3.2.45 \(\int \frac {x^9 (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=184 \[ -\frac {5 b^2 (7 b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{9/2}}+\frac {5 b \sqrt {b x^2+c x^4} (7 b B-6 A c)}{16 c^4}-\frac {5 x^2 \sqrt {b x^2+c x^4} (7 b B-6 A c)}{24 c^3}+\frac {x^4 \sqrt {b x^2+c x^4} (7 b B-6 A c)}{6 b c^2}-\frac {x^8 (b B-A c)}{b c \sqrt {b x^2+c x^4}} \]

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Rubi [A]  time = 0.33, antiderivative size = 184, normalized size of antiderivative = 1.00, 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, 788, 670, 640, 620, 206} \begin {gather*} -\frac {5 b^2 (7 b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{9/2}}+\frac {x^4 \sqrt {b x^2+c x^4} (7 b B-6 A c)}{6 b c^2}-\frac {5 x^2 \sqrt {b x^2+c x^4} (7 b B-6 A c)}{24 c^3}+\frac {5 b \sqrt {b x^2+c x^4} (7 b B-6 A c)}{16 c^4}-\frac {x^8 (b B-A c)}{b c \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^9*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(((b*B - A*c)*x^8)/(b*c*Sqrt[b*x^2 + c*x^4])) + (5*b*(7*b*B - 6*A*c)*Sqrt[b*x^2 + c*x^4])/(16*c^4) - (5*(7*b*
B - 6*A*c)*x^2*Sqrt[b*x^2 + c*x^4])/(24*c^3) + ((7*b*B - 6*A*c)*x^4*Sqrt[b*x^2 + c*x^4])/(6*b*c^2) - (5*b^2*(7
*b*B - 6*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(16*c^(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 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 788

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

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]

Rubi steps

\begin {align*} \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^4 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {(b B-A c) x^8}{b c \sqrt {b x^2+c x^4}}+\frac {1}{2} \left (-\frac {6 A}{b}+\frac {7 B}{c}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac {(b B-A c) x^8}{b c \sqrt {b x^2+c x^4}}+\frac {(7 b B-6 A c) x^4 \sqrt {b x^2+c x^4}}{6 b c^2}-\frac {(5 (7 b B-6 A c)) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{12 c^2}\\ &=-\frac {(b B-A c) x^8}{b c \sqrt {b x^2+c x^4}}-\frac {5 (7 b B-6 A c) x^2 \sqrt {b x^2+c x^4}}{24 c^3}+\frac {(7 b B-6 A c) x^4 \sqrt {b x^2+c x^4}}{6 b c^2}+\frac {(5 b (7 b B-6 A c)) \operatorname {Subst}\left (\int \frac {x}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{16 c^3}\\ &=-\frac {(b B-A c) x^8}{b c \sqrt {b x^2+c x^4}}+\frac {5 b (7 b B-6 A c) \sqrt {b x^2+c x^4}}{16 c^4}-\frac {5 (7 b B-6 A c) x^2 \sqrt {b x^2+c x^4}}{24 c^3}+\frac {(7 b B-6 A c) x^4 \sqrt {b x^2+c x^4}}{6 b c^2}-\frac {\left (5 b^2 (7 b B-6 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{32 c^4}\\ &=-\frac {(b B-A c) x^8}{b c \sqrt {b x^2+c x^4}}+\frac {5 b (7 b B-6 A c) \sqrt {b x^2+c x^4}}{16 c^4}-\frac {5 (7 b B-6 A c) x^2 \sqrt {b x^2+c x^4}}{24 c^3}+\frac {(7 b B-6 A c) x^4 \sqrt {b x^2+c x^4}}{6 b c^2}-\frac {\left (5 b^2 (7 b B-6 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 )}{16 c^4}\\ &=-\frac {(b B-A c) x^8}{b c \sqrt {b x^2+c x^4}}+\frac {5 b (7 b B-6 A c) \sqrt {b x^2+c x^4}}{16 c^4}-\frac {5 (7 b B-6 A c) x^2 \sqrt {b x^2+c x^4}}{24 c^3}+\frac {(7 b B-6 A c) x^4 \sqrt {b x^2+c x^4}}{6 b c^2}-\frac {5 b^2 (7 b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 136, normalized size = 0.74 \begin {gather*} \frac {x \left (\sqrt {c} x \left (b^2 \left (35 B c x^2-90 A c\right )-2 b c^2 x^2 \left (15 A+7 B x^2\right )+4 c^3 x^4 \left (3 A+2 B x^2\right )+105 b^3 B\right )-15 b^{5/2} \sqrt {\frac {c x^2}{b}+1} (7 b B-6 A c) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )\right )}{48 c^{9/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^9*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x*(Sqrt[c]*x*(105*b^3*B + 4*c^3*x^4*(3*A + 2*B*x^2) - 2*b*c^2*x^2*(15*A + 7*B*x^2) + b^2*(-90*A*c + 35*B*c*x^
2)) - 15*b^(5/2)*(7*b*B - 6*A*c)*Sqrt[1 + (c*x^2)/b]*ArcSinh[(Sqrt[c]*x)/Sqrt[b]]))/(48*c^(9/2)*Sqrt[x^2*(b +
c*x^2)])

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IntegrateAlgebraic [A]  time = 0.67, size = 148, normalized size = 0.80 \begin {gather*} \frac {5 \left (7 b^3 B-6 A b^2 c\right ) \log \left (-2 \sqrt {c} \sqrt {b x^2+c x^4}+b+2 c x^2\right )}{32 c^{9/2}}+\frac {\sqrt {b x^2+c x^4} \left (-90 A b^2 c-30 A b c^2 x^2+12 A c^3 x^4+105 b^3 B+35 b^2 B c x^2-14 b B c^2 x^4+8 B c^3 x^6\right )}{48 c^4 \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^9*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(105*b^3*B - 90*A*b^2*c + 35*b^2*B*c*x^2 - 30*A*b*c^2*x^2 - 14*b*B*c^2*x^4 + 12*A*c^3*x^4
+ 8*B*c^3*x^6))/(48*c^4*(b + c*x^2)) + (5*(7*b^3*B - 6*A*b^2*c)*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[b*x^2 + c*x^
4]])/(32*c^(9/2))

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fricas [A]  time = 0.44, size = 340, normalized size = 1.85 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B b^{4} - 6 \, A b^{3} c + {\left (7 \, B b^{3} c - 6 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{4} x^{6} + 105 \, B b^{3} c - 90 \, A b^{2} c^{2} - 2 \, {\left (7 \, B b c^{3} - 6 \, A c^{4}\right )} x^{4} + 5 \, {\left (7 \, B b^{2} c^{2} - 6 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, {\left (c^{6} x^{2} + b c^{5}\right )}}, \frac {15 \, {\left (7 \, B b^{4} - 6 \, A b^{3} c + {\left (7 \, B b^{3} c - 6 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (8 \, B c^{4} x^{6} + 105 \, B b^{3} c - 90 \, A b^{2} c^{2} - 2 \, {\left (7 \, B b c^{3} - 6 \, A c^{4}\right )} x^{4} + 5 \, {\left (7 \, B b^{2} c^{2} - 6 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, {\left (c^{6} x^{2} + b c^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/96*(15*(7*B*b^4 - 6*A*b^3*c + (7*B*b^3*c - 6*A*b^2*c^2)*x^2)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x
^2)*sqrt(c)) - 2*(8*B*c^4*x^6 + 105*B*b^3*c - 90*A*b^2*c^2 - 2*(7*B*b*c^3 - 6*A*c^4)*x^4 + 5*(7*B*b^2*c^2 - 6*
A*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/(c^6*x^2 + b*c^5), 1/48*(15*(7*B*b^4 - 6*A*b^3*c + (7*B*b^3*c - 6*A*b^2*c^2
)*x^2)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + (8*B*c^4*x^6 + 105*B*b^3*c - 90*A*b^2*c^2 -
2*(7*B*b*c^3 - 6*A*c^4)*x^4 + 5*(7*B*b^2*c^2 - 6*A*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/(c^6*x^2 + b*c^5)]

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giac [A]  time = 0.31, size = 176, normalized size = 0.96 \begin {gather*} \frac {1}{48} \, \sqrt {c x^{4} + b x^{2}} {\left (2 \, x^{2} {\left (\frac {4 \, B x^{2}}{c^{2}} - \frac {11 \, B b c^{10} - 6 \, A c^{11}}{c^{13}}\right )} + \frac {3 \, {\left (19 \, B b^{2} c^{9} - 14 \, A b c^{10}\right )}}{c^{13}}\right )} + \frac {5 \, {\left (7 \, B b^{3} - 6 \, A b^{2} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} \sqrt {c} - b \right |}\right )}{32 \, c^{\frac {9}{2}}} + \frac {B b^{4} - A b^{3} c}{{\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} c + b \sqrt {c}\right )} c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

1/48*sqrt(c*x^4 + b*x^2)*(2*x^2*(4*B*x^2/c^2 - (11*B*b*c^10 - 6*A*c^11)/c^13) + 3*(19*B*b^2*c^9 - 14*A*b*c^10)
/c^13) + 5/32*(7*B*b^3 - 6*A*b^2*c)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))*sqrt(c) - b))/c^(9/2) + (B*
b^4 - A*b^3*c)/(((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))*c + b*sqrt(c))*c^4)

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maple [A]  time = 0.06, size = 166, normalized size = 0.90 \begin {gather*} \frac {\left (c \,x^{2}+b \right ) \left (8 B \,c^{\frac {9}{2}} x^{7}+12 A \,c^{\frac {9}{2}} x^{5}-14 B b \,c^{\frac {7}{2}} x^{5}-30 A b \,c^{\frac {7}{2}} x^{3}+35 B \,b^{2} c^{\frac {5}{2}} x^{3}-90 A \,b^{2} c^{\frac {5}{2}} x +105 B \,b^{3} c^{\frac {3}{2}} x +90 \sqrt {c \,x^{2}+b}\, A \,b^{2} c^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-105 \sqrt {c \,x^{2}+b}\, B \,b^{3} c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )\right ) x^{3}}{48 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)

[Out]

1/48*x^3*(c*x^2+b)*(8*B*c^(9/2)*x^7+12*A*c^(9/2)*x^5-14*B*c^(7/2)*x^5*b-30*A*c^(7/2)*x^3*b+35*B*c^(5/2)*x^3*b^
2-90*A*c^(5/2)*x*b^2+105*B*c^(3/2)*x*b^3+90*A*(c*x^2+b)^(1/2)*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b^2*c^2-105*B*(c*x
^2+b)^(1/2)*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b^3*c)/(c*x^4+b*x^2)^(3/2)/c^(11/2)

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maxima [A]  time = 1.49, size = 237, normalized size = 1.29 \begin {gather*} \frac {1}{16} \, {\left (\frac {4 \, x^{6}}{\sqrt {c x^{4} + b x^{2}} c} - \frac {10 \, b x^{4}}{\sqrt {c x^{4} + b x^{2}} c^{2}} - \frac {30 \, b^{2} x^{2}}{\sqrt {c x^{4} + b x^{2}} c^{3}} + \frac {15 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}}\right )} A + \frac {1}{96} \, {\left (\frac {16 \, x^{8}}{\sqrt {c x^{4} + b x^{2}} c} - \frac {28 \, b x^{6}}{\sqrt {c x^{4} + b x^{2}} c^{2}} + \frac {70 \, b^{2} x^{4}}{\sqrt {c x^{4} + b x^{2}} c^{3}} + \frac {210 \, b^{3} x^{2}}{\sqrt {c x^{4} + b x^{2}} c^{4}} - \frac {105 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {9}{2}}}\right )} B \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

1/16*(4*x^6/(sqrt(c*x^4 + b*x^2)*c) - 10*b*x^4/(sqrt(c*x^4 + b*x^2)*c^2) - 30*b^2*x^2/(sqrt(c*x^4 + b*x^2)*c^3
) + 15*b^2*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(7/2))*A + 1/96*(16*x^8/(sqrt(c*x^4 + b*x^2)*c)
- 28*b*x^6/(sqrt(c*x^4 + b*x^2)*c^2) + 70*b^2*x^4/(sqrt(c*x^4 + b*x^2)*c^3) + 210*b^3*x^2/(sqrt(c*x^4 + b*x^2)
*c^4) - 105*b^3*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(9/2))*B

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^9\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^9*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)

[Out]

int((x^9*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{9} \left (A + B x^{2}\right )}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**9*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)

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