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

Optimal. Leaf size=165 \[ \frac {3 c^5 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{256 b^{7/2}}-\frac {3 c^4 \sqrt {b x^2+c x^4}}{256 b^3 x^3}+\frac {c^3 \sqrt {b x^2+c x^4}}{128 b^2 x^5}-\frac {c^2 \sqrt {b x^2+c x^4}}{160 b x^7}-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}-\frac {3 c \sqrt {b x^2+c x^4}}{80 x^9} \]

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Rubi [A]  time = 0.25, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2020, 2025, 2008, 206} \begin {gather*} -\frac {3 c^4 \sqrt {b x^2+c x^4}}{256 b^3 x^3}+\frac {c^3 \sqrt {b x^2+c x^4}}{128 b^2 x^5}+\frac {3 c^5 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{256 b^{7/2}}-\frac {c^2 \sqrt {b x^2+c x^4}}{160 b x^7}-\frac {3 c \sqrt {b x^2+c x^4}}{80 x^9}-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^14,x]

[Out]

(-3*c*Sqrt[b*x^2 + c*x^4])/(80*x^9) - (c^2*Sqrt[b*x^2 + c*x^4])/(160*b*x^7) + (c^3*Sqrt[b*x^2 + c*x^4])/(128*b
^2*x^5) - (3*c^4*Sqrt[b*x^2 + c*x^4])/(256*b^3*x^3) - (b*x^2 + c*x^4)^(3/2)/(10*x^13) + (3*c^5*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]

Rubi steps

\begin {align*} \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{14}} \, dx &=-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}+\frac {1}{10} (3 c) \int \frac {\sqrt {b x^2+c x^4}}{x^{10}} \, dx\\ &=-\frac {3 c \sqrt {b x^2+c x^4}}{80 x^9}-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}+\frac {1}{80} \left (3 c^2\right ) \int \frac {1}{x^6 \sqrt {b x^2+c x^4}} \, dx\\ &=-\frac {3 c \sqrt {b x^2+c x^4}}{80 x^9}-\frac {c^2 \sqrt {b x^2+c x^4}}{160 b x^7}-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}-\frac {c^3 \int \frac {1}{x^4 \sqrt {b x^2+c x^4}} \, dx}{32 b}\\ &=-\frac {3 c \sqrt {b x^2+c x^4}}{80 x^9}-\frac {c^2 \sqrt {b x^2+c x^4}}{160 b x^7}+\frac {c^3 \sqrt {b x^2+c x^4}}{128 b^2 x^5}-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}+\frac {\left (3 c^4\right ) \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx}{128 b^2}\\ &=-\frac {3 c \sqrt {b x^2+c x^4}}{80 x^9}-\frac {c^2 \sqrt {b x^2+c x^4}}{160 b x^7}+\frac {c^3 \sqrt {b x^2+c x^4}}{128 b^2 x^5}-\frac {3 c^4 \sqrt {b x^2+c x^4}}{256 b^3 x^3}-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}-\frac {\left (3 c^5\right ) \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{256 b^3}\\ &=-\frac {3 c \sqrt {b x^2+c x^4}}{80 x^9}-\frac {c^2 \sqrt {b x^2+c x^4}}{160 b x^7}+\frac {c^3 \sqrt {b x^2+c x^4}}{128 b^2 x^5}-\frac {3 c^4 \sqrt {b x^2+c x^4}}{256 b^3 x^3}-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}+\frac {\left (3 c^5\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 {3 c \sqrt {b x^2+c x^4}}{80 x^9}-\frac {c^2 \sqrt {b x^2+c x^4}}{160 b x^7}+\frac {c^3 \sqrt {b x^2+c x^4}}{128 b^2 x^5}-\frac {3 c^4 \sqrt {b x^2+c x^4}}{256 b^3 x^3}-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}+\frac {3 c^5 \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 = 0.02, size = 46, normalized size = 0.28 \begin {gather*} \frac {c^5 \left (x^2 \left (b+c x^2\right )\right )^{5/2} \, _2F_1\left (\frac {5}{2},6;\frac {7}{2};\frac {c x^2}{b}+1\right )}{5 b^6 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c^5*(x^2*(b + c*x^2))^(5/2)*Hypergeometric2F1[5/2, 6, 7/2, 1 + (c*x^2)/b])/(5*b^6*x^5)

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IntegrateAlgebraic [A]  time = 0.80, size = 104, normalized size = 0.63 \begin {gather*} \frac {3 c^5 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{256 b^{7/2}}+\frac {\sqrt {b x^2+c x^4} \left (-128 b^4-176 b^3 c x^2-8 b^2 c^2 x^4+10 b c^3 x^6-15 c^4 x^8\right )}{1280 b^3 x^{11}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x^2 + c*x^4]*(-128*b^4 - 176*b^3*c*x^2 - 8*b^2*c^2*x^4 + 10*b*c^3*x^6 - 15*c^4*x^8))/(1280*b^3*x^11) +
(3*c^5*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(256*b^(7/2))

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fricas [A]  time = 1.17, size = 229, normalized size = 1.39 \begin {gather*} \left [\frac {15 \, \sqrt {b} c^{5} 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 \, b c^{4} x^{8} - 10 \, b^{2} c^{3} x^{6} + 8 \, b^{3} c^{2} x^{4} + 176 \, b^{4} c x^{2} + 128 \, b^{5}\right )} \sqrt {c x^{4} + b x^{2}}}{2560 \, b^{4} x^{11}}, -\frac {15 \, \sqrt {-b} c^{5} x^{11} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + {\left (15 \, b c^{4} x^{8} - 10 \, b^{2} c^{3} x^{6} + 8 \, b^{3} c^{2} x^{4} + 176 \, b^{4} c x^{2} + 128 \, b^{5}\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((c*x^4+b*x^2)^(3/2)/x^14,x, algorithm="fricas")

[Out]

[1/2560*(15*sqrt(b)*c^5*x^11*log(-(c*x^3 + 2*b*x + 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) - 2*(15*b*c^4*x^8 - 10*
b^2*c^3*x^6 + 8*b^3*c^2*x^4 + 176*b^4*c*x^2 + 128*b^5)*sqrt(c*x^4 + b*x^2))/(b^4*x^11), -1/1280*(15*sqrt(-b)*c
^5*x^11*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + (15*b*c^4*x^8 - 10*b^2*c^3*x^6 + 8*b^3*c^2*x^4 +
176*b^4*c*x^2 + 128*b^5)*sqrt(c*x^4 + b*x^2))/(b^4*x^11)]

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giac [A]  time = 0.38, size = 138, normalized size = 0.84 \begin {gather*} -\frac {\frac {15 \, c^{6} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-b} b^{3}} + \frac {15 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} c^{6} \mathrm {sgn}\relax (x) - 70 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} b c^{6} \mathrm {sgn}\relax (x) + 128 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} b^{2} c^{6} \mathrm {sgn}\relax (x) + 70 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} b^{3} c^{6} \mathrm {sgn}\relax (x) - 15 \, \sqrt {c x^{2} + b} 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((c*x^4+b*x^2)^(3/2)/x^14,x, algorithm="giac")

[Out]

-1/1280*(15*c^6*arctan(sqrt(c*x^2 + b)/sqrt(-b))*sgn(x)/(sqrt(-b)*b^3) + (15*(c*x^2 + b)^(9/2)*c^6*sgn(x) - 70
*(c*x^2 + b)^(7/2)*b*c^6*sgn(x) + 128*(c*x^2 + b)^(5/2)*b^2*c^6*sgn(x) + 70*(c*x^2 + b)^(3/2)*b^3*c^6*sgn(x) -
15*sqrt(c*x^2 + b)*b^4*c^6*sgn(x))/(b^3*c^5*x^10))/c

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maple [A]  time = 0.03, size = 186, normalized size = 1.13 \begin {gather*} \frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (15 b^{\frac {3}{2}} c^{5} x^{10} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-15 \sqrt {c \,x^{2}+b}\, b \,c^{5} x^{10}-5 \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{5} x^{10}+5 \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{4} x^{8}+10 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,c^{3} x^{6}-40 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2} c^{2} x^{4}+80 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{3} c \,x^{2}-128 \left (c \,x^{2}+b \right )^{\frac {5}{2}} 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((c*x^4+b*x^2)^(3/2)/x^14,x)

[Out]

1/1280*(c*x^4+b*x^2)^(3/2)*(-5*(c*x^2+b)^(3/2)*x^10*c^5+15*b^(3/2)*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^10*c^
5+5*(c*x^2+b)^(5/2)*x^8*c^4-15*(c*x^2+b)^(1/2)*x^10*b*c^5+10*(c*x^2+b)^(5/2)*x^6*b*c^3-40*(c*x^2+b)^(5/2)*x^4*
b^2*c^2+80*(c*x^2+b)^(5/2)*x^2*b^3*c-128*(c*x^2+b)^(5/2)*b^4)/x^13/(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}}}{x^{14}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\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((b*x^2 + c*x^4)^(3/2)/x^14,x)

[Out]

int((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}}}{x^{14}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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