3.2.46 \(\int \frac {x^7 (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=147 \[ \frac {3 b (5 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{7/2}}-\frac {3 \sqrt {b x^2+c x^4} (5 b B-4 A c)}{8 c^3}+\frac {x^2 \sqrt {b x^2+c x^4} (5 b B-4 A c)}{4 b c^2}-\frac {x^6 (b B-A c)}{b c \sqrt {b x^2+c x^4}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.28, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {x^2 \sqrt {b x^2+c x^4} (5 b B-4 A c)}{4 b c^2}-\frac {3 \sqrt {b x^2+c x^4} (5 b B-4 A c)}{8 c^3}+\frac {3 b (5 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{7/2}}-\frac {x^6 (b B-A c)}{b c \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^7*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 113, normalized size = 0.77 \begin {gather*} \frac {x \left (3 b^{3/2} \sqrt {\frac {c x^2}{b}+1} (5 b B-4 A c) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )+\sqrt {c} x \left (b c \left (12 A-5 B x^2\right )+2 c^2 x^2 \left (2 A+B x^2\right )-15 b^2 B\right )\right )}{8 c^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x*(Sqrt[c]*x*(-15*b^2*B + b*c*(12*A - 5*B*x^2) + 2*c^2*x^2*(2*A + B*x^2)) + 3*b^(3/2)*(5*b*B - 4*A*c)*Sqrt[1
+ (c*x^2)/b]*ArcSinh[(Sqrt[c]*x)/Sqrt[b]]))/(8*c^(7/2)*Sqrt[x^2*(b + c*x^2)])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.58, size = 122, normalized size = 0.83 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (12 A b c+4 A c^2 x^2-15 b^2 B-5 b B c x^2+2 B c^2 x^4\right )}{8 c^3 \left (b+c x^2\right )}-\frac {3 \left (5 b^2 B-4 A b c\right ) \log \left (-2 \sqrt {c} \sqrt {b x^2+c x^4}+b+2 c x^2\right )}{16 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^7*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-15*b^2*B + 12*A*b*c - 5*b*B*c*x^2 + 4*A*c^2*x^2 + 2*B*c^2*x^4))/(8*c^3*(b + c*x^2)) - (
3*(5*b^2*B - 4*A*b*c)*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[b*x^2 + c*x^4]])/(16*c^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 289, normalized size = 1.97 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B b^{3} - 4 \, A b^{2} c + {\left (5 \, B b^{2} c - 4 \, A b 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 (2 \, B c^{3} x^{4} - 15 \, B b^{2} c + 12 \, A b c^{2} - {\left (5 \, B b c^{2} - 4 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{16 \, {\left (c^{5} x^{2} + b c^{4}\right )}}, -\frac {3 \, {\left (5 \, B b^{3} - 4 \, A b^{2} c + {\left (5 \, B b^{2} c - 4 \, A b 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 (2 \, B c^{3} x^{4} - 15 \, B b^{2} c + 12 \, A b c^{2} - {\left (5 \, B b c^{2} - 4 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{8 \, {\left (c^{5} x^{2} + b c^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(3*(5*B*b^3 - 4*A*b^2*c + (5*B*b^2*c - 4*A*b*c^2)*x^2)*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)
*sqrt(c)) - 2*(2*B*c^3*x^4 - 15*B*b^2*c + 12*A*b*c^2 - (5*B*b*c^2 - 4*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/(c^5*x^
2 + b*c^4), -1/8*(3*(5*B*b^3 - 4*A*b^2*c + (5*B*b^2*c - 4*A*b*c^2)*x^2)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sq
rt(-c)/(c*x^2 + b)) - (2*B*c^3*x^4 - 15*B*b^2*c + 12*A*b*c^2 - (5*B*b*c^2 - 4*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))
/(c^5*x^2 + b*c^4)]

________________________________________________________________________________________

giac [A]  time = 0.28, size = 147, normalized size = 1.00 \begin {gather*} \frac {1}{8} \, \sqrt {c x^{4} + b x^{2}} {\left (\frac {2 \, B x^{2}}{c^{2}} - \frac {7 \, B b c^{5} - 4 \, A c^{6}}{c^{8}}\right )} - \frac {3 \, {\left (5 \, B b^{2} - 4 \, A b c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} \sqrt {c} - b \right |}\right )}{16 \, c^{\frac {7}{2}}} - \frac {B b^{3} - A b^{2} c}{{\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} c + b \sqrt {c}\right )} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

1/8*sqrt(c*x^4 + b*x^2)*(2*B*x^2/c^2 - (7*B*b*c^5 - 4*A*c^6)/c^8) - 3/16*(5*B*b^2 - 4*A*b*c)*log(abs(-2*(sqrt(
c)*x^2 - sqrt(c*x^4 + b*x^2))*sqrt(c) - b))/c^(7/2) - (B*b^3 - A*b^2*c)/(((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))*
c + b*sqrt(c))*c^3)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 140, normalized size = 0.95 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-2 B \,c^{\frac {7}{2}} x^{5}-4 A \,c^{\frac {7}{2}} x^{3}+5 B b \,c^{\frac {5}{2}} x^{3}-12 A b \,c^{\frac {5}{2}} x +15 B \,b^{2} c^{\frac {3}{2}} x +12 \sqrt {c \,x^{2}+b}\, A b \,c^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-15 \sqrt {c \,x^{2}+b}\, B \,b^{2} c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )\right ) x^{3}}{8 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)

[Out]

-1/8*x^3*(c*x^2+b)*(-2*B*c^(7/2)*x^5-4*A*c^(7/2)*x^3+5*B*c^(5/2)*x^3*b-12*A*c^(5/2)*x*b+15*B*c^(3/2)*x*b^2+12*
A*(c*x^2+b)^(1/2)*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b*c^2-15*B*(c*x^2+b)^(1/2)*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b^2*c
)/(c*x^4+b*x^2)^(3/2)/c^(9/2)

________________________________________________________________________________________

maxima [A]  time = 1.54, size = 187, normalized size = 1.27 \begin {gather*} \frac {1}{4} \, {\left (\frac {2 \, x^{4}}{\sqrt {c x^{4} + b x^{2}} c} + \frac {6 \, b x^{2}}{\sqrt {c x^{4} + b x^{2}} c^{2}} - \frac {3 \, b \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}}\right )} A + \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 )} B \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

1/4*(2*x^4/(sqrt(c*x^4 + b*x^2)*c) + 6*b*x^2/(sqrt(c*x^4 + b*x^2)*c^2) - 3*b*log(2*c*x^2 + b + 2*sqrt(c*x^4 +
b*x^2)*sqrt(c))/c^(5/2))*A + 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))*B

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^7\,\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^7*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)

[Out]

int((x^7*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7} \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**7*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**7*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)

________________________________________________________________________________________