3.2.28 \(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{14}} \, dx\)

Optimal. Leaf size=214 \[ -\frac {3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{256 b^{7/2}}+\frac {3 c^3 \sqrt {b x^2+c x^4} (2 b B-A c)}{256 b^3 x^3}-\frac {c^2 \sqrt {b x^2+c x^4} (2 b B-A c)}{128 b^2 x^5}-\frac {\left (b x^2+c x^4\right )^{3/2} (2 b B-A c)}{16 b x^{11}}-\frac {c \sqrt {b x^2+c x^4} (2 b B-A c)}{32 b x^7}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}} \]

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Rubi [A]  time = 0.34, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2038, 2020, 2025, 2008, 206} \begin {gather*} \frac {3 c^3 \sqrt {b x^2+c x^4} (2 b B-A c)}{256 b^3 x^3}-\frac {c^2 \sqrt {b x^2+c x^4} (2 b B-A c)}{128 b^2 x^5}-\frac {3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{256 b^{7/2}}-\frac {c \sqrt {b x^2+c x^4} (2 b B-A c)}{32 b x^7}-\frac {\left (b x^2+c x^4\right )^{3/2} (2 b B-A c)}{16 b x^{11}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^14,x]

[Out]

-(c*(2*b*B - A*c)*Sqrt[b*x^2 + c*x^4])/(32*b*x^7) - (c^2*(2*b*B - A*c)*Sqrt[b*x^2 + c*x^4])/(128*b^2*x^5) + (3
*c^3*(2*b*B - A*c)*Sqrt[b*x^2 + c*x^4])/(256*b^3*x^3) - ((2*b*B - A*c)*(b*x^2 + c*x^4)^(3/2))/(16*b*x^11) - (A
*(b*x^2 + c*x^4)^(5/2))/(10*b*x^15) - (3*c^4*(2*b*B - A*c)*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(256*b^(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 2008

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 2020

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b*
x^n)^p)/(c*(m + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2038

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(c*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*(m + j*p + 1)), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, -(n*p) - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{14}} \, dx &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}-\frac {(-10 b B+5 A c) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{12}} \, dx}{10 b}\\ &=-\frac {(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}+\frac {(3 c (2 b B-A c)) \int \frac {\sqrt {b x^2+c x^4}}{x^8} \, dx}{16 b}\\ &=-\frac {c (2 b B-A c) \sqrt {b x^2+c x^4}}{32 b x^7}-\frac {(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}+\frac {\left (c^2 (2 b B-A c)\right ) \int \frac {1}{x^4 \sqrt {b x^2+c x^4}} \, dx}{32 b}\\ &=-\frac {c (2 b B-A c) \sqrt {b x^2+c x^4}}{32 b x^7}-\frac {c^2 (2 b B-A c) \sqrt {b x^2+c x^4}}{128 b^2 x^5}-\frac {(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}-\frac {\left (3 c^3 (2 b B-A c)\right ) \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx}{128 b^2}\\ &=-\frac {c (2 b B-A c) \sqrt {b x^2+c x^4}}{32 b x^7}-\frac {c^2 (2 b B-A c) \sqrt {b x^2+c x^4}}{128 b^2 x^5}+\frac {3 c^3 (2 b B-A c) \sqrt {b x^2+c x^4}}{256 b^3 x^3}-\frac {(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}+\frac {\left (3 c^4 (2 b B-A c)\right ) \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{256 b^3}\\ &=-\frac {c (2 b B-A c) \sqrt {b x^2+c x^4}}{32 b x^7}-\frac {c^2 (2 b B-A c) \sqrt {b x^2+c x^4}}{128 b^2 x^5}+\frac {3 c^3 (2 b B-A c) \sqrt {b x^2+c x^4}}{256 b^3 x^3}-\frac {(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}-\frac {\left (3 c^4 (2 b B-A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+c x^4}}\right )}{256 b^3}\\ &=-\frac {c (2 b B-A c) \sqrt {b x^2+c x^4}}{32 b x^7}-\frac {c^2 (2 b B-A c) \sqrt {b x^2+c x^4}}{128 b^2 x^5}+\frac {3 c^3 (2 b B-A c) \sqrt {b x^2+c x^4}}{256 b^3 x^3}-\frac {(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}-\frac {3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{256 b^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 65, normalized size = 0.30 \begin {gather*} -\frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (A b^5+c^4 x^{10} (2 b B-A c) \, _2F_1\left (\frac {5}{2},5;\frac {7}{2};\frac {c x^2}{b}+1\right )\right )}{10 b^6 x^{15}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^14,x]

[Out]

-1/10*((x^2*(b + c*x^2))^(5/2)*(A*b^5 + c^4*(2*b*B - A*c)*x^10*Hypergeometric2F1[5/2, 5, 7/2, 1 + (c*x^2)/b]))
/(b^6*x^15)

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IntegrateAlgebraic [A]  time = 1.50, size = 161, normalized size = 0.75 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-128 A b^4-176 A b^3 c x^2-8 A b^2 c^2 x^4+10 A b c^3 x^6-15 A c^4 x^8-160 b^4 B x^2-240 b^3 B c x^4-20 b^2 B c^2 x^6+30 b B c^3 x^8\right )}{1280 b^3 x^{11}}-\frac {3 \left (2 b B c^4-A c^5\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{256 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^14,x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-128*A*b^4 - 160*b^4*B*x^2 - 176*A*b^3*c*x^2 - 240*b^3*B*c*x^4 - 8*A*b^2*c^2*x^4 - 20*b^
2*B*c^2*x^6 + 10*A*b*c^3*x^6 + 30*b*B*c^3*x^8 - 15*A*c^4*x^8))/(1280*b^3*x^11) - (3*(2*b*B*c^4 - A*c^5)*ArcTan
h[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(256*b^(7/2))

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fricas [A]  time = 0.51, size = 345, normalized size = 1.61 \begin {gather*} \left [-\frac {15 \, {\left (2 \, B b c^{4} - A c^{5}\right )} \sqrt {b} x^{11} \log \left (-\frac {c x^{3} + 2 \, b x + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) - 2 \, {\left (15 \, {\left (2 \, B b^{2} c^{3} - A b c^{4}\right )} x^{8} - 10 \, {\left (2 \, B b^{3} c^{2} - A b^{2} c^{3}\right )} x^{6} - 128 \, A b^{5} - 8 \, {\left (30 \, B b^{4} c + A b^{3} c^{2}\right )} x^{4} - 16 \, {\left (10 \, B b^{5} + 11 \, A b^{4} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{2560 \, b^{4} x^{11}}, \frac {15 \, {\left (2 \, B b c^{4} - A c^{5}\right )} \sqrt {-b} x^{11} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + {\left (15 \, {\left (2 \, B b^{2} c^{3} - A b c^{4}\right )} x^{8} - 10 \, {\left (2 \, B b^{3} c^{2} - A b^{2} c^{3}\right )} x^{6} - 128 \, A b^{5} - 8 \, {\left (30 \, B b^{4} c + A b^{3} c^{2}\right )} x^{4} - 16 \, {\left (10 \, B b^{5} + 11 \, A b^{4} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{1280 \, b^{4} x^{11}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^14,x, algorithm="fricas")

[Out]

[-1/2560*(15*(2*B*b*c^4 - A*c^5)*sqrt(b)*x^11*log(-(c*x^3 + 2*b*x + 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) - 2*(1
5*(2*B*b^2*c^3 - A*b*c^4)*x^8 - 10*(2*B*b^3*c^2 - A*b^2*c^3)*x^6 - 128*A*b^5 - 8*(30*B*b^4*c + A*b^3*c^2)*x^4
- 16*(10*B*b^5 + 11*A*b^4*c)*x^2)*sqrt(c*x^4 + b*x^2))/(b^4*x^11), 1/1280*(15*(2*B*b*c^4 - A*c^5)*sqrt(-b)*x^1
1*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + (15*(2*B*b^2*c^3 - A*b*c^4)*x^8 - 10*(2*B*b^3*c^2 - A*b
^2*c^3)*x^6 - 128*A*b^5 - 8*(30*B*b^4*c + A*b^3*c^2)*x^4 - 16*(10*B*b^5 + 11*A*b^4*c)*x^2)*sqrt(c*x^4 + b*x^2)
)/(b^4*x^11)]

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giac [A]  time = 0.31, size = 234, normalized size = 1.09 \begin {gather*} \frac {\frac {15 \, {\left (2 \, B b c^{5} \mathrm {sgn}\relax (x) - A c^{6} \mathrm {sgn}\relax (x)\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} + \frac {30 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} B b c^{5} \mathrm {sgn}\relax (x) - 140 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B b^{2} c^{5} \mathrm {sgn}\relax (x) + 140 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b^{4} c^{5} \mathrm {sgn}\relax (x) - 30 \, \sqrt {c x^{2} + b} B b^{5} c^{5} \mathrm {sgn}\relax (x) - 15 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} A c^{6} \mathrm {sgn}\relax (x) + 70 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} A b c^{6} \mathrm {sgn}\relax (x) - 128 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A b^{2} c^{6} \mathrm {sgn}\relax (x) - 70 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A b^{3} c^{6} \mathrm {sgn}\relax (x) + 15 \, \sqrt {c x^{2} + b} A b^{4} c^{6} \mathrm {sgn}\relax (x)}{b^{3} c^{5} x^{10}}}{1280 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^14,x, algorithm="giac")

[Out]

1/1280*(15*(2*B*b*c^5*sgn(x) - A*c^6*sgn(x))*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^3) + (30*(c*x^2 + b)
^(9/2)*B*b*c^5*sgn(x) - 140*(c*x^2 + b)^(7/2)*B*b^2*c^5*sgn(x) + 140*(c*x^2 + b)^(3/2)*B*b^4*c^5*sgn(x) - 30*s
qrt(c*x^2 + b)*B*b^5*c^5*sgn(x) - 15*(c*x^2 + b)^(9/2)*A*c^6*sgn(x) + 70*(c*x^2 + b)^(7/2)*A*b*c^6*sgn(x) - 12
8*(c*x^2 + b)^(5/2)*A*b^2*c^6*sgn(x) - 70*(c*x^2 + b)^(3/2)*A*b^3*c^6*sgn(x) + 15*sqrt(c*x^2 + b)*A*b^4*c^6*sg
n(x))/(b^3*c^5*x^10))/c

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maple [A]  time = 0.09, size = 344, normalized size = 1.61 \begin {gather*} \frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (15 A \,b^{\frac {3}{2}} c^{5} x^{10} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-30 B \,b^{\frac {5}{2}} c^{4} x^{10} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-15 \sqrt {c \,x^{2}+b}\, A b \,c^{5} x^{10}+30 \sqrt {c \,x^{2}+b}\, B \,b^{2} c^{4} x^{10}-5 \left (c \,x^{2}+b \right )^{\frac {3}{2}} A \,c^{5} x^{10}+10 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B b \,c^{4} x^{10}+5 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A \,c^{4} x^{8}-10 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B b \,c^{3} x^{8}+10 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A b \,c^{3} x^{6}-20 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B \,b^{2} c^{2} x^{6}-40 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A \,b^{2} c^{2} x^{4}+80 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B \,b^{3} c \,x^{4}+80 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A \,b^{3} c \,x^{2}-160 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B \,b^{4} x^{2}-128 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A \,b^{4}\right )}{1280 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{5} x^{13}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^14,x)

[Out]

1/1280*(c*x^4+b*x^2)^(3/2)*(-5*A*(c*x^2+b)^(3/2)*x^10*c^5+15*A*b^(3/2)*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^1
0*c^5+10*B*(c*x^2+b)^(3/2)*x^10*b*c^4-30*B*b^(5/2)*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^10*c^4+5*A*(c*x^2+b)^
(5/2)*x^8*c^4-15*A*(c*x^2+b)^(1/2)*x^10*b*c^5-10*B*(c*x^2+b)^(5/2)*x^8*b*c^3+30*B*(c*x^2+b)^(1/2)*x^10*b^2*c^4
+10*A*(c*x^2+b)^(5/2)*x^6*b*c^3-20*B*(c*x^2+b)^(5/2)*x^6*b^2*c^2-40*A*(c*x^2+b)^(5/2)*x^4*b^2*c^2+80*B*(c*x^2+
b)^(5/2)*x^4*b^3*c+80*A*(c*x^2+b)^(5/2)*x^2*b^3*c-160*B*(c*x^2+b)^(5/2)*x^2*b^4-128*A*(c*x^2+b)^(5/2)*b^4)/x^1
3/(c*x^2+b)^(3/2)/b^5

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{14}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^14,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^14, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{14}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^14,x)

[Out]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^14, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{14}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**14,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**14, x)

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