3.3.49 \(\int \frac {(a x^2+b x^3)^{3/2}}{x^7} \, dx\) [249]

Optimal. Leaf size=109 \[ -\frac {b \sqrt {a x^2+b x^3}}{4 x^3}-\frac {b^2 \sqrt {a x^2+b x^3}}{8 a x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}+\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{3/2}} \]

[Out]

-1/3*(b*x^3+a*x^2)^(3/2)/x^6+1/8*b^3*arctanh(x*a^(1/2)/(b*x^3+a*x^2)^(1/2))/a^(3/2)-1/4*b*(b*x^3+a*x^2)^(1/2)/
x^3-1/8*b^2*(b*x^3+a*x^2)^(1/2)/a/x^2

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2045, 2050, 2033, 212} \begin {gather*} \frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{3/2}}-\frac {b^2 \sqrt {a x^2+b x^3}}{8 a x^2}-\frac {b \sqrt {a x^2+b x^3}}{4 x^3}-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x^7,x]

[Out]

-1/4*(b*Sqrt[a*x^2 + b*x^3])/x^3 - (b^2*Sqrt[a*x^2 + b*x^3])/(8*a*x^2) - (a*x^2 + b*x^3)^(3/2)/(3*x^6) + (b^3*
ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(8*a^(3/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2033

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 2045

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 2050

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 (a x^2+b x^3\right )^{3/2}}{x^7} \, dx &=-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}+\frac {1}{2} b \int \frac {\sqrt {a x^2+b x^3}}{x^4} \, dx\\ &=-\frac {b \sqrt {a x^2+b x^3}}{4 x^3}-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}+\frac {1}{8} b^2 \int \frac {1}{x \sqrt {a x^2+b x^3}} \, dx\\ &=-\frac {b \sqrt {a x^2+b x^3}}{4 x^3}-\frac {b^2 \sqrt {a x^2+b x^3}}{8 a x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}-\frac {b^3 \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx}{16 a}\\ &=-\frac {b \sqrt {a x^2+b x^3}}{4 x^3}-\frac {b^2 \sqrt {a x^2+b x^3}}{8 a x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}+\frac {b^3 \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {x}{\sqrt {a x^2+b x^3}}\right )}{8 a}\\ &=-\frac {b \sqrt {a x^2+b x^3}}{4 x^3}-\frac {b^2 \sqrt {a x^2+b x^3}}{8 a x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}+\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.12, size = 94, normalized size = 0.86 \begin {gather*} \frac {\sqrt {x^2 (a+b x)} \left (-\sqrt {a} \sqrt {a+b x} \left (8 a^2+14 a b x+3 b^2 x^2\right )+3 b^3 x^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{24 a^{3/2} x^4 \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(a + b*x)]*(-(Sqrt[a]*Sqrt[a + b*x]*(8*a^2 + 14*a*b*x + 3*b^2*x^2)) + 3*b^3*x^3*ArcTanh[Sqrt[a + b*x
]/Sqrt[a]]))/(24*a^(3/2)*x^4*Sqrt[a + b*x])

________________________________________________________________________________________

Maple [A]
time = 0.37, size = 87, normalized size = 0.80

method	result	size
risch	\(-\frac {\left (3 b^{2} x^{2}+14 a b x +8 a^{2}\right ) \sqrt {x^{2} \left (b x +a \right )}}{24 x^{4} a}+\frac {b^{3} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{8 a^{\frac {3}{2}} x \sqrt {b x +a}}\)	\(81\)
default	\(-\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (3 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {3}{2}}-3 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{3} x^{3}+8 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {5}{2}}-3 \sqrt {b x +a}\, a^{\frac {7}{2}}\right )}{24 x^{6} \left (b x +a \right )^{\frac {3}{2}} a^{\frac {5}{2}}}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a*x^2)^(3/2)/x^7,x,method=_RETURNVERBOSE)

[Out]

-1/24*(b*x^3+a*x^2)^(3/2)*(3*(b*x+a)^(5/2)*a^(3/2)-3*arctanh((b*x+a)^(1/2)/a^(1/2))*a*b^3*x^3+8*(b*x+a)^(3/2)*
a^(5/2)-3*(b*x+a)^(1/2)*a^(7/2))/x^6/(b*x+a)^(3/2)/a^(5/2)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 1.81, size = 175, normalized size = 1.61 \begin {gather*} \left [\frac {3 \, \sqrt {a} b^{3} x^{4} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, {\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, a^{2} x^{4}}, -\frac {3 \, \sqrt {-a} b^{3} x^{4} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + {\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, a^{2} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(3*sqrt(a)*b^3*x^4*log((b*x^2 + 2*a*x + 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) - 2*(3*a*b^2*x^2 + 14*a^2*b*
x + 8*a^3)*sqrt(b*x^3 + a*x^2))/(a^2*x^4), -1/24*(3*sqrt(-a)*b^3*x^4*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(a*x)
) + (3*a*b^2*x^2 + 14*a^2*b*x + 8*a^3)*sqrt(b*x^3 + a*x^2))/(a^2*x^4)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 1.56, size = 92, normalized size = 0.84 \begin {gather*} -\frac {\frac {3 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a} a} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} \mathrm {sgn}\left (x\right ) + 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {b x + a} a^{2} b^{4} \mathrm {sgn}\left (x\right )}{a b^{3} x^{3}}}{24 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*b^4*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/(sqrt(-a)*a) + (3*(b*x + a)^(5/2)*b^4*sgn(x) + 8*(b*x + a)^
(3/2)*a*b^4*sgn(x) - 3*sqrt(b*x + a)*a^2*b^4*sgn(x))/(a*b^3*x^3))/b

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3)^(3/2)/x^7,x)

[Out]

int((a*x^2 + b*x^3)^(3/2)/x^7, x)

________________________________________________________________________________________