3.4.58 \(\int \frac {1}{x^3 (a+b x)^{5/2}} \, dx\) [358]

Optimal. Leaf size=106 \[ \frac {35 b^2}{12 a^3 (a+b x)^{3/2}}-\frac {1}{2 a x^2 (a+b x)^{3/2}}+\frac {7 b}{4 a^2 x (a+b x)^{3/2}}+\frac {35 b^2}{4 a^4 \sqrt {a+b x}}-\frac {35 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{9/2}} \]

[Out]

35/12*b^2/a^3/(b*x+a)^(3/2)-1/2/a/x^2/(b*x+a)^(3/2)+7/4*b/a^2/x/(b*x+a)^(3/2)-35/4*b^2*arctanh((b*x+a)^(1/2)/a
^(1/2))/a^(9/2)+35/4*b^2/a^4/(b*x+a)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {44, 53, 65, 214} \begin {gather*} -\frac {35 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{9/2}}+\frac {35 b^2}{4 a^4 \sqrt {a+b x}}+\frac {35 b^2}{12 a^3 (a+b x)^{3/2}}+\frac {7 b}{4 a^2 x (a+b x)^{3/2}}-\frac {1}{2 a x^2 (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x)^(5/2)),x]

[Out]

(35*b^2)/(12*a^3*(a + b*x)^(3/2)) - 1/(2*a*x^2*(a + b*x)^(3/2)) + (7*b)/(4*a^2*x*(a + b*x)^(3/2)) + (35*b^2)/(
4*a^4*Sqrt[a + b*x]) - (35*b^2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(4*a^(9/2))

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rubi steps

\begin {align*} \int \frac {1}{x^3 (a+b x)^{5/2}} \, dx &=\frac {2}{3 a x^2 (a+b x)^{3/2}}+\frac {7 \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx}{3 a}\\ &=\frac {2}{3 a x^2 (a+b x)^{3/2}}+\frac {14}{3 a^2 x^2 \sqrt {a+b x}}+\frac {35 \int \frac {1}{x^3 \sqrt {a+b x}} \, dx}{3 a^2}\\ &=\frac {2}{3 a x^2 (a+b x)^{3/2}}+\frac {14}{3 a^2 x^2 \sqrt {a+b x}}-\frac {35 \sqrt {a+b x}}{6 a^3 x^2}-\frac {(35 b) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{4 a^3}\\ &=\frac {2}{3 a x^2 (a+b x)^{3/2}}+\frac {14}{3 a^2 x^2 \sqrt {a+b x}}-\frac {35 \sqrt {a+b x}}{6 a^3 x^2}+\frac {35 b \sqrt {a+b x}}{4 a^4 x}+\frac {\left (35 b^2\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{8 a^4}\\ &=\frac {2}{3 a x^2 (a+b x)^{3/2}}+\frac {14}{3 a^2 x^2 \sqrt {a+b x}}-\frac {35 \sqrt {a+b x}}{6 a^3 x^2}+\frac {35 b \sqrt {a+b x}}{4 a^4 x}+\frac {(35 b) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{4 a^4}\\ &=\frac {2}{3 a x^2 (a+b x)^{3/2}}+\frac {14}{3 a^2 x^2 \sqrt {a+b x}}-\frac {35 \sqrt {a+b x}}{6 a^3 x^2}+\frac {35 b \sqrt {a+b x}}{4 a^4 x}-\frac {35 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 78, normalized size = 0.74 \begin {gather*} \frac {-6 a^3+21 a^2 b x+140 a b^2 x^2+105 b^3 x^3}{12 a^4 x^2 (a+b x)^{3/2}}-\frac {35 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x)^(5/2)),x]

[Out]

(-6*a^3 + 21*a^2*b*x + 140*a*b^2*x^2 + 105*b^3*x^3)/(12*a^4*x^2*(a + b*x)^(3/2)) - (35*b^2*ArcTanh[Sqrt[a + b*
x]/Sqrt[a]])/(4*a^(9/2))

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Maple [A]
time = 0.11, size = 81, normalized size = 0.76

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (-11 b x +2 a \right )}{4 a^{4} x^{2}}+\frac {b^{2} \left (-\frac {70 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {48}{\sqrt {b x +a}}+\frac {16 a}{3 \left (b x +a \right )^{\frac {3}{2}}}\right )}{8 a^{4}}\)	\(70\)
derivativedivides	\(2 b^{2} \left (-\frac {\frac {-\frac {11 \left (b x +a \right )^{\frac {3}{2}}}{8}+\frac {13 a \sqrt {b x +a}}{8}}{b^{2} x^{2}}+\frac {35 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{4}}+\frac {3}{a^{4} \sqrt {b x +a}}+\frac {1}{3 a^{3} \left (b x +a \right )^{\frac {3}{2}}}\right )\)	\(81\)
default	\(2 b^{2} \left (-\frac {\frac {-\frac {11 \left (b x +a \right )^{\frac {3}{2}}}{8}+\frac {13 a \sqrt {b x +a}}{8}}{b^{2} x^{2}}+\frac {35 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{4}}+\frac {3}{a^{4} \sqrt {b x +a}}+\frac {1}{3 a^{3} \left (b x +a \right )^{\frac {3}{2}}}\right )\)	\(81\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2*b^2*(-1/a^4*((-11/8*(b*x+a)^(3/2)+13/8*a*(b*x+a)^(1/2))/b^2/x^2+35/8*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(1/2))
+3/a^4/(b*x+a)^(1/2)+1/3/a^3/(b*x+a)^(3/2))

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Maxima [A]
time = 0.48, size = 123, normalized size = 1.16 \begin {gather*} \frac {105 \, {\left (b x + a\right )}^{3} b^{2} - 175 \, {\left (b x + a\right )}^{2} a b^{2} + 56 \, {\left (b x + a\right )} a^{2} b^{2} + 8 \, a^{3} b^{2}}{12 \, {\left ({\left (b x + a\right )}^{\frac {7}{2}} a^{4} - 2 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{5} + {\left (b x + a\right )}^{\frac {3}{2}} a^{6}\right )}} + \frac {35 \, b^{2} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{8 \, a^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

1/12*(105*(b*x + a)^3*b^2 - 175*(b*x + a)^2*a*b^2 + 56*(b*x + a)*a^2*b^2 + 8*a^3*b^2)/((b*x + a)^(7/2)*a^4 - 2
*(b*x + a)^(5/2)*a^5 + (b*x + a)^(3/2)*a^6) + 35/8*b^2*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a))
)/a^(9/2)

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Fricas [A]
time = 0.49, size = 255, normalized size = 2.41 \begin {gather*} \left [\frac {105 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (105 \, a b^{3} x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a^{3} b x - 6 \, a^{4}\right )} \sqrt {b x + a}}{24 \, {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}}, \frac {105 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (105 \, a b^{3} x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a^{3} b x - 6 \, a^{4}\right )} \sqrt {b x + a}}{12 \, {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[1/24*(105*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*sqrt(a)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(105
*a*b^3*x^3 + 140*a^2*b^2*x^2 + 21*a^3*b*x - 6*a^4)*sqrt(b*x + a))/(a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2), 1/12*
(105*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (105*a*b^3*x^3 + 140*a^
2*b^2*x^2 + 21*a^3*b*x - 6*a^4)*sqrt(b*x + a))/(a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 464 vs. \(2 (99) = 198\).
time = 10.56, size = 464, normalized size = 4.38 \begin {gather*} - \frac {6 a^{\frac {89}{2}} b^{75} x^{75}}{12 a^{\frac {93}{2}} b^{\frac {151}{2}} x^{\frac {155}{2}} \sqrt {\frac {a}{b x} + 1} + 12 a^{\frac {91}{2}} b^{\frac {153}{2}} x^{\frac {157}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {21 a^{\frac {87}{2}} b^{76} x^{76}}{12 a^{\frac {93}{2}} b^{\frac {151}{2}} x^{\frac {155}{2}} \sqrt {\frac {a}{b x} + 1} + 12 a^{\frac {91}{2}} b^{\frac {153}{2}} x^{\frac {157}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {140 a^{\frac {85}{2}} b^{77} x^{77}}{12 a^{\frac {93}{2}} b^{\frac {151}{2}} x^{\frac {155}{2}} \sqrt {\frac {a}{b x} + 1} + 12 a^{\frac {91}{2}} b^{\frac {153}{2}} x^{\frac {157}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {105 a^{\frac {83}{2}} b^{78} x^{78}}{12 a^{\frac {93}{2}} b^{\frac {151}{2}} x^{\frac {155}{2}} \sqrt {\frac {a}{b x} + 1} + 12 a^{\frac {91}{2}} b^{\frac {153}{2}} x^{\frac {157}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {105 a^{42} b^{\frac {155}{2}} x^{\frac {155}{2}} \sqrt {\frac {a}{b x} + 1} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{12 a^{\frac {93}{2}} b^{\frac {151}{2}} x^{\frac {155}{2}} \sqrt {\frac {a}{b x} + 1} + 12 a^{\frac {91}{2}} b^{\frac {153}{2}} x^{\frac {157}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {105 a^{41} b^{\frac {157}{2}} x^{\frac {157}{2}} \sqrt {\frac {a}{b x} + 1} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{12 a^{\frac {93}{2}} b^{\frac {151}{2}} x^{\frac {155}{2}} \sqrt {\frac {a}{b x} + 1} + 12 a^{\frac {91}{2}} b^{\frac {153}{2}} x^{\frac {157}{2}} \sqrt {\frac {a}{b x} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x+a)**(5/2),x)

[Out]

-6*a**(89/2)*b**75*x**75/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) + 1) + 12*a**(91/2)*b**(153/2)*x**(1
57/2)*sqrt(a/(b*x) + 1)) + 21*a**(87/2)*b**76*x**76/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) + 1) + 12
*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1)) + 140*a**(85/2)*b**77*x**77/(12*a**(93/2)*b**(151/2)*x**(1
55/2)*sqrt(a/(b*x) + 1) + 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1)) + 105*a**(83/2)*b**78*x**78/(1
2*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) + 1) + 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1)) -
105*a**42*b**(155/2)*x**(155/2)*sqrt(a/(b*x) + 1)*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(12*a**(93/2)*b**(151/2)*x*
*(155/2)*sqrt(a/(b*x) + 1) + 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1)) - 105*a**41*b**(157/2)*x**(
157/2)*sqrt(a/(b*x) + 1)*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) + 1
) + 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1))

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Giac [A]
time = 0.91, size = 93, normalized size = 0.88 \begin {gather*} \frac {35 \, b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{4}} + \frac {2 \, {\left (9 \, {\left (b x + a\right )} b^{2} + a b^{2}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4}} + \frac {11 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 13 \, \sqrt {b x + a} a b^{2}}{4 \, a^{4} b^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

35/4*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^4) + 2/3*(9*(b*x + a)*b^2 + a*b^2)/((b*x + a)^(3/2)*a^4) +
1/4*(11*(b*x + a)^(3/2)*b^2 - 13*sqrt(b*x + a)*a*b^2)/(a^4*b^2*x^2)

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Mupad [B]
time = 0.12, size = 105, normalized size = 0.99 \begin {gather*} \frac {\frac {2\,b^2}{3\,a}-\frac {175\,b^2\,{\left (a+b\,x\right )}^2}{12\,a^3}+\frac {35\,b^2\,{\left (a+b\,x\right )}^3}{4\,a^4}+\frac {14\,b^2\,\left (a+b\,x\right )}{3\,a^2}}{{\left (a+b\,x\right )}^{7/2}-2\,a\,{\left (a+b\,x\right )}^{5/2}+a^2\,{\left (a+b\,x\right )}^{3/2}}-\frac {35\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{4\,a^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x)^(5/2)),x)

[Out]

((2*b^2)/(3*a) - (175*b^2*(a + b*x)^2)/(12*a^3) + (35*b^2*(a + b*x)^3)/(4*a^4) + (14*b^2*(a + b*x))/(3*a^2))/(
(a + b*x)^(7/2) - 2*a*(a + b*x)^(5/2) + a^2*(a + b*x)^(3/2)) - (35*b^2*atanh((a + b*x)^(1/2)/a^(1/2)))/(4*a^(9
/2))

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