3.24.99 \(\int \frac {1}{\sqrt {a+b \sqrt {\frac {c}{x}}} x^4} \, dx\)

Optimal. Leaf size=172 \[ \frac {4 a^5 \sqrt {a+b \sqrt {\frac {c}{x}}}}{b^6 c^3}-\frac {20 a^4 \left (a+b \sqrt {\frac {c}{x}}\right )^{3/2}}{3 b^6 c^3}+\frac {8 a^3 \left (a+b \sqrt {\frac {c}{x}}\right )^{5/2}}{b^6 c^3}-\frac {40 a^2 \left (a+b \sqrt {\frac {c}{x}}\right )^{7/2}}{7 b^6 c^3}-\frac {4 \left (a+b \sqrt {\frac {c}{x}}\right )^{11/2}}{11 b^6 c^3}+\frac {20 a \left (a+b \sqrt {\frac {c}{x}}\right )^{9/2}}{9 b^6 c^3} \]

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Rubi [A]  time = 0.10, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {369, 266, 43} \begin {gather*} -\frac {40 a^2 \left (a+b \sqrt {\frac {c}{x}}\right )^{7/2}}{7 b^6 c^3}+\frac {8 a^3 \left (a+b \sqrt {\frac {c}{x}}\right )^{5/2}}{b^6 c^3}-\frac {20 a^4 \left (a+b \sqrt {\frac {c}{x}}\right )^{3/2}}{3 b^6 c^3}+\frac {4 a^5 \sqrt {a+b \sqrt {\frac {c}{x}}}}{b^6 c^3}-\frac {4 \left (a+b \sqrt {\frac {c}{x}}\right )^{11/2}}{11 b^6 c^3}+\frac {20 a \left (a+b \sqrt {\frac {c}{x}}\right )^{9/2}}{9 b^6 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b*Sqrt[c/x]]*x^4),x]

[Out]

(4*a^5*Sqrt[a + b*Sqrt[c/x]])/(b^6*c^3) - (20*a^4*(a + b*Sqrt[c/x])^(3/2))/(3*b^6*c^3) + (8*a^3*(a + b*Sqrt[c/
x])^(5/2))/(b^6*c^3) - (40*a^2*(a + b*Sqrt[c/x])^(7/2))/(7*b^6*c^3) + (20*a*(a + b*Sqrt[c/x])^(9/2))/(9*b^6*c^
3) - (4*(a + b*Sqrt[c/x])^(11/2))/(11*b^6*c^3)

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 369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Su
bst[Int[(d*x)^m*(a + b*c^n*x^(n*q))^p, x], x^(1/k), (c*x^q)^(1/k)/(c^(1/k)*(x^(1/k))^(q - 1))]] /; FreeQ[{a, b
, c, d, m, p, q}, x] && FractionQ[n]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a+b \sqrt {\frac {c}{x}}} x^4} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b \sqrt {c}}{\sqrt {x}}} x^4} \, dx,\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=-\operatorname {Subst}\left (2 \operatorname {Subst}\left (\int \frac {x^5}{\sqrt {a+b \sqrt {c} x}} \, dx,x,\frac {1}{\sqrt {x}}\right ),\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=-\operatorname {Subst}\left (2 \operatorname {Subst}\left (\int \left (-\frac {a^5}{b^5 c^{5/2} \sqrt {a+b \sqrt {c} x}}+\frac {5 a^4 \sqrt {a+b \sqrt {c} x}}{b^5 c^{5/2}}-\frac {10 a^3 \left (a+b \sqrt {c} x\right )^{3/2}}{b^5 c^{5/2}}+\frac {10 a^2 \left (a+b \sqrt {c} x\right )^{5/2}}{b^5 c^{5/2}}-\frac {5 a \left (a+b \sqrt {c} x\right )^{7/2}}{b^5 c^{5/2}}+\frac {\left (a+b \sqrt {c} x\right )^{9/2}}{b^5 c^{5/2}}\right ) \, dx,x,\frac {1}{\sqrt {x}}\right ),\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=\frac {4 a^5 \sqrt {a+b \sqrt {\frac {c}{x}}}}{b^6 c^3}-\frac {20 a^4 \left (a+b \sqrt {\frac {c}{x}}\right )^{3/2}}{3 b^6 c^3}+\frac {8 a^3 \left (a+b \sqrt {\frac {c}{x}}\right )^{5/2}}{b^6 c^3}-\frac {40 a^2 \left (a+b \sqrt {\frac {c}{x}}\right )^{7/2}}{7 b^6 c^3}+\frac {20 a \left (a+b \sqrt {\frac {c}{x}}\right )^{9/2}}{9 b^6 c^3}-\frac {4 \left (a+b \sqrt {\frac {c}{x}}\right )^{11/2}}{11 b^6 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 111, normalized size = 0.65 \begin {gather*} \frac {4 \sqrt {a+b \sqrt {\frac {c}{x}}} \left (256 a^5 x^2-128 a^4 b x^2 \sqrt {\frac {c}{x}}+96 a^3 b^2 c x-80 a^2 b^3 c x \sqrt {\frac {c}{x}}+70 a b^4 c^2-63 b^5 c x \left (\frac {c}{x}\right )^{3/2}\right )}{693 b^6 c^3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b*Sqrt[c/x]]*x^4),x]

[Out]

(4*Sqrt[a + b*Sqrt[c/x]]*(70*a*b^4*c^2 + 96*a^3*b^2*c*x - 80*a^2*b^3*c*Sqrt[c/x]*x - 63*b^5*c*(c/x)^(3/2)*x +
256*a^5*x^2 - 128*a^4*b*Sqrt[c/x]*x^2))/(693*b^6*c^3*x^2)

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IntegrateAlgebraic [A]  time = 0.05, size = 103, normalized size = 0.60 \begin {gather*} \frac {4 \sqrt {a+b \sqrt {\frac {c}{x}}} \left (256 a^5-128 a^4 b \sqrt {\frac {c}{x}}+\frac {96 a^3 b^2 c}{x}-80 a^2 b^3 \left (\frac {c}{x}\right )^{3/2}+\frac {70 a b^4 c^2}{x^2}-63 b^5 \left (\frac {c}{x}\right )^{5/2}\right )}{693 b^6 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(Sqrt[a + b*Sqrt[c/x]]*x^4),x]

[Out]

(4*Sqrt[a + b*Sqrt[c/x]]*(256*a^5 - 128*a^4*b*Sqrt[c/x] - 80*a^2*b^3*(c/x)^(3/2) - 63*b^5*(c/x)^(5/2) + (70*a*
b^4*c^2)/x^2 + (96*a^3*b^2*c)/x))/(693*b^6*c^3)

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fricas [A]  time = 1.32, size = 89, normalized size = 0.52 \begin {gather*} \frac {4 \, {\left (70 \, a b^{4} c^{2} + 96 \, a^{3} b^{2} c x + 256 \, a^{5} x^{2} - {\left (63 \, b^{5} c^{2} + 80 \, a^{2} b^{3} c x + 128 \, a^{4} b x^{2}\right )} \sqrt {\frac {c}{x}}\right )} \sqrt {b \sqrt {\frac {c}{x}} + a}}{693 \, b^{6} c^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*(c/x)^(1/2))^(1/2),x, algorithm="fricas")

[Out]

4/693*(70*a*b^4*c^2 + 96*a^3*b^2*c*x + 256*a^5*x^2 - (63*b^5*c^2 + 80*a^2*b^3*c*x + 128*a^4*b*x^2)*sqrt(c/x))*
sqrt(b*sqrt(c/x) + a)/(b^6*c^3*x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b \sqrt {\frac {c}{x}} + a} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*(c/x)^(1/2))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(b*sqrt(c/x) + a)*x^4), x)

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maple [C]  time = 0.06, size = 400, normalized size = 2.33 \begin {gather*} -\frac {\sqrt {a +\sqrt {\frac {c}{x}}\, b}\, \left (-693 \sqrt {\frac {c}{x}}\, a^{6} b \,x^{4} \ln \left (\frac {2 a \sqrt {x}+\sqrt {\frac {c}{x}}\, b \sqrt {x}+2 \sqrt {\left (a +\sqrt {\frac {c}{x}}\, b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )+693 \sqrt {\frac {c}{x}}\, a^{6} b \,x^{4} \ln \left (\frac {2 a \sqrt {x}+\sqrt {\frac {c}{x}}\, b \sqrt {x}+2 \sqrt {a x +\sqrt {\frac {c}{x}}\, b x}\, \sqrt {a}}{2 \sqrt {a}}\right )+1386 \sqrt {a x +\sqrt {\frac {c}{x}}\, b x}\, a^{\frac {13}{2}} x^{\frac {7}{2}}+1386 \sqrt {\left (a +\sqrt {\frac {c}{x}}\, b \right ) x}\, a^{\frac {13}{2}} x^{\frac {7}{2}}-2772 \left (a x +\sqrt {\frac {c}{x}}\, b x \right )^{\frac {3}{2}} a^{\frac {11}{2}} x^{\frac {5}{2}}+1748 \left (a x +\sqrt {\frac {c}{x}}\, b x \right )^{\frac {3}{2}} \sqrt {\frac {c}{x}}\, a^{\frac {9}{2}} b \,x^{\frac {5}{2}}-1236 \left (a x +\sqrt {\frac {c}{x}}\, b x \right )^{\frac {3}{2}} a^{\frac {7}{2}} b^{2} c \,x^{\frac {3}{2}}+852 \left (a x +\sqrt {\frac {c}{x}}\, b x \right )^{\frac {3}{2}} \left (\frac {c}{x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{3} x^{\frac {5}{2}}-532 \left (a x +\sqrt {\frac {c}{x}}\, b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{4} c^{2} \sqrt {x}+252 \left (a x +\sqrt {\frac {c}{x}}\, b x \right )^{\frac {3}{2}} \left (\frac {c}{x}\right )^{\frac {5}{2}} \sqrt {a}\, b^{5} x^{\frac {5}{2}}\right )}{693 \sqrt {\left (a +\sqrt {\frac {c}{x}}\, b \right ) x}\, \left (\frac {c}{x}\right )^{\frac {7}{2}} \sqrt {a}\, b^{7} x^{\frac {13}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(a+(c/x)^(1/2)*b)^(1/2),x)

[Out]

-1/693*(a+(c/x)^(1/2)*b)^(1/2)*(1386*x^(7/2)*(a*x+(c/x)^(1/2)*b*x)^(1/2)*a^(13/2)+1386*x^(7/2)*((a+(c/x)^(1/2)
*b)*x)^(1/2)*a^(13/2)+252*x^(5/2)*(a*x+(c/x)^(1/2)*b*x)^(3/2)*(c/x)^(5/2)*a^(1/2)*b^5+852*x^(5/2)*(a*x+(c/x)^(
1/2)*b*x)^(3/2)*(c/x)^(3/2)*a^(5/2)*b^3+1748*x^(5/2)*(a*x+(c/x)^(1/2)*b*x)^(3/2)*(c/x)^(1/2)*a^(9/2)*b-2772*x^
(5/2)*(a*x+(c/x)^(1/2)*b*x)^(3/2)*a^(11/2)-1236*x^(3/2)*(a*x+(c/x)^(1/2)*b*x)^(3/2)*a^(7/2)*b^2*c-532*x^(1/2)*
(a*x+(c/x)^(1/2)*b*x)^(3/2)*a^(3/2)*b^4*c^2-693*ln(1/2*(2*a*x^(1/2)+(c/x)^(1/2)*b*x^(1/2)+2*((a+(c/x)^(1/2)*b)
*x)^(1/2)*a^(1/2))/a^(1/2))*(c/x)^(1/2)*x^4*a^6*b+693*ln(1/2*(2*a*x^(1/2)+(c/x)^(1/2)*b*x^(1/2)+2*(a*x+(c/x)^(
1/2)*b*x)^(1/2)*a^(1/2))/a^(1/2))*(c/x)^(1/2)*x^4*a^6*b)/x^(13/2)/((a+(c/x)^(1/2)*b)*x)^(1/2)/b^7/(c/x)^(7/2)/
a^(1/2)

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maxima [A]  time = 0.48, size = 127, normalized size = 0.74 \begin {gather*} -\frac {4 \, {\left (\frac {63 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {11}{2}}}{b^{6}} - \frac {385 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {9}{2}} a}{b^{6}} + \frac {990 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {7}{2}} a^{2}}{b^{6}} - \frac {1386 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {5}{2}} a^{3}}{b^{6}} + \frac {1155 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {3}{2}} a^{4}}{b^{6}} - \frac {693 \, \sqrt {b \sqrt {\frac {c}{x}} + a} a^{5}}{b^{6}}\right )}}{693 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*(c/x)^(1/2))^(1/2),x, algorithm="maxima")

[Out]

-4/693*(63*(b*sqrt(c/x) + a)^(11/2)/b^6 - 385*(b*sqrt(c/x) + a)^(9/2)*a/b^6 + 990*(b*sqrt(c/x) + a)^(7/2)*a^2/
b^6 - 1386*(b*sqrt(c/x) + a)^(5/2)*a^3/b^6 + 1155*(b*sqrt(c/x) + a)^(3/2)*a^4/b^6 - 693*sqrt(b*sqrt(c/x) + a)*
a^5/b^6)/c^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^4\,\sqrt {a+b\,\sqrt {\frac {c}{x}}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*(c/x)^(1/2))^(1/2)),x)

[Out]

int(1/(x^4*(a + b*(c/x)^(1/2))^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{4} \sqrt {a + b \sqrt {\frac {c}{x}}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(a+b*(c/x)**(1/2))**(1/2),x)

[Out]

Integral(1/(x**4*sqrt(a + b*sqrt(c/x))), x)

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