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

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

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Rubi [A]  time = 0.33, antiderivative size = 176, 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, 794, 670, 640, 620, 206} \begin {gather*} -\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} \end {gather*}

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 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 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 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 )}{\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*}

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Mathematica [A]  time = 0.26, size = 145, normalized size = 0.82 \begin {gather*} \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 )}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.57, size = 139, normalized size = 0.79 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (120 A b^2 c-80 A b c^2 x^2+64 A c^3 x^4-105 b^3 B+70 b^2 B c x^2-56 b B c^2 x^4+48 B c^3 x^6\right )}{384 c^4}-\frac {5 \left (7 b^4 B-8 A b^3 c\right ) \log \left (-2 \sqrt {c} \sqrt {b x^2+c x^4}+b+2 c x^2\right )}{256 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x^2 + c*x^4]*(-105*b^3*B + 120*A*b^2*c + 70*b^2*B*c*x^2 - 80*A*b*c^2*x^2 - 56*b*B*c^2*x^4 + 64*A*c^3*x
^4 + 48*B*c^3*x^6))/(384*c^4) - (5*(7*b^4*B - 8*A*b^3*c)*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[b*x^2 + c*x^4]])/(25
6*c^(9/2))

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fricas [A]  time = 0.46, size = 275, normalized size = 1.56 \begin {gather*} \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 {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)^(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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giac [A]  time = 0.22, size = 149, normalized size = 0.85 \begin {gather*} \frac {1}{384} \, \sqrt {c x^{4} + b x^{2}} {\left (2 \, {\left (4 \, {\left (\frac {6 \, B x^{2}}{c} - \frac {7 \, B b c^{2} - 8 \, A c^{3}}{c^{4}}\right )} x^{2} + \frac {5 \, {\left (7 \, B b^{2} c - 8 \, A b c^{2}\right )}}{c^{4}}\right )} x^{2} - \frac {15 \, {\left (7 \, B b^{3} - 8 \, A b^{2} c\right )}}{c^{4}}\right )} - \frac {5 \, {\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} \sqrt {c} - b \right |}\right )}{256 \, c^{\frac {9}{2}}} \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)^(1/2),x, algorithm="giac")

[Out]

1/384*sqrt(c*x^4 + b*x^2)*(2*(4*(6*B*x^2/c - (7*B*b*c^2 - 8*A*c^3)/c^4)*x^2 + 5*(7*B*b^2*c - 8*A*b*c^2)/c^4)*x
^2 - 15*(7*B*b^3 - 8*A*b^2*c)/c^4) - 5/256*(7*B*b^4 - 8*A*b^3*c)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2)
)*sqrt(c) - b))/c^(9/2)

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maple [A]  time = 0.06, size = 211, normalized size = 1.20 \begin {gather*} \frac {\sqrt {c \,x^{2}+b}\, \left (48 \sqrt {c \,x^{2}+b}\, B \,c^{\frac {9}{2}} x^{7}+64 \sqrt {c \,x^{2}+b}\, A \,c^{\frac {9}{2}} x^{5}-56 \sqrt {c \,x^{2}+b}\, B b \,c^{\frac {7}{2}} x^{5}-80 \sqrt {c \,x^{2}+b}\, A b \,c^{\frac {7}{2}} x^{3}+70 \sqrt {c \,x^{2}+b}\, B \,b^{2} c^{\frac {5}{2}} x^{3}-120 A \,b^{3} c^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+105 B \,b^{4} c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+120 \sqrt {c \,x^{2}+b}\, A \,b^{2} c^{\frac {5}{2}} x -105 \sqrt {c \,x^{2}+b}\, B \,b^{3} c^{\frac {3}{2}} x \right ) x}{384 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{\frac {11}{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)^(1/2),x)

[Out]

1/384*x*(c*x^2+b)^(1/2)*(48*B*(c*x^2+b)^(1/2)*c^(9/2)*x^7+64*A*(c*x^2+b)^(1/2)*c^(9/2)*x^5-56*B*(c*x^2+b)^(1/2
)*c^(7/2)*x^5*b-80*A*(c*x^2+b)^(1/2)*c^(7/2)*x^3*b+70*B*(c*x^2+b)^(1/2)*c^(5/2)*x^3*b^2+120*A*(c*x^2+b)^(1/2)*
c^(5/2)*x*b^2-105*B*(c*x^2+b)^(1/2)*c^(3/2)*x*b^3-120*A*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b^3*c^2+105*B*ln(c^(1/2)
*x+(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 [A]  time = 1.38, size = 231, normalized size = 1.31 \begin {gather*} \frac {1}{96} \, {\left (\frac {16 \, \sqrt {c x^{4} + b x^{2}} x^{4}}{c} - \frac {20 \, \sqrt {c x^{4} + b x^{2}} b x^{2}}{c^{2}} - \frac {15 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}} + \frac {30 \, \sqrt {c x^{4} + b x^{2}} b^{2}}{c^{3}}\right )} A + \frac {1}{768} \, {\left (\frac {96 \, \sqrt {c x^{4} + b x^{2}} x^{6}}{c} - \frac {112 \, \sqrt {c x^{4} + b x^{2}} b x^{4}}{c^{2}} + \frac {140 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}}{c^{3}} + \frac {105 \, b^{4} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {9}{2}}} - \frac {210 \, \sqrt {c x^{4} + b x^{2}} b^{3}}{c^{4}}\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)^(1/2),x, algorithm="maxima")

[Out]

1/96*(16*sqrt(c*x^4 + b*x^2)*x^4/c - 20*sqrt(c*x^4 + b*x^2)*b*x^2/c^2 - 15*b^3*log(2*c*x^2 + b + 2*sqrt(c*x^4
+ b*x^2)*sqrt(c))/c^(7/2) + 30*sqrt(c*x^4 + b*x^2)*b^2/c^3)*A + 1/768*(96*sqrt(c*x^4 + b*x^2)*x^6/c - 112*sqrt
(c*x^4 + b*x^2)*b*x^4/c^2 + 140*sqrt(c*x^4 + b*x^2)*b^2*x^2/c^3 + 105*b^4*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x
^2)*sqrt(c))/c^(9/2) - 210*sqrt(c*x^4 + b*x^2)*b^3/c^4)*B

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7} \left (A + B x^{2}\right )}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, 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)**(1/2),x)

[Out]

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

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