3.2.30 \(\int \frac {x^7 (A+B x^2)}{\sqrt {b x^2+c x^4}} \, dx\) [130]

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

[Out]

5/128*b^3*(-8*A*c+7*B*b)*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2))/c^(9/2)-5/128*b^2*(-8*A*c+7*B*b)*(c*x^4+b*x^
2)^(1/2)/c^4+5/192*b*(-8*A*c+7*B*b)*x^2*(c*x^4+b*x^2)^(1/2)/c^3-1/48*(-8*A*c+7*B*b)*x^4*(c*x^4+b*x^2)^(1/2)/c^
2+1/8*B*x^6*(c*x^4+b*x^2)^(1/2)/c

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Rubi [A]
time = 0.21, 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 = {2059, 808, 684, 654, 634, 212} \begin {gather*} \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} \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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 634

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 654

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 684

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 808

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 2059

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} \text {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 ) \text {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)) \text {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 ) \text {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 ) \text {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 ) \text {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 = 0.18, size = 147, normalized size = 0.84 \begin {gather*} \frac {x \left (-\sqrt {c} x \left (b+c x^2\right ) \left (105 b^3 B-16 c^3 x^4 \left (4 A+3 B x^2\right )+8 b c^2 x^2 \left (10 A+7 B x^2\right )-10 b^2 c \left (12 A+7 B x^2\right )\right )-15 b^3 (7 b B-8 A c) \sqrt {b+c x^2} \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\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]*Log[-(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 = 0.39, size = 211, normalized size = 1.20

method	result	size
risch	\(\frac {x^{2} \left (48 B \,c^{3} x^{6}+64 A \,c^{3} x^{4}-56 B b \,c^{2} x^{4}-80 A b \,c^{2} x^{2}+70 B \,b^{2} c \,x^{2}+120 A \,b^{2} c -105 B \,b^{3}\right ) \left (c \,x^{2}+b \right )}{384 c^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {\left (-\frac {5 b^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) A}{16 c^{\frac {7}{2}}}+\frac {35 b^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) B}{128 c^{\frac {9}{2}}}\right ) x \sqrt {c \,x^{2}+b}}{\sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(167\)
default	\(\frac {x \sqrt {c \,x^{2}+b}\, \left (48 B \sqrt {c \,x^{2}+b}\, c^{\frac {9}{2}} x^{7}+64 A \sqrt {c \,x^{2}+b}\, c^{\frac {9}{2}} x^{5}-56 B \sqrt {c \,x^{2}+b}\, c^{\frac {7}{2}} b \,x^{5}-80 A \sqrt {c \,x^{2}+b}\, c^{\frac {7}{2}} b \,x^{3}+70 B \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}} b^{2} x^{3}+120 A \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}} b^{2} x -105 B \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} b^{3} x -120 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{3} c^{2}+105 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{4} c \right )}{384 \sqrt {x^{4} c +b \,x^{2}}\, c^{\frac {11}{2}}}\)	\(211\)

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,method=_RETURNVERBOSE)

[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)*b*x^5-80*A*(c*x^2+b)^(1/2)*c^(7/2)*b*x^3+70*B*(c*x^2+b)^(1/2)*c^(5/2)*b^2*x^3+120*A*(c*x^2+b)^(1/2)*
c^(5/2)*b^2*x-105*B*(c*x^2+b)^(1/2)*c^(3/2)*b^3*x-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 = 0.29, 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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Fricas [A]
time = 2.42, 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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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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Giac [A]
time = 0.69, size = 184, normalized size = 1.05 \begin {gather*} \frac {1}{384} \, {\left (2 \, {\left (4 \, x^{2} {\left (\frac {6 \, B x^{2}}{c \mathrm {sgn}\left (x\right )} - \frac {7 \, B b c^{5} \mathrm {sgn}\left (x\right ) - 8 \, A c^{6} \mathrm {sgn}\left (x\right )}{c^{7}}\right )} + \frac {5 \, {\left (7 \, B b^{2} c^{4} \mathrm {sgn}\left (x\right ) - 8 \, A b c^{5} \mathrm {sgn}\left (x\right )\right )}}{c^{7}}\right )} x^{2} - \frac {15 \, {\left (7 \, B b^{3} c^{3} \mathrm {sgn}\left (x\right ) - 8 \, A b^{2} c^{4} \mathrm {sgn}\left (x\right )\right )}}{c^{7}}\right )} \sqrt {c x^{2} + b} x + \frac {5 \, {\left (7 \, B b^{4} \log \left ({\left | b \right |}\right ) - 8 \, A b^{3} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{256 \, c^{\frac {9}{2}}} - \frac {5 \, {\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{128 \, c^{\frac {9}{2}} \mathrm {sgn}\left (x\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="giac")

[Out]

1/384*(2*(4*x^2*(6*B*x^2/(c*sgn(x)) - (7*B*b*c^5*sgn(x) - 8*A*c^6*sgn(x))/c^7) + 5*(7*B*b^2*c^4*sgn(x) - 8*A*b
*c^5*sgn(x))/c^7)*x^2 - 15*(7*B*b^3*c^3*sgn(x) - 8*A*b^2*c^4*sgn(x))/c^7)*sqrt(c*x^2 + b)*x + 5/256*(7*B*b^4*l
og(abs(b)) - 8*A*b^3*c*log(abs(b)))*sgn(x)/c^(9/2) - 5/128*(7*B*b^4 - 8*A*b^3*c)*log(abs(-sqrt(c)*x + sqrt(c*x
^2 + b)))/(c^(9/2)*sgn(x))

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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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