3.4.94 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{3/2}}{x^7} \, dx\)

Optimal. Leaf size=163 \[ -\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^4 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {b^3 \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.05, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1112, 266, 43} \begin {gather*} -\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^4 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {b^3 \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(6*x^6*(a + b*x^2)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*x^4*
(a + b*x^2)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*x^2*(a + b*x^2)) + (b^3*Sqrt[a^2 + 2*a*b*x^2 + b^2
*x^4]*Log[x])/(a + b*x^2)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (a b+b^2 x^2\right )^3}{x^7} \, dx}{b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \operatorname {Subst}\left (\int \frac {\left (a b+b^2 x\right )^3}{x^4} \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \operatorname {Subst}\left (\int \left (\frac {a^3 b^3}{x^4}+\frac {3 a^2 b^4}{x^3}+\frac {3 a b^5}{x^2}+\frac {b^6}{x}\right ) \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )}\\ &=-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^4 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 63, normalized size = 0.39 \begin {gather*} -\frac {\sqrt {\left (a+b x^2\right )^2} \left (a \left (2 a^2+9 a b x^2+18 b^2 x^4\right )-12 b^3 x^6 \log (x)\right )}{12 x^6 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(Sqrt[(a + b*x^2)^2]*(a*(2*a^2 + 9*a*b*x^2 + 18*b^2*x^4) - 12*b^3*x^6*Log[x]))/(x^6*(a + b*x^2))

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IntegrateAlgebraic [B]  time = 4.94, size = 944, normalized size = 5.79 \begin {gather*} \frac {1}{2} \tanh ^{-1}\left (\frac {\sqrt {b^2} x^2}{a}-\frac {\sqrt {b^2 x^4+2 a b x^2+a^2}}{a}\right ) b^3+\frac {\sqrt {b^2 x^4+2 a b x^2+a^2} \left (-65536 b^{20} x^{34}-598016 a b^{19} x^{32}-2560000 a^2 b^{18} x^{30}-6836224 a^3 b^{17} x^{28}-12769280 a^4 b^{16} x^{26}-17724928 a^5 b^{15} x^{24}-18952960 a^6 b^{14} x^{22}-15961088 a^7 b^{13} x^{20}-10726144 a^8 b^{12} x^{18}-5788640 a^9 b^{11} x^{16}-2509936 a^{10} b^{10} x^{14}-869648 a^{11} b^9 x^{12}-237848 a^{12} b^8 x^{10}-50266 a^{13} b^7 x^8-7925 a^{14} b^6 x^6-878 a^{15} b^5 x^4-61 a^{16} b^4 x^2-2 a^{17} b^3\right )+\sqrt {b^2} \left (65536 b^{20} x^{36}+663552 a b^{19} x^{34}+3158016 a^2 b^{18} x^{32}+9396224 a^3 b^{17} x^{30}[phone] a^4 b^{16} x^{28}[phone] a^5 b^{15} x^{26}[phone] a^6 b^{14} x^{24}[phone] a^7 b^{13} x^{22}[phone] a^8 b^{12} x^{20}[phone] a^9 b^{11} x^{18}+8298576 a^{10} b^{10} x^{16}+3379584 a^{11} b^9 x^{14}+1107496 a^{12} b^8 x^{12}+288114 a^{13} b^7 x^{10}+58191 a^{14} b^6 x^8+8803 a^{15} b^5 x^6+939 a^{16} b^4 x^4+63 a^{17} b^3 x^2+2 a^{18} b^2\right )}{3 \sqrt {b^2} \sqrt {b^2 x^4+2 a b x^2+a^2} \left (-16384 b^{16} x^{28}-131072 a b^{15} x^{26}-483328 a^2 b^{14} x^{24}-1089536 a^3 b^{13} x^{22}-1678336 a^4 b^{12} x^{20}-1869824 a^5 b^{11} x^{18}-1554432 a^6 b^{10} x^{16}-979968 a^7 b^9 x^{14}-470976 a^8 b^8 x^{12}-171776 a^9 b^7 x^{10}-46816 a^{10} b^6 x^8-9248 a^{11} b^5 x^6-1252 a^{12} b^4 x^4-104 a^{13} b^3 x^2-4 a^{14} b^2\right ) x^6+3 \left (16384 b^{18} x^{30}+147456 a b^{17} x^{28}+614400 a^2 b^{16} x^{26}+1572864 a^3 b^{15} x^{24}+2767872 a^4 b^{14} x^{22}+3548160 a^5 b^{13} x^{20}+3424256 a^6 b^{12} x^{18}+2534400 a^7 b^{11} x^{16}+1450944 a^8 b^{10} x^{14}+642752 a^9 b^9 x^{12}+218592 a^{10} b^8 x^{10}+56064 a^{11} b^7 x^8+10500 a^{12} b^6 x^6+1356 a^{13} b^5 x^4+108 a^{14} b^4 x^2+4 a^{15} b^3\right ) x^6}-\frac {1}{4} \left (b^2\right )^{3/2} \log \left (-\sqrt {b^2} x^2-a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right )-\frac {1}{4} \left (b^2\right )^{3/2} \log \left (-\sqrt {b^2} x^2+a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-2*a^17*b^3 - 61*a^16*b^4*x^2 - 878*a^15*b^5*x^4 - 7925*a^14*b^6*x^6 - 50266
*a^13*b^7*x^8 - 237848*a^12*b^8*x^10 - 869648*a^11*b^9*x^12 - 2509936*a^10*b^10*x^14 - 5788640*a^9*b^11*x^16 -
10726144*a^8*b^12*x^18 - 15961088*a^7*b^13*x^20 - 18952960*a^6*b^14*x^22 - 17724928*a^5*b^15*x^24 - 12769280*
a^4*b^16*x^26 - 6836224*a^3*b^17*x^28 - 2560000*a^2*b^18*x^30 - 598016*a*b^19*x^32 - 65536*b^20*x^34) + Sqrt[b
^2]*(2*a^18*b^2 + 63*a^17*b^3*x^2 + 939*a^16*b^4*x^4 + 8803*a^15*b^5*x^6 + 58191*a^14*b^6*x^8 + 288114*a^13*b^
7*x^10 + 1107496*a^12*b^8*x^12 + 3379584*a^11*b^9*x^14 + 8298576*a^10*b^10*x^16 + 16514784*a^9*b^11*x^18 + 266
87232*a^8*b^12*x^20 + 34914048*a^7*b^13*x^22 + 36677888*a^6*b^14*x^24 + 30494208*a^5*b^15*x^26 + 19605504*a^4*
b^16*x^28 + 9396224*a^3*b^17*x^30 + 3158016*a^2*b^18*x^32 + 663552*a*b^19*x^34 + 65536*b^20*x^36))/(3*Sqrt[b^2
]*x^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-4*a^14*b^2 - 104*a^13*b^3*x^2 - 1252*a^12*b^4*x^4 - 9248*a^11*b^5*x^6
- 46816*a^10*b^6*x^8 - 171776*a^9*b^7*x^10 - 470976*a^8*b^8*x^12 - 979968*a^7*b^9*x^14 - 1554432*a^6*b^10*x^16
- 1869824*a^5*b^11*x^18 - 1678336*a^4*b^12*x^20 - 1089536*a^3*b^13*x^22 - 483328*a^2*b^14*x^24 - 131072*a*b^1
5*x^26 - 16384*b^16*x^28) + 3*x^6*(4*a^15*b^3 + 108*a^14*b^4*x^2 + 1356*a^13*b^5*x^4 + 10500*a^12*b^6*x^6 + 56
064*a^11*b^7*x^8 + 218592*a^10*b^8*x^10 + 642752*a^9*b^9*x^12 + 1450944*a^8*b^10*x^14 + 2534400*a^7*b^11*x^16
+ 3424256*a^6*b^12*x^18 + 3548160*a^5*b^13*x^20 + 2767872*a^4*b^14*x^22 + 1572864*a^3*b^15*x^24 + 614400*a^2*b
^16*x^26 + 147456*a*b^17*x^28 + 16384*b^18*x^30)) + (b^3*ArcTanh[(Sqrt[b^2]*x^2)/a - Sqrt[a^2 + 2*a*b*x^2 + b^
2*x^4]/a])/2 - ((b^2)^(3/2)*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/4 - ((b^2)^(3/2)*Log[a
- Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/4

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fricas [A]  time = 0.63, size = 39, normalized size = 0.24 \begin {gather*} \frac {12 \, b^{3} x^{6} \log \relax (x) - 18 \, a b^{2} x^{4} - 9 \, a^{2} b x^{2} - 2 \, a^{3}}{12 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*b^3*x^6*log(x) - 18*a*b^2*x^4 - 9*a^2*b*x^2 - 2*a^3)/x^6

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giac [A]  time = 0.19, size = 87, normalized size = 0.53 \begin {gather*} \frac {1}{2} \, b^{3} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {11 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 18 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 9 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{12 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^3*log(x^2)*sgn(b*x^2 + a) - 1/12*(11*b^3*x^6*sgn(b*x^2 + a) + 18*a*b^2*x^4*sgn(b*x^2 + a) + 9*a^2*b*x^2*
sgn(b*x^2 + a) + 2*a^3*sgn(b*x^2 + a))/x^6

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maple [A]  time = 0.01, size = 60, normalized size = 0.37 \begin {gather*} \frac {\left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} \left (12 b^{3} x^{6} \ln \relax (x )-18 a \,b^{2} x^{4}-9 a^{2} b \,x^{2}-2 a^{3}\right )}{12 \left (b \,x^{2}+a \right )^{3} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^7,x)

[Out]

1/12*((b*x^2+a)^2)^(3/2)*(12*b^3*ln(x)*x^6-18*a*b^2*x^4-9*a^2*b*x^2-2*a^3)/(b*x^2+a)^3/x^6

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maxima [A]  time = 1.26, size = 33, normalized size = 0.20 \begin {gather*} b^{3} \log \relax (x) - \frac {3 \, a b^{2}}{2 \, x^{2}} - \frac {3 \, a^{2} b}{4 \, x^{4}} - \frac {a^{3}}{6 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^3*log(x) - 3/2*a*b^2/x^2 - 3/4*a^2*b/x^4 - 1/6*a^3/x^6

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^7,x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^7, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{7}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/x**7,x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/x**7, x)

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