### 3.60 $$\int \frac{x^6}{(a x+b x^2)^{5/2}} \, dx$$

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

[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))

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Rubi [A]  time = 0.0582204, antiderivative size = 122, normalized size of antiderivative = 1., 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} $\frac{35 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{4 b^{9/2}}-\frac{14 x^3}{3 b^2 \sqrt{a x+b x^2}}+\frac{35 x \sqrt{a x+b x^2}}{6 b^3}-\frac{35 a \sqrt{a x+b x^2}}{4 b^4}-\frac{2 x^5}{3 b \left (a x+b x^2\right )^{3/2}}$

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 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]

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 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 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])

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*}

Mathematica [C]  time = 0.0145022, size = 50, normalized size = 0.41 $\frac{2 x^5 \sqrt{\frac{b x}{a}+1} \, _2F_1\left (\frac{5}{2},\frac{9}{2};\frac{11}{2};-\frac{b x}{a}\right )}{9 a^2 \sqrt{x (a+b x)}}$

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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Maple [A]  time = 0.049, size = 176, normalized size = 1.4 \begin{align*}{\frac{{x}^{5}}{2\,b} \left ( b{x}^{2}+ax \right ) ^{-{\frac{3}{2}}}}-{\frac{7\,a{x}^{4}}{4\,{b}^{2}} \left ( b{x}^{2}+ax \right ) ^{-{\frac{3}{2}}}}-{\frac{35\,{a}^{2}{x}^{3}}{24\,{b}^{3}} \left ( b{x}^{2}+ax \right ) ^{-{\frac{3}{2}}}}+{\frac{35\,{x}^{2}{a}^{3}}{16\,{b}^{4}} \left ( b{x}^{2}+ax \right ) ^{-{\frac{3}{2}}}}+{\frac{35\,{a}^{4}x}{48\,{b}^{5}} \left ( b{x}^{2}+ax \right ) ^{-{\frac{3}{2}}}}-{\frac{245\,{a}^{2}x}{24\,{b}^{4}}{\frac{1}{\sqrt{b{x}^{2}+ax}}}}-{\frac{35\,{a}^{3}}{48\,{b}^{5}}{\frac{1}{\sqrt{b{x}^{2}+ax}}}}+{\frac{35\,{a}^{2}}{8}\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ){b}^{-{\frac{9}{2}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.98898, size = 556, normalized size = 4.56 \begin{align*} \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{align*}

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

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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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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]

Exception raised: TypeError