3.1.80 \(\int \frac {1}{x^{7/2} \sqrt {b \sqrt {x}+a x}} \, dx\)

Optimal. Leaf size=170 \[ \frac {1024 a^5 \sqrt {a x+b \sqrt {x}}}{693 b^6 \sqrt {x}}-\frac {512 a^4 \sqrt {a x+b \sqrt {x}}}{693 b^5 x}+\frac {128 a^3 \sqrt {a x+b \sqrt {x}}}{231 b^4 x^{3/2}}-\frac {320 a^2 \sqrt {a x+b \sqrt {x}}}{693 b^3 x^2}+\frac {40 a \sqrt {a x+b \sqrt {x}}}{99 b^2 x^{5/2}}-\frac {4 \sqrt {a x+b \sqrt {x}}}{11 b x^3} \]

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Rubi [A]  time = 0.24, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2016, 2014} \begin {gather*} \frac {128 a^3 \sqrt {a x+b \sqrt {x}}}{231 b^4 x^{3/2}}-\frac {320 a^2 \sqrt {a x+b \sqrt {x}}}{693 b^3 x^2}+\frac {1024 a^5 \sqrt {a x+b \sqrt {x}}}{693 b^6 \sqrt {x}}-\frac {512 a^4 \sqrt {a x+b \sqrt {x}}}{693 b^5 x}+\frac {40 a \sqrt {a x+b \sqrt {x}}}{99 b^2 x^{5/2}}-\frac {4 \sqrt {a x+b \sqrt {x}}}{11 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*Sqrt[b*Sqrt[x] + a*x]),x]

[Out]

(-4*Sqrt[b*Sqrt[x] + a*x])/(11*b*x^3) + (40*a*Sqrt[b*Sqrt[x] + a*x])/(99*b^2*x^(5/2)) - (320*a^2*Sqrt[b*Sqrt[x
] + a*x])/(693*b^3*x^2) + (128*a^3*Sqrt[b*Sqrt[x] + a*x])/(231*b^4*x^(3/2)) - (512*a^4*Sqrt[b*Sqrt[x] + a*x])/
(693*b^5*x) + (1024*a^5*Sqrt[b*Sqrt[x] + a*x])/(693*b^6*Sqrt[x])

Rule 2014

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*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2016

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, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps

\begin {align*} \int \frac {1}{x^{7/2} \sqrt {b \sqrt {x}+a x}} \, dx &=-\frac {4 \sqrt {b \sqrt {x}+a x}}{11 b x^3}-\frac {(10 a) \int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx}{11 b}\\ &=-\frac {4 \sqrt {b \sqrt {x}+a x}}{11 b x^3}+\frac {40 a \sqrt {b \sqrt {x}+a x}}{99 b^2 x^{5/2}}+\frac {\left (80 a^2\right ) \int \frac {1}{x^{5/2} \sqrt {b \sqrt {x}+a x}} \, dx}{99 b^2}\\ &=-\frac {4 \sqrt {b \sqrt {x}+a x}}{11 b x^3}+\frac {40 a \sqrt {b \sqrt {x}+a x}}{99 b^2 x^{5/2}}-\frac {320 a^2 \sqrt {b \sqrt {x}+a x}}{693 b^3 x^2}-\frac {\left (160 a^3\right ) \int \frac {1}{x^2 \sqrt {b \sqrt {x}+a x}} \, dx}{231 b^3}\\ &=-\frac {4 \sqrt {b \sqrt {x}+a x}}{11 b x^3}+\frac {40 a \sqrt {b \sqrt {x}+a x}}{99 b^2 x^{5/2}}-\frac {320 a^2 \sqrt {b \sqrt {x}+a x}}{693 b^3 x^2}+\frac {128 a^3 \sqrt {b \sqrt {x}+a x}}{231 b^4 x^{3/2}}+\frac {\left (128 a^4\right ) \int \frac {1}{x^{3/2} \sqrt {b \sqrt {x}+a x}} \, dx}{231 b^4}\\ &=-\frac {4 \sqrt {b \sqrt {x}+a x}}{11 b x^3}+\frac {40 a \sqrt {b \sqrt {x}+a x}}{99 b^2 x^{5/2}}-\frac {320 a^2 \sqrt {b \sqrt {x}+a x}}{693 b^3 x^2}+\frac {128 a^3 \sqrt {b \sqrt {x}+a x}}{231 b^4 x^{3/2}}-\frac {512 a^4 \sqrt {b \sqrt {x}+a x}}{693 b^5 x}-\frac {\left (256 a^5\right ) \int \frac {1}{x \sqrt {b \sqrt {x}+a x}} \, dx}{693 b^5}\\ &=-\frac {4 \sqrt {b \sqrt {x}+a x}}{11 b x^3}+\frac {40 a \sqrt {b \sqrt {x}+a x}}{99 b^2 x^{5/2}}-\frac {320 a^2 \sqrt {b \sqrt {x}+a x}}{693 b^3 x^2}+\frac {128 a^3 \sqrt {b \sqrt {x}+a x}}{231 b^4 x^{3/2}}-\frac {512 a^4 \sqrt {b \sqrt {x}+a x}}{693 b^5 x}+\frac {1024 a^5 \sqrt {b \sqrt {x}+a x}}{693 b^6 \sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 83, normalized size = 0.49 \begin {gather*} \frac {4 \sqrt {a x+b \sqrt {x}} \left (256 a^5 x^{5/2}-128 a^4 b x^2+96 a^3 b^2 x^{3/2}-80 a^2 b^3 x+70 a b^4 \sqrt {x}-63 b^5\right )}{693 b^6 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*Sqrt[b*Sqrt[x] + a*x]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.20, size = 83, normalized size = 0.49 \begin {gather*} \frac {4 \sqrt {a x+b \sqrt {x}} \left (256 a^5 x^{5/2}-128 a^4 b x^2+96 a^3 b^2 x^{3/2}-80 a^2 b^3 x+70 a b^4 \sqrt {x}-63 b^5\right )}{693 b^6 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^(7/2)*Sqrt[b*Sqrt[x] + a*x]),x]

[Out]

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

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fricas [A]  time = 0.71, size = 72, normalized size = 0.42 \begin {gather*} -\frac {4 \, {\left (128 \, a^{4} b x^{2} + 80 \, a^{2} b^{3} x + 63 \, b^{5} - 2 \, {\left (128 \, a^{5} x^{2} + 48 \, a^{3} b^{2} x + 35 \, a b^{4}\right )} \sqrt {x}\right )} \sqrt {a x + b \sqrt {x}}}{693 \, b^{6} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x^(1/2)+a*x)^(1/2),x, algorithm="fricas")

[Out]

-4/693*(128*a^4*b*x^2 + 80*a^2*b^3*x + 63*b^5 - 2*(128*a^5*x^2 + 48*a^3*b^2*x + 35*a*b^4)*sqrt(x))*sqrt(a*x +
b*sqrt(x))/(b^6*x^3)

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giac [A]  time = 0.24, size = 177, normalized size = 1.04 \begin {gather*} \frac {4 \, {\left (3696 \, a^{\frac {5}{2}} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{5} + 7920 \, a^{2} b {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{4} + 6930 \, a^{\frac {3}{2}} b^{2} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{3} + 3080 \, a b^{3} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{2} + 693 \, \sqrt {a} b^{4} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} + 63 \, b^{5}\right )}}{693 \, {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{11}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x^(1/2)+a*x)^(1/2),x, algorithm="giac")

[Out]

4/693*(3696*a^(5/2)*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^5 + 7920*a^2*b*(sqrt(a)*sqrt(x) - sqrt(a*x + b*s
qrt(x)))^4 + 6930*a^(3/2)*b^2*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^3 + 3080*a*b^3*(sqrt(a)*sqrt(x) - sqrt
(a*x + b*sqrt(x)))^2 + 693*sqrt(a)*b^4*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))) + 63*b^5)/(sqrt(a)*sqrt(x) -
sqrt(a*x + b*sqrt(x)))^11

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maple [C]  time = 0.06, size = 284, normalized size = 1.67 \begin {gather*} \frac {\sqrt {a x +b \sqrt {x}}\, \left (693 a^{6} b \,x^{\frac {13}{2}} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-693 a^{6} b \,x^{\frac {13}{2}} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-1386 \sqrt {a x +b \sqrt {x}}\, a^{\frac {13}{2}} x^{\frac {13}{2}}-1386 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {13}{2}} x^{\frac {13}{2}}+2772 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {11}{2}} x^{\frac {11}{2}}-1748 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {9}{2}} b \,x^{5}+1236 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {7}{2}} b^{2} x^{\frac {9}{2}}-852 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{3} x^{4}+532 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{4} x^{\frac {7}{2}}-252 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} \sqrt {a}\, b^{5} x^{3}\right )}{693 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}\, b^{7} x^{\frac {13}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(7/2)/(a*x+b*x^(1/2))^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a x + b \sqrt {x}} x^{\frac {7}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x^(1/2)+a*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(a*x + b*sqrt(x))*x^(7/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*(a*x + b*x^(1/2))^(1/2)),x)

[Out]

int(1/(x^(7/2)*(a*x + b*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^{\frac {7}{2}} \sqrt {a x + b \sqrt {x}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(7/2)/(b*x**(1/2)+a*x)**(1/2),x)

[Out]

Integral(1/(x**(7/2)*sqrt(a*x + b*sqrt(x))), x)

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