3.130 \(\int \frac{x^7 (A+B x^2)}{\sqrt{b x^2+c x^4}} \, dx\)

Optimal. Leaf size=176 \[ -\frac{5 b^2 \sqrt{b x^2+c x^4} (7 b B-8 A c)}{128 c^4}+\frac{5 b^3 (7 b B-8 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{128 c^{9/2}}-\frac{x^4 \sqrt{b x^2+c x^4} (7 b B-8 A c)}{48 c^2}+\frac{5 b x^2 \sqrt{b x^2+c x^4} (7 b B-8 A c)}{192 c^3}+\frac{B x^6 \sqrt{b x^2+c x^4}}{8 c} \]

[Out]

(-5*b^2*(7*b*B - 8*A*c)*Sqrt[b*x^2 + c*x^4])/(128*c^4) + (5*b*(7*b*B - 8*A*c)*x^2*Sqrt[b*x^2 + c*x^4])/(192*c^
3) - ((7*b*B - 8*A*c)*x^4*Sqrt[b*x^2 + c*x^4])/(48*c^2) + (B*x^6*Sqrt[b*x^2 + c*x^4])/(8*c) + (5*b^3*(7*b*B -
8*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(128*c^(9/2))

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Rubi [A]  time = 0.328448, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2034, 794, 670, 640, 620, 206} \[ -\frac{5 b^2 \sqrt{b x^2+c x^4} (7 b B-8 A c)}{128 c^4}+\frac{5 b^3 (7 b B-8 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{128 c^{9/2}}-\frac{x^4 \sqrt{b x^2+c x^4} (7 b B-8 A c)}{48 c^2}+\frac{5 b x^2 \sqrt{b x^2+c x^4} (7 b B-8 A c)}{192 c^3}+\frac{B x^6 \sqrt{b x^2+c x^4}}{8 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^7*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]

[Out]

(-5*b^2*(7*b*B - 8*A*c)*Sqrt[b*x^2 + c*x^4])/(128*c^4) + (5*b*(7*b*B - 8*A*c)*x^2*Sqrt[b*x^2 + c*x^4])/(192*c^
3) - ((7*b*B - 8*A*c)*x^4*Sqrt[b*x^2 + c*x^4])/(48*c^2) + (B*x^6*Sqrt[b*x^2 + c*x^4])/(8*c) + (5*b^3*(7*b*B -
8*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(128*c^(9/2))

Rule 2034

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rule 794

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

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^7 \left (A+B x^2\right )}{\sqrt{b x^2+c x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^3 (A+B x)}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )\\ &=\frac{B x^6 \sqrt{b x^2+c x^4}}{8 c}+\frac{\left (3 (-b B+A c)+\frac{1}{2} (-b B+2 A c)\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{8 c}\\ &=-\frac{(7 b B-8 A c) x^4 \sqrt{b x^2+c x^4}}{48 c^2}+\frac{B x^6 \sqrt{b x^2+c x^4}}{8 c}+\frac{(5 b (7 b B-8 A c)) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{96 c^2}\\ &=\frac{5 b (7 b B-8 A c) x^2 \sqrt{b x^2+c x^4}}{192 c^3}-\frac{(7 b B-8 A c) x^4 \sqrt{b x^2+c x^4}}{48 c^2}+\frac{B x^6 \sqrt{b x^2+c x^4}}{8 c}-\frac{\left (5 b^2 (7 b B-8 A c)\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{128 c^3}\\ &=-\frac{5 b^2 (7 b B-8 A c) \sqrt{b x^2+c x^4}}{128 c^4}+\frac{5 b (7 b B-8 A c) x^2 \sqrt{b x^2+c x^4}}{192 c^3}-\frac{(7 b B-8 A c) x^4 \sqrt{b x^2+c x^4}}{48 c^2}+\frac{B x^6 \sqrt{b x^2+c x^4}}{8 c}+\frac{\left (5 b^3 (7 b B-8 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{256 c^4}\\ &=-\frac{5 b^2 (7 b B-8 A c) \sqrt{b x^2+c x^4}}{128 c^4}+\frac{5 b (7 b B-8 A c) x^2 \sqrt{b x^2+c x^4}}{192 c^3}-\frac{(7 b B-8 A c) x^4 \sqrt{b x^2+c x^4}}{48 c^2}+\frac{B x^6 \sqrt{b x^2+c x^4}}{8 c}+\frac{\left (5 b^3 (7 b B-8 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x^2}{\sqrt{b x^2+c x^4}}\right )}{128 c^4}\\ &=-\frac{5 b^2 (7 b B-8 A c) \sqrt{b x^2+c x^4}}{128 c^4}+\frac{5 b (7 b B-8 A c) x^2 \sqrt{b x^2+c x^4}}{192 c^3}-\frac{(7 b B-8 A c) x^4 \sqrt{b x^2+c x^4}}{48 c^2}+\frac{B x^6 \sqrt{b x^2+c x^4}}{8 c}+\frac{5 b^3 (7 b B-8 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{128 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.167758, size = 145, normalized size = 0.82 \[ \frac{x \left (15 b^3 \sqrt{b+c x^2} (7 b B-8 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b+c x^2}}\right )-\sqrt{c} x \left (b+c x^2\right ) \left (-10 b^2 c \left (12 A+7 B x^2\right )+8 b c^2 x^2 \left (10 A+7 B x^2\right )-16 c^3 x^4 \left (4 A+3 B x^2\right )+105 b^3 B\right )\right )}{384 c^{9/2} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]

[Out]

(x*(-(Sqrt[c]*x*(b + c*x^2)*(105*b^3*B - 16*c^3*x^4*(4*A + 3*B*x^2) + 8*b*c^2*x^2*(10*A + 7*B*x^2) - 10*b^2*c*
(12*A + 7*B*x^2))) + 15*b^3*(7*b*B - 8*A*c)*Sqrt[b + c*x^2]*ArcTanh[(Sqrt[c]*x)/Sqrt[b + c*x^2]]))/(384*c^(9/2
)*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.013, size = 211, normalized size = 1.2 \begin{align*}{\frac{x}{384}\sqrt{c{x}^{2}+b} \left ( 48\,B{c}^{9/2}\sqrt{c{x}^{2}+b}{x}^{7}+64\,A{c}^{9/2}\sqrt{c{x}^{2}+b}{x}^{5}-56\,B{c}^{7/2}\sqrt{c{x}^{2}+b}{x}^{5}b-80\,A{c}^{7/2}\sqrt{c{x}^{2}+b}{x}^{3}b+70\,B{c}^{5/2}\sqrt{c{x}^{2}+b}{x}^{3}{b}^{2}+120\,A{c}^{5/2}\sqrt{c{x}^{2}+b}x{b}^{2}-105\,B{c}^{3/2}\sqrt{c{x}^{2}+b}x{b}^{3}-120\,A\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{3}{c}^{2}+105\,B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{4}c \right ){\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}}}}{c}^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x)

[Out]

1/384*x*(c*x^2+b)^(1/2)*(48*B*c^(9/2)*(c*x^2+b)^(1/2)*x^7+64*A*c^(9/2)*(c*x^2+b)^(1/2)*x^5-56*B*c^(7/2)*(c*x^2
+b)^(1/2)*x^5*b-80*A*c^(7/2)*(c*x^2+b)^(1/2)*x^3*b+70*B*c^(5/2)*(c*x^2+b)^(1/2)*x^3*b^2+120*A*c^(5/2)*(c*x^2+b
)^(1/2)*x*b^2-105*B*c^(3/2)*(c*x^2+b)^(1/2)*x*b^3-120*A*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*b^3*c^2+105*B*ln(x*c^(1/
2)+(c*x^2+b)^(1/2))*b^4*c)/(c*x^4+b*x^2)^(1/2)/c^(11/2)

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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^7*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.21276, size = 626, normalized size = 3.56 \begin{align*} \left [-\frac{15 \,{\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt{c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{c}\right ) - 2 \,{\left (48 \, B c^{4} x^{6} - 105 \, B b^{3} c + 120 \, A b^{2} c^{2} - 8 \,{\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{4} + 10 \,{\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{768 \, c^{5}}, -\frac{15 \,{\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-c}}{c x^{2} + b}\right ) -{\left (48 \, B c^{4} x^{6} - 105 \, B b^{3} c + 120 \, A b^{2} c^{2} - 8 \,{\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{4} + 10 \,{\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{384 \, c^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/768*(15*(7*B*b^4 - 8*A*b^3*c)*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(48*B*c^4*x^6
- 105*B*b^3*c + 120*A*b^2*c^2 - 8*(7*B*b*c^3 - 8*A*c^4)*x^4 + 10*(7*B*b^2*c^2 - 8*A*b*c^3)*x^2)*sqrt(c*x^4 + b
*x^2))/c^5, -1/384*(15*(7*B*b^4 - 8*A*b^3*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - (48*B
*c^4*x^6 - 105*B*b^3*c + 120*A*b^2*c^2 - 8*(7*B*b*c^3 - 8*A*c^4)*x^4 + 10*(7*B*b^2*c^2 - 8*A*b*c^3)*x^2)*sqrt(
c*x^4 + b*x^2))/c^5]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{7} \left (A + B x^{2}\right )}{\sqrt{x^{2} \left (b + c x^{2}\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(B*x**2+A)/(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral(x**7*(A + B*x**2)/sqrt(x**2*(b + c*x**2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} x^{7}}{\sqrt{c x^{4} + b x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*x^7/sqrt(c*x^4 + b*x^2), x)