^{9} x^{6} + 6 \, a^{2} b^{8} x^{4} + 4 \, a^{3} b^{7} x^{2} + a^{4} b^{6}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(29/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

[1/140*(315*(a*b^4*x^8 + 4*a^2*b^3*x^6 + 6*a^3*b^2*x^4 + 4*a^4*b*x^2 + a^5)*sqrt(b)*log(2*b*x^2 - 2*sqrt(b*x^3
+ a*x)*sqrt(b)*sqrt(x) + a) + 2*(35*b^5*x^8 + 528*a*b^4*x^6 + 1218*a^2*b^3*x^4 + 1050*a^3*b^2*x^2 + 315*a^4*b
)*sqrt(b*x^3 + a*x)*sqrt(x))/(b^10*x^8 + 4*a*b^9*x^6 + 6*a^2*b^8*x^4 + 4*a^3*b^7*x^2 + a^4*b^6), 1/70*(315*(a*
b^4*x^8 + 4*a^2*b^3*x^6 + 6*a^3*b^2*x^4 + 4*a^4*b*x^2 + a^5)*sqrt(-b)*arctan(sqrt(b*x^3 + a*x)*sqrt(-b)/(b*x^(
3/2))) + (35*b^5*x^8 + 528*a*b^4*x^6 + 1218*a^2*b^3*x^4 + 1050*a^3*b^2*x^2 + 315*a^4*b)*sqrt(b*x^3 + a*x)*sqrt
(x))/(b^10*x^8 + 4*a*b^9*x^6 + 6*a^2*b^8*x^4 + 4*a^3*b^7*x^2 + a^4*b^6)]

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Sympy [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**(29/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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Giac [A]
time = 1.51, size = 100, normalized size = 0.63 \begin {gather*} \frac {{\left ({\left ({\left (x^{2} {\left (\frac {35 \, x^{2}}{b} + \frac {528 \, a}{b^{2}}\right )} + \frac {1218 \, a^{2}}{b^{3}}\right )} x^{2} + \frac {1050 \, a^{3}}{b^{4}}\right )} x^{2} + \frac {315 \, a^{4}}{b^{5}}\right )} x}{70 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} + \frac {9 \, a \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {11}{2}}} - \frac {9 \, a \log \left ({\left | a \right |}\right )}{4 \, b^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(29/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

1/70*(((x^2*(35*x^2/b + 528*a/b^2) + 1218*a^2/b^3)*x^2 + 1050*a^3/b^4)*x^2 + 315*a^4/b^5)*x/(b*x^2 + a)^(7/2)
+ 9/2*a*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(11/2) - 9/4*a*log(abs(a))/b^(11/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{29/2}}{{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(29/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(29/2)/(a*x + b*x^3)^(9/2), x)

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