3.4.70 \(\int \frac {\sqrt {a+b x} (A+B x)}{x^4} \, dx\)

Optimal. Leaf size=112 \[ -\frac {b^2 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{5/2}}+\frac {b \sqrt {a+b x} (A b-2 a B)}{8 a^2 x}+\frac {\sqrt {a+b x} (A b-2 a B)}{4 a x^2}-\frac {A (a+b x)^{3/2}}{3 a x^3} \]

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Rubi [A]  time = 0.05, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {78, 47, 51, 63, 208} \begin {gather*} -\frac {b^2 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{5/2}}+\frac {b \sqrt {a+b x} (A b-2 a B)}{8 a^2 x}+\frac {\sqrt {a+b x} (A b-2 a B)}{4 a x^2}-\frac {A (a+b x)^{3/2}}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x]*(A + B*x))/x^4,x]

[Out]

((A*b - 2*a*B)*Sqrt[a + b*x])/(4*a*x^2) + (b*(A*b - 2*a*B)*Sqrt[a + b*x])/(8*a^2*x) - (A*(a + b*x)^(3/2))/(3*a
*x^3) - (b^2*(A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(5/2))

Rule 47

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

Rule 51

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 63

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 78

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

Rule 208

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

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x} (A+B x)}{x^4} \, dx &=-\frac {A (a+b x)^{3/2}}{3 a x^3}+\frac {\left (-\frac {3 A b}{2}+3 a B\right ) \int \frac {\sqrt {a+b x}}{x^3} \, dx}{3 a}\\ &=\frac {(A b-2 a B) \sqrt {a+b x}}{4 a x^2}-\frac {A (a+b x)^{3/2}}{3 a x^3}-\frac {(b (A b-2 a B)) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{8 a}\\ &=\frac {(A b-2 a B) \sqrt {a+b x}}{4 a x^2}+\frac {b (A b-2 a B) \sqrt {a+b x}}{8 a^2 x}-\frac {A (a+b x)^{3/2}}{3 a x^3}+\frac {\left (b^2 (A b-2 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{16 a^2}\\ &=\frac {(A b-2 a B) \sqrt {a+b x}}{4 a x^2}+\frac {b (A b-2 a B) \sqrt {a+b x}}{8 a^2 x}-\frac {A (a+b x)^{3/2}}{3 a x^3}+\frac {(b (A b-2 a B)) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{8 a^2}\\ &=\frac {(A b-2 a B) \sqrt {a+b x}}{4 a x^2}+\frac {b (A b-2 a B) \sqrt {a+b x}}{8 a^2 x}-\frac {A (a+b x)^{3/2}}{3 a x^3}-\frac {b^2 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 57, normalized size = 0.51 \begin {gather*} -\frac {(a+b x)^{3/2} \left (a^3 A+b^2 x^3 (2 a B-A b) \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {b x}{a}+1\right )\right )}{3 a^4 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x]*(A + B*x))/x^4,x]

[Out]

-1/3*((a + b*x)^(3/2)*(a^3*A + b^2*(-(A*b) + 2*a*B)*x^3*Hypergeometric2F1[3/2, 3, 5/2, 1 + (b*x)/a]))/(a^4*x^3
)

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IntegrateAlgebraic [A]  time = 0.20, size = 108, normalized size = 0.96 \begin {gather*} \frac {\left (2 a b^2 B-A b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{5/2}}+\frac {\sqrt {a+b x} \left (6 a^3 B-3 a^2 A b-8 a A b (a+b x)+3 A b (a+b x)^2-6 a B (a+b x)^2\right )}{24 a^2 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[a + b*x]*(A + B*x))/x^4,x]

[Out]

(Sqrt[a + b*x]*(-3*a^2*A*b + 6*a^3*B - 8*a*A*b*(a + b*x) + 3*A*b*(a + b*x)^2 - 6*a*B*(a + b*x)^2))/(24*a^2*b*x
^3) + ((-(A*b^3) + 2*a*b^2*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(5/2))

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fricas [A]  time = 1.32, size = 209, normalized size = 1.87 \begin {gather*} \left [-\frac {3 \, {\left (2 \, B a b^{2} - A b^{3}\right )} \sqrt {a} x^{3} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (8 \, A a^{3} + 3 \, {\left (2 \, B a^{2} b - A a b^{2}\right )} x^{2} + 2 \, {\left (6 \, B a^{3} + A a^{2} b\right )} x\right )} \sqrt {b x + a}}{48 \, a^{3} x^{3}}, -\frac {3 \, {\left (2 \, B a b^{2} - A b^{3}\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (8 \, A a^{3} + 3 \, {\left (2 \, B a^{2} b - A a b^{2}\right )} x^{2} + 2 \, {\left (6 \, B a^{3} + A a^{2} b\right )} x\right )} \sqrt {b x + a}}{24 \, a^{3} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(3*(2*B*a*b^2 - A*b^3)*sqrt(a)*x^3*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(8*A*a^3 + 3*(2*B*a
^2*b - A*a*b^2)*x^2 + 2*(6*B*a^3 + A*a^2*b)*x)*sqrt(b*x + a))/(a^3*x^3), -1/24*(3*(2*B*a*b^2 - A*b^3)*sqrt(-a)
*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (8*A*a^3 + 3*(2*B*a^2*b - A*a*b^2)*x^2 + 2*(6*B*a^3 + A*a^2*b)*x)*sqrt
(b*x + a))/(a^3*x^3)]

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giac [A]  time = 1.22, size = 128, normalized size = 1.14 \begin {gather*} -\frac {\frac {3 \, {\left (2 \, B a b^{3} - A b^{4}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {6 \, {\left (b x + a\right )}^{\frac {5}{2}} B a b^{3} - 6 \, \sqrt {b x + a} B a^{3} b^{3} - 3 \, {\left (b x + a\right )}^{\frac {5}{2}} A b^{4} + 8 \, {\left (b x + a\right )}^{\frac {3}{2}} A a b^{4} + 3 \, \sqrt {b x + a} A a^{2} b^{4}}{a^{2} b^{3} x^{3}}}{24 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*(2*B*a*b^3 - A*b^4)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (6*(b*x + a)^(5/2)*B*a*b^3 - 6*sq
rt(b*x + a)*B*a^3*b^3 - 3*(b*x + a)^(5/2)*A*b^4 + 8*(b*x + a)^(3/2)*A*a*b^4 + 3*sqrt(b*x + a)*A*a^2*b^4)/(a^2*
b^3*x^3))/b

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maple [A]  time = 0.02, size = 91, normalized size = 0.81 \begin {gather*} 2 \left (-\frac {\left (A b -2 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 a^{\frac {5}{2}}}+\frac {-\frac {\left (b x +a \right )^{\frac {3}{2}} A b}{6 a}+\frac {\left (A b -2 B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{16 a^{2}}+\left (-\frac {A b}{16}+\frac {B a}{8}\right ) \sqrt {b x +a}}{b^{3} x^{3}}\right ) b^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^2*((1/16*(A*b-2*B*a)/a^2*(b*x+a)^(5/2)-1/6*A*b/a*(b*x+a)^(3/2)+(-1/16*A*b+1/8*B*a)*(b*x+a)^(1/2))/x^3/b^3-
1/16*(A*b-2*B*a)/a^(5/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))

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maxima [A]  time = 1.90, size = 152, normalized size = 1.36 \begin {gather*} -\frac {1}{48} \, b^{3} {\left (\frac {2 \, {\left (8 \, {\left (b x + a\right )}^{\frac {3}{2}} A a b + 3 \, {\left (2 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {5}{2}} - 3 \, {\left (2 \, B a^{3} - A a^{2} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{3} a^{2} b - 3 \, {\left (b x + a\right )}^{2} a^{3} b + 3 \, {\left (b x + a\right )} a^{4} b - a^{5} b} + \frac {3 \, {\left (2 \, B a - A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}} b}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*b^3*(2*(8*(b*x + a)^(3/2)*A*a*b + 3*(2*B*a - A*b)*(b*x + a)^(5/2) - 3*(2*B*a^3 - A*a^2*b)*sqrt(b*x + a))
/((b*x + a)^3*a^2*b - 3*(b*x + a)^2*a^3*b + 3*(b*x + a)*a^4*b - a^5*b) + 3*(2*B*a - A*b)*log((sqrt(b*x + a) -
sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(5/2)*b))

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mupad [B]  time = 0.44, size = 129, normalized size = 1.15 \begin {gather*} \frac {\left (\frac {A\,b^3}{8}-\frac {B\,a\,b^2}{4}\right )\,\sqrt {a+b\,x}-\frac {\left (A\,b^3-2\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^{5/2}}{8\,a^2}+\frac {A\,b^3\,{\left (a+b\,x\right )}^{3/2}}{3\,a}}{3\,a\,{\left (a+b\,x\right )}^2-3\,a^2\,\left (a+b\,x\right )-{\left (a+b\,x\right )}^3+a^3}-\frac {b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (A\,b-2\,B\,a\right )}{8\,a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((A*b^3)/8 - (B*a*b^2)/4)*(a + b*x)^(1/2) - ((A*b^3 - 2*B*a*b^2)*(a + b*x)^(5/2))/(8*a^2) + (A*b^3*(a + b*x)^
(3/2))/(3*a))/(3*a*(a + b*x)^2 - 3*a^2*(a + b*x) - (a + b*x)^3 + a^3) - (b^2*atanh((a + b*x)^(1/2)/a^(1/2))*(A
*b - 2*B*a))/(8*a^(5/2))

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sympy [B]  time = 38.71, size = 666, normalized size = 5.95 \begin {gather*} - \frac {66 A a^{3} b^{3} \sqrt {a + b x}}{96 a^{6} + 144 a^{5} b x - 144 a^{4} \left (a + b x\right )^{2} + 48 a^{3} \left (a + b x\right )^{3}} + \frac {80 A a^{2} b^{3} \left (a + b x\right )^{\frac {3}{2}}}{96 a^{6} + 144 a^{5} b x - 144 a^{4} \left (a + b x\right )^{2} + 48 a^{3} \left (a + b x\right )^{3}} - \frac {30 A a b^{3} \left (a + b x\right )^{\frac {5}{2}}}{96 a^{6} + 144 a^{5} b x - 144 a^{4} \left (a + b x\right )^{2} + 48 a^{3} \left (a + b x\right )^{3}} - \frac {10 A a b^{3} \sqrt {a + b x}}{- 8 a^{4} - 16 a^{3} b x + 8 a^{2} \left (a + b x\right )^{2}} - \frac {5 A a b^{3} \sqrt {\frac {1}{a^{7}}} \log {\left (- a^{4} \sqrt {\frac {1}{a^{7}}} + \sqrt {a + b x} \right )}}{16} + \frac {5 A a b^{3} \sqrt {\frac {1}{a^{7}}} \log {\left (a^{4} \sqrt {\frac {1}{a^{7}}} + \sqrt {a + b x} \right )}}{16} + \frac {6 A b^{3} \left (a + b x\right )^{\frac {3}{2}}}{- 8 a^{4} - 16 a^{3} b x + 8 a^{2} \left (a + b x\right )^{2}} + \frac {3 A b^{3} \sqrt {\frac {1}{a^{5}}} \log {\left (- a^{3} \sqrt {\frac {1}{a^{5}}} + \sqrt {a + b x} \right )}}{8} - \frac {3 A b^{3} \sqrt {\frac {1}{a^{5}}} \log {\left (a^{3} \sqrt {\frac {1}{a^{5}}} + \sqrt {a + b x} \right )}}{8} - \frac {10 B a^{2} b^{2} \sqrt {a + b x}}{- 8 a^{4} - 16 a^{3} b x + 8 a^{2} \left (a + b x\right )^{2}} + \frac {6 B a b^{2} \left (a + b x\right )^{\frac {3}{2}}}{- 8 a^{4} - 16 a^{3} b x + 8 a^{2} \left (a + b x\right )^{2}} + \frac {3 B a b^{2} \sqrt {\frac {1}{a^{5}}} \log {\left (- a^{3} \sqrt {\frac {1}{a^{5}}} + \sqrt {a + b x} \right )}}{8} - \frac {3 B a b^{2} \sqrt {\frac {1}{a^{5}}} \log {\left (a^{3} \sqrt {\frac {1}{a^{5}}} + \sqrt {a + b x} \right )}}{8} - \frac {B b^{2} \sqrt {\frac {1}{a^{3}}} \log {\left (- a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} + \frac {B b^{2} \sqrt {\frac {1}{a^{3}}} \log {\left (a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} - \frac {B b \sqrt {a + b x}}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-66*A*a**3*b**3*sqrt(a + b*x)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) + 80*A*a
**2*b**3*(a + b*x)**(3/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) - 30*A*a*b**
3*(a + b*x)**(5/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) - 10*A*a*b**3*sqrt(
a + b*x)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) - 5*A*a*b**3*sqrt(a**(-7))*log(-a**4*sqrt(a**(-7)) + sq
rt(a + b*x))/16 + 5*A*a*b**3*sqrt(a**(-7))*log(a**4*sqrt(a**(-7)) + sqrt(a + b*x))/16 + 6*A*b**3*(a + b*x)**(3
/2)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 3*A*b**3*sqrt(a**(-5))*log(-a**3*sqrt(a**(-5)) + sqrt(a +
b*x))/8 - 3*A*b**3*sqrt(a**(-5))*log(a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 10*B*a**2*b**2*sqrt(a + b*x)/(-8*
a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 6*B*a*b**2*(a + b*x)**(3/2)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b
*x)**2) + 3*B*a*b**2*sqrt(a**(-5))*log(-a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 3*B*a*b**2*sqrt(a**(-5))*log(a
**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - B*b**2*sqrt(a**(-3))*log(-a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + B*b**
2*sqrt(a**(-3))*log(a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 - B*b*sqrt(a + b*x)/(a*x)

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