3.1.60 \(\int \frac {x^6}{(a x+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=122 \[ \frac {35 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{4 b^{9/2}}-\frac {35 a \sqrt {a x+b x^2}}{4 b^4}+\frac {35 x \sqrt {a x+b x^2}}{6 b^3}-\frac {14 x^3}{3 b^2 \sqrt {a x+b x^2}}-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {668, 670, 640, 620, 206} \begin {gather*} \frac {35 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{4 b^{9/2}}-\frac {14 x^3}{3 b^2 \sqrt {a x+b x^2}}+\frac {35 x \sqrt {a x+b x^2}}{6 b^3}-\frac {35 a \sqrt {a x+b x^2}}{4 b^4}-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^6/(a*x + b*x^2)^(5/2),x]

[Out]

(-2*x^5)/(3*b*(a*x + b*x^2)^(3/2)) - (14*x^3)/(3*b^2*Sqrt[a*x + b*x^2]) - (35*a*Sqrt[a*x + b*x^2])/(4*b^4) + (
35*x*Sqrt[a*x + b*x^2])/(6*b^3) + (35*a^2*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(4*b^(9/2))

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 620

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]

Rule 640

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

Rule 668

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

Rule 670

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

Rubi steps

\begin {align*} \int \frac {x^6}{\left (a x+b x^2\right )^{5/2}} \, dx &=-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}}+\frac {7 \int \frac {x^4}{\left (a x+b x^2\right )^{3/2}} \, dx}{3 b}\\ &=-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}}-\frac {14 x^3}{3 b^2 \sqrt {a x+b x^2}}+\frac {35 \int \frac {x^2}{\sqrt {a x+b x^2}} \, dx}{3 b^2}\\ &=-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}}-\frac {14 x^3}{3 b^2 \sqrt {a x+b x^2}}+\frac {35 x \sqrt {a x+b x^2}}{6 b^3}-\frac {(35 a) \int \frac {x}{\sqrt {a x+b x^2}} \, dx}{4 b^3}\\ &=-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}}-\frac {14 x^3}{3 b^2 \sqrt {a x+b x^2}}-\frac {35 a \sqrt {a x+b x^2}}{4 b^4}+\frac {35 x \sqrt {a x+b x^2}}{6 b^3}+\frac {\left (35 a^2\right ) \int \frac {1}{\sqrt {a x+b x^2}} \, dx}{8 b^4}\\ &=-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}}-\frac {14 x^3}{3 b^2 \sqrt {a x+b x^2}}-\frac {35 a \sqrt {a x+b x^2}}{4 b^4}+\frac {35 x \sqrt {a x+b x^2}}{6 b^3}+\frac {\left (35 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right )}{4 b^4}\\ &=-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}}-\frac {14 x^3}{3 b^2 \sqrt {a x+b x^2}}-\frac {35 a \sqrt {a x+b x^2}}{4 b^4}+\frac {35 x \sqrt {a x+b x^2}}{6 b^3}+\frac {35 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{4 b^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.41 \begin {gather*} \frac {2 x^5 \sqrt {\frac {b x}{a}+1} \, _2F_1\left (\frac {5}{2},\frac {9}{2};\frac {11}{2};-\frac {b x}{a}\right )}{9 a^2 \sqrt {x (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a*x + b*x^2)^(5/2),x]

[Out]

(2*x^5*Sqrt[1 + (b*x)/a]*Hypergeometric2F1[5/2, 9/2, 11/2, -((b*x)/a)])/(9*a^2*Sqrt[x*(a + b*x)])

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IntegrateAlgebraic [A]  time = 0.46, size = 103, normalized size = 0.84 \begin {gather*} \frac {\sqrt {a x+b x^2} \left (-105 a^3-140 a^2 b x-21 a b^2 x^2+6 b^3 x^3\right )}{12 b^4 (a+b x)^2}-\frac {35 a^2 \log \left (-2 b^{9/2} \sqrt {a x+b x^2}+a b^4+2 b^5 x\right )}{8 b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^6/(a*x + b*x^2)^(5/2),x]

[Out]

(Sqrt[a*x + b*x^2]*(-105*a^3 - 140*a^2*b*x - 21*a*b^2*x^2 + 6*b^3*x^3))/(12*b^4*(a + b*x)^2) - (35*a^2*Log[a*b
^4 + 2*b^5*x - 2*b^(9/2)*Sqrt[a*x + b*x^2]])/(8*b^(9/2))

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fricas [A]  time = 0.43, size = 248, normalized size = 2.03 \begin {gather*} \left [\frac {105 \, {\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (6 \, b^{4} x^{3} - 21 \, a b^{3} x^{2} - 140 \, a^{2} b^{2} x - 105 \, a^{3} b\right )} \sqrt {b x^{2} + a x}}{24 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}, -\frac {105 \, {\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) - {\left (6 \, b^{4} x^{3} - 21 \, a b^{3} x^{2} - 140 \, a^{2} b^{2} x - 105 \, a^{3} b\right )} \sqrt {b x^{2} + a x}}{12 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a*x)^(5/2),x, algorithm="fricas")

[Out]

[1/24*(105*(a^2*b^2*x^2 + 2*a^3*b*x + a^4)*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(6*b^4*x^3
- 21*a*b^3*x^2 - 140*a^2*b^2*x - 105*a^3*b)*sqrt(b*x^2 + a*x))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5), -1/12*(105*(a
^2*b^2*x^2 + 2*a^3*b*x + a^4)*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) - (6*b^4*x^3 - 21*a*b^3*x^2 -
140*a^2*b^2*x - 105*a^3*b)*sqrt(b*x^2 + a*x))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)]

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giac [A]  time = 0.27, size = 162, normalized size = 1.33 \begin {gather*} \frac {1}{4} \, \sqrt {b x^{2} + a x} {\left (\frac {2 \, x}{b^{3}} - \frac {11 \, a}{b^{4}}\right )} - \frac {35 \, a^{2} \log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right )}{8 \, b^{\frac {9}{2}}} - \frac {2 \, {\left (12 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a^{3} b^{\frac {3}{2}} + 21 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a^{4} b + 10 \, a^{5} \sqrt {b}\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} b + a \sqrt {b}\right )}^{3} b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a*x)^(5/2),x, algorithm="giac")

[Out]

1/4*sqrt(b*x^2 + a*x)*(2*x/b^3 - 11*a/b^4) - 35/8*a^2*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a))
/b^(9/2) - 2/3*(12*(sqrt(b)*x - sqrt(b*x^2 + a*x))^2*a^3*b^(3/2) + 21*(sqrt(b)*x - sqrt(b*x^2 + a*x))*a^4*b +
10*a^5*sqrt(b))/(((sqrt(b)*x - sqrt(b*x^2 + a*x))*b + a*sqrt(b))^3*b^(7/2))

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maple [A]  time = 0.05, size = 176, normalized size = 1.44 \begin {gather*} \frac {x^{5}}{2 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b}-\frac {7 a \,x^{4}}{4 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{2}}-\frac {35 a^{2} x^{3}}{24 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{3}}+\frac {35 a^{3} x^{2}}{16 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{4}}+\frac {35 a^{4} x}{48 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{5}}-\frac {245 a^{2} x}{24 \sqrt {b \,x^{2}+a x}\, b^{4}}+\frac {35 a^{2} \ln \left (\frac {b x +\frac {a}{2}}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {9}{2}}}-\frac {35 a^{3}}{48 \sqrt {b \,x^{2}+a x}\, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b*x^2+a*x)^(5/2),x)

[Out]

1/2*x^5/b/(b*x^2+a*x)^(3/2)-7/4*a/b^2*x^4/(b*x^2+a*x)^(3/2)-35/24*a^2/b^3*x^3/(b*x^2+a*x)^(3/2)+35/16*a^3/b^4*
x^2/(b*x^2+a*x)^(3/2)+35/48*a^4/b^5/(b*x^2+a*x)^(3/2)*x-245/24*a^2/b^4/(b*x^2+a*x)^(1/2)*x-35/48*a^3/b^5/(b*x^
2+a*x)^(1/2)+35/8*a^2/b^(9/2)*ln((b*x+1/2*a)/b^(1/2)+(b*x^2+a*x)^(1/2))

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maxima [A]  time = 1.49, size = 190, normalized size = 1.56 \begin {gather*} \frac {x^{5}}{2 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} b} - \frac {35 \, a^{2} x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a x\right )}^{\frac {3}{2}} b} + \frac {a x}{{\left (b x^{2} + a x\right )}^{\frac {3}{2}} b^{2}} - \frac {2 \, x}{\sqrt {b x^{2} + a x} a b} - \frac {1}{\sqrt {b x^{2} + a x} b^{2}}\right )}}{24 \, b^{2}} - \frac {7 \, a x^{4}}{4 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} b^{2}} - \frac {35 \, a^{2} x}{6 \, \sqrt {b x^{2} + a x} b^{4}} + \frac {35 \, a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {9}{2}}} - \frac {35 \, \sqrt {b x^{2} + a x} a}{12 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a*x)^(5/2),x, algorithm="maxima")

[Out]

1/2*x^5/((b*x^2 + a*x)^(3/2)*b) - 35/24*a^2*x*(3*x^2/((b*x^2 + a*x)^(3/2)*b) + a*x/((b*x^2 + a*x)^(3/2)*b^2) -
2*x/(sqrt(b*x^2 + a*x)*a*b) - 1/(sqrt(b*x^2 + a*x)*b^2))/b^2 - 7/4*a*x^4/((b*x^2 + a*x)^(3/2)*b^2) - 35/6*a^2
*x/(sqrt(b*x^2 + a*x)*b^4) + 35/8*a^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(9/2) - 35/12*sqrt(b*x^2
+ a*x)*a/b^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^6}{{\left (b\,x^2+a\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(a*x + b*x^2)^(5/2),x)

[Out]

int(x^6/(a*x + b*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{6}}{\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b*x**2+a*x)**(5/2),x)

[Out]

Integral(x**6/(x*(a + b*x))**(5/2), x)

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