3.407 \(\int \frac {(a+b x)^{3/2} (A+B x)}{x^5} \, dx\)

Optimal. Leaf size=146 \[ -\frac {b^3 (3 A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{5/2}}+\frac {b^2 \sqrt {a+b x} (3 A b-8 a B)}{64 a^2 x}+\frac {(a+b x)^{3/2} (3 A b-8 a B)}{24 a x^3}+\frac {b \sqrt {a+b x} (3 A b-8 a B)}{32 a x^2}-\frac {A (a+b x)^{5/2}}{4 a x^4} \]

[Out]

1/24*(3*A*b-8*B*a)*(b*x+a)^(3/2)/x^3/a-1/4*A*(b*x+a)^(5/2)/a/x^4-1/64*b^3*(3*A*b-8*B*a)*arctanh((b*x+a)^(1/2)/
a^(1/2))/a^(5/2)+1/32*b*(3*A*b-8*B*a)*(b*x+a)^(1/2)/a/x^2+1/64*b^2*(3*A*b-8*B*a)*(b*x+a)^(1/2)/a^2/x

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Rubi [A]  time = 0.06, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {b^2 \sqrt {a+b x} (3 A b-8 a B)}{64 a^2 x}-\frac {b^3 (3 A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{5/2}}+\frac {b \sqrt {a+b x} (3 A b-8 a B)}{32 a x^2}+\frac {(a+b x)^{3/2} (3 A b-8 a B)}{24 a x^3}-\frac {A (a+b x)^{5/2}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(3/2)*(A + B*x))/x^5,x]

[Out]

(b*(3*A*b - 8*a*B)*Sqrt[a + b*x])/(32*a*x^2) + (b^2*(3*A*b - 8*a*B)*Sqrt[a + b*x])/(64*a^2*x) + ((3*A*b - 8*a*
B)*(a + b*x)^(3/2))/(24*a*x^3) - (A*(a + b*x)^(5/2))/(4*a*x^4) - (b^3*(3*A*b - 8*a*B)*ArcTanh[Sqrt[a + b*x]/Sq
rt[a]])/(64*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 {(a+b x)^{3/2} (A+B x)}{x^5} \, dx &=-\frac {A (a+b x)^{5/2}}{4 a x^4}+\frac {\left (-\frac {3 A b}{2}+4 a B\right ) \int \frac {(a+b x)^{3/2}}{x^4} \, dx}{4 a}\\ &=\frac {(3 A b-8 a B) (a+b x)^{3/2}}{24 a x^3}-\frac {A (a+b x)^{5/2}}{4 a x^4}-\frac {(b (3 A b-8 a B)) \int \frac {\sqrt {a+b x}}{x^3} \, dx}{16 a}\\ &=\frac {b (3 A b-8 a B) \sqrt {a+b x}}{32 a x^2}+\frac {(3 A b-8 a B) (a+b x)^{3/2}}{24 a x^3}-\frac {A (a+b x)^{5/2}}{4 a x^4}-\frac {\left (b^2 (3 A b-8 a B)\right ) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{64 a}\\ &=\frac {b (3 A b-8 a B) \sqrt {a+b x}}{32 a x^2}+\frac {b^2 (3 A b-8 a B) \sqrt {a+b x}}{64 a^2 x}+\frac {(3 A b-8 a B) (a+b x)^{3/2}}{24 a x^3}-\frac {A (a+b x)^{5/2}}{4 a x^4}+\frac {\left (b^3 (3 A b-8 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{128 a^2}\\ &=\frac {b (3 A b-8 a B) \sqrt {a+b x}}{32 a x^2}+\frac {b^2 (3 A b-8 a B) \sqrt {a+b x}}{64 a^2 x}+\frac {(3 A b-8 a B) (a+b x)^{3/2}}{24 a x^3}-\frac {A (a+b x)^{5/2}}{4 a x^4}+\frac {\left (b^2 (3 A b-8 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{64 a^2}\\ &=\frac {b (3 A b-8 a B) \sqrt {a+b x}}{32 a x^2}+\frac {b^2 (3 A b-8 a B) \sqrt {a+b x}}{64 a^2 x}+\frac {(3 A b-8 a B) (a+b x)^{3/2}}{24 a x^3}-\frac {A (a+b x)^{5/2}}{4 a x^4}-\frac {b^3 (3 A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 58, normalized size = 0.40 \[ -\frac {(a+b x)^{5/2} \left (5 a^4 A+b^3 x^4 (3 A b-8 a B) \, _2F_1\left (\frac {5}{2},4;\frac {7}{2};\frac {b x}{a}+1\right )\right )}{20 a^5 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(3/2)*(A + B*x))/x^5,x]

[Out]

-1/20*((a + b*x)^(5/2)*(5*a^4*A + b^3*(3*A*b - 8*a*B)*x^4*Hypergeometric2F1[5/2, 4, 7/2, 1 + (b*x)/a]))/(a^5*x
^4)

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fricas [A]  time = 0.67, size = 259, normalized size = 1.77 \[ \left [-\frac {3 \, {\left (8 \, B a b^{3} - 3 \, A b^{4}\right )} \sqrt {a} x^{4} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (48 \, A a^{4} + 3 \, {\left (8 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{3} + 2 \, {\left (56 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} + 9 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{384 \, a^{3} x^{4}}, -\frac {3 \, {\left (8 \, B a b^{3} - 3 \, A b^{4}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (48 \, A a^{4} + 3 \, {\left (8 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{3} + 2 \, {\left (56 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} + 9 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{192 \, a^{3} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(3*(8*B*a*b^3 - 3*A*b^4)*sqrt(a)*x^4*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(48*A*a^4 + 3*(8
*B*a^2*b^2 - 3*A*a*b^3)*x^3 + 2*(56*B*a^3*b + 3*A*a^2*b^2)*x^2 + 8*(8*B*a^4 + 9*A*a^3*b)*x)*sqrt(b*x + a))/(a^
3*x^4), -1/192*(3*(8*B*a*b^3 - 3*A*b^4)*sqrt(-a)*x^4*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (48*A*a^4 + 3*(8*B*a^2
*b^2 - 3*A*a*b^3)*x^3 + 2*(56*B*a^3*b + 3*A*a^2*b^2)*x^2 + 8*(8*B*a^4 + 9*A*a^3*b)*x)*sqrt(b*x + a))/(a^3*x^4)
]

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giac [A]  time = 1.30, size = 176, normalized size = 1.21 \[ -\frac {\frac {3 \, {\left (8 \, B a b^{4} - 3 \, A b^{5}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {24 \, {\left (b x + a\right )}^{\frac {7}{2}} B a b^{4} + 40 \, {\left (b x + a\right )}^{\frac {5}{2}} B a^{2} b^{4} - 88 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{3} b^{4} + 24 \, \sqrt {b x + a} B a^{4} b^{4} - 9 \, {\left (b x + a\right )}^{\frac {7}{2}} A b^{5} + 33 \, {\left (b x + a\right )}^{\frac {5}{2}} A a b^{5} + 33 \, {\left (b x + a\right )}^{\frac {3}{2}} A a^{2} b^{5} - 9 \, \sqrt {b x + a} A a^{3} b^{5}}{a^{2} b^{4} x^{4}}}{192 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(3*(8*B*a*b^4 - 3*A*b^5)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (24*(b*x + a)^(7/2)*B*a*b^4 +
40*(b*x + a)^(5/2)*B*a^2*b^4 - 88*(b*x + a)^(3/2)*B*a^3*b^4 + 24*sqrt(b*x + a)*B*a^4*b^4 - 9*(b*x + a)^(7/2)*A
*b^5 + 33*(b*x + a)^(5/2)*A*a*b^5 + 33*(b*x + a)^(3/2)*A*a^2*b^5 - 9*sqrt(b*x + a)*A*a^3*b^5)/(a^2*b^4*x^4))/b

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maple [A]  time = 0.02, size = 119, normalized size = 0.82 \[ 2 \left (-\frac {\left (3 A b -8 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {5}{2}}}+\frac {\frac {\left (3 A b -8 B a \right ) \sqrt {b x +a}\, a}{128}-\frac {\left (33 A b +40 B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{384 a}+\frac {\left (3 A b -8 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{128 a^{2}}+\left (-\frac {11 A b}{128}+\frac {11 B a}{48}\right ) \left (b x +a \right )^{\frac {3}{2}}}{b^{4} x^{4}}\right ) b^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(B*x+A)/x^5,x)

[Out]

2*b^3*((1/128*(3*A*b-8*B*a)/a^2*(b*x+a)^(7/2)-1/384*(33*A*b+40*B*a)/a*(b*x+a)^(5/2)+(-11/128*A*b+11/48*B*a)*(b
*x+a)^(3/2)+1/128*a*(3*A*b-8*B*a)*(b*x+a)^(1/2))/x^4/b^4-1/128*(3*A*b-8*B*a)/a^(5/2)*arctanh((b*x+a)^(1/2)/a^(
1/2)))

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maxima [A]  time = 2.06, size = 195, normalized size = 1.34 \[ -\frac {1}{384} \, b^{4} {\left (\frac {2 \, {\left (3 \, {\left (8 \, B a - 3 \, A b\right )} {\left (b x + a\right )}^{\frac {7}{2}} + {\left (40 \, B a^{2} + 33 \, A a b\right )} {\left (b x + a\right )}^{\frac {5}{2}} - 11 \, {\left (8 \, B a^{3} - 3 \, A a^{2} b\right )} {\left (b x + a\right )}^{\frac {3}{2}} + 3 \, {\left (8 \, B a^{4} - 3 \, A a^{3} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{4} a^{2} b - 4 \, {\left (b x + a\right )}^{3} a^{3} b + 6 \, {\left (b x + a\right )}^{2} a^{4} b - 4 \, {\left (b x + a\right )} a^{5} b + a^{6} b} + \frac {3 \, {\left (8 \, B a - 3 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}} b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*b^4*(2*(3*(8*B*a - 3*A*b)*(b*x + a)^(7/2) + (40*B*a^2 + 33*A*a*b)*(b*x + a)^(5/2) - 11*(8*B*a^3 - 3*A*a
^2*b)*(b*x + a)^(3/2) + 3*(8*B*a^4 - 3*A*a^3*b)*sqrt(b*x + a))/((b*x + a)^4*a^2*b - 4*(b*x + a)^3*a^3*b + 6*(b
*x + a)^2*a^4*b - 4*(b*x + a)*a^5*b + a^6*b) + 3*(8*B*a - 3*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 = 177, normalized size = 1.21 \[ -\frac {\left (\frac {11\,A\,b^4}{64}-\frac {11\,B\,a\,b^3}{24}\right )\,{\left (a+b\,x\right )}^{3/2}+\left (\frac {B\,a^2\,b^3}{8}-\frac {3\,A\,a\,b^4}{64}\right )\,\sqrt {a+b\,x}-\frac {\left (3\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{7/2}}{64\,a^2}+\frac {\left (33\,A\,b^4+40\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{5/2}}{192\,a}}{{\left (a+b\,x\right )}^4-4\,a^3\,\left (a+b\,x\right )-4\,a\,{\left (a+b\,x\right )}^3+6\,a^2\,{\left (a+b\,x\right )}^2+a^4}-\frac {b^3\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (3\,A\,b-8\,B\,a\right )}{64\,a^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(3/2))/x^5,x)

[Out]

- (((11*A*b^4)/64 - (11*B*a*b^3)/24)*(a + b*x)^(3/2) + ((B*a^2*b^3)/8 - (3*A*a*b^4)/64)*(a + b*x)^(1/2) - ((3*
A*b^4 - 8*B*a*b^3)*(a + b*x)^(7/2))/(64*a^2) + ((33*A*b^4 + 40*B*a*b^3)*(a + b*x)^(5/2))/(192*a))/((a + b*x)^4
- 4*a^3*(a + b*x) - 4*a*(a + b*x)^3 + 6*a^2*(a + b*x)^2 + a^4) - (b^3*atanh((a + b*x)^(1/2)/a^(1/2))*(3*A*b -
8*B*a))/(64*a^(5/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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