3.408 \(\int \frac {(a+b x)^{3/2} (A+B x)}{x^6} \, dx\)

Optimal. Leaf size=172 \[ \frac {3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{7/2}}-\frac {3 b^3 \sqrt {a+b x} (A b-2 a B)}{128 a^3 x}+\frac {b^2 \sqrt {a+b x} (A b-2 a B)}{64 a^2 x^2}+\frac {(a+b x)^{3/2} (A b-2 a B)}{8 a x^4}+\frac {b \sqrt {a+b x} (A b-2 a B)}{16 a x^3}-\frac {A (a+b x)^{5/2}}{5 a x^5} \]

[Out]

1/8*(A*b-2*B*a)*(b*x+a)^(3/2)/a/x^4-1/5*A*(b*x+a)^(5/2)/a/x^5+3/128*b^4*(A*b-2*B*a)*arctanh((b*x+a)^(1/2)/a^(1
/2))/a^(7/2)+1/16*b*(A*b-2*B*a)*(b*x+a)^(1/2)/x^3/a+1/64*b^2*(A*b-2*B*a)*(b*x+a)^(1/2)/a^2/x^2-3/128*b^3*(A*b-
2*B*a)*(b*x+a)^(1/2)/a^3/x

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Rubi [A]  time = 0.08, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {b^2 \sqrt {a+b x} (A b-2 a B)}{64 a^2 x^2}-\frac {3 b^3 \sqrt {a+b x} (A b-2 a B)}{128 a^3 x}+\frac {3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{7/2}}+\frac {b \sqrt {a+b x} (A b-2 a B)}{16 a x^3}+\frac {(a+b x)^{3/2} (A b-2 a B)}{8 a x^4}-\frac {A (a+b x)^{5/2}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(3/2)*(A + B*x))/x^6,x]

[Out]

(b*(A*b - 2*a*B)*Sqrt[a + b*x])/(16*a*x^3) + (b^2*(A*b - 2*a*B)*Sqrt[a + b*x])/(64*a^2*x^2) - (3*b^3*(A*b - 2*
a*B)*Sqrt[a + b*x])/(128*a^3*x) + ((A*b - 2*a*B)*(a + b*x)^(3/2))/(8*a*x^4) - (A*(a + b*x)^(5/2))/(5*a*x^5) +
(3*b^4*(A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(128*a^(7/2))

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^6} \, dx &=-\frac {A (a+b x)^{5/2}}{5 a x^5}+\frac {\left (-\frac {5 A b}{2}+5 a B\right ) \int \frac {(a+b x)^{3/2}}{x^5} \, dx}{5 a}\\ &=\frac {(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac {A (a+b x)^{5/2}}{5 a x^5}-\frac {(3 b (A b-2 a B)) \int \frac {\sqrt {a+b x}}{x^4} \, dx}{16 a}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x}}{16 a x^3}+\frac {(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac {A (a+b x)^{5/2}}{5 a x^5}-\frac {\left (b^2 (A b-2 a B)\right ) \int \frac {1}{x^3 \sqrt {a+b x}} \, dx}{32 a}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x}}{16 a x^3}+\frac {b^2 (A b-2 a B) \sqrt {a+b x}}{64 a^2 x^2}+\frac {(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac {A (a+b x)^{5/2}}{5 a x^5}+\frac {\left (3 b^3 (A b-2 a B)\right ) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{128 a^2}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x}}{16 a x^3}+\frac {b^2 (A b-2 a B) \sqrt {a+b x}}{64 a^2 x^2}-\frac {3 b^3 (A b-2 a B) \sqrt {a+b x}}{128 a^3 x}+\frac {(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac {A (a+b x)^{5/2}}{5 a x^5}-\frac {\left (3 b^4 (A b-2 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{256 a^3}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x}}{16 a x^3}+\frac {b^2 (A b-2 a B) \sqrt {a+b x}}{64 a^2 x^2}-\frac {3 b^3 (A b-2 a B) \sqrt {a+b x}}{128 a^3 x}+\frac {(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac {A (a+b x)^{5/2}}{5 a x^5}-\frac {\left (3 b^3 (A b-2 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{128 a^3}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x}}{16 a x^3}+\frac {b^2 (A b-2 a B) \sqrt {a+b x}}{64 a^2 x^2}-\frac {3 b^3 (A b-2 a B) \sqrt {a+b x}}{128 a^3 x}+\frac {(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac {A (a+b x)^{5/2}}{5 a x^5}+\frac {3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 57, normalized size = 0.33 \[ -\frac {(a+b x)^{5/2} \left (a^5 A+b^4 x^5 (2 a B-A b) \, _2F_1\left (\frac {5}{2},5;\frac {7}{2};\frac {b x}{a}+1\right )\right )}{5 a^6 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(3/2)*(A + B*x))/x^6,x]

[Out]

-1/5*((a + b*x)^(5/2)*(a^5*A + b^4*(-(A*b) + 2*a*B)*x^5*Hypergeometric2F1[5/2, 5, 7/2, 1 + (b*x)/a]))/(a^6*x^5
)

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fricas [A]  time = 0.76, size = 306, normalized size = 1.78 \[ \left [-\frac {15 \, {\left (2 \, B a b^{4} - A b^{5}\right )} \sqrt {a} x^{5} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (128 \, A a^{5} - 15 \, {\left (2 \, B a^{2} b^{3} - A a b^{4}\right )} x^{4} + 10 \, {\left (2 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{3} + 8 \, {\left (30 \, B a^{4} b + A a^{3} b^{2}\right )} x^{2} + 16 \, {\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{1280 \, a^{4} x^{5}}, \frac {15 \, {\left (2 \, B a b^{4} - A b^{5}\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (128 \, A a^{5} - 15 \, {\left (2 \, B a^{2} b^{3} - A a b^{4}\right )} x^{4} + 10 \, {\left (2 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{3} + 8 \, {\left (30 \, B a^{4} b + A a^{3} b^{2}\right )} x^{2} + 16 \, {\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{640 \, a^{4} x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^6,x, algorithm="fricas")

[Out]

[-1/1280*(15*(2*B*a*b^4 - A*b^5)*sqrt(a)*x^5*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(128*A*a^5 - 15*
(2*B*a^2*b^3 - A*a*b^4)*x^4 + 10*(2*B*a^3*b^2 - A*a^2*b^3)*x^3 + 8*(30*B*a^4*b + A*a^3*b^2)*x^2 + 16*(10*B*a^5
+ 11*A*a^4*b)*x)*sqrt(b*x + a))/(a^4*x^5), 1/640*(15*(2*B*a*b^4 - A*b^5)*sqrt(-a)*x^5*arctan(sqrt(b*x + a)*sq
rt(-a)/a) - (128*A*a^5 - 15*(2*B*a^2*b^3 - A*a*b^4)*x^4 + 10*(2*B*a^3*b^2 - A*a^2*b^3)*x^3 + 8*(30*B*a^4*b + A
*a^3*b^2)*x^2 + 16*(10*B*a^5 + 11*A*a^4*b)*x)*sqrt(b*x + a))/(a^4*x^5)]

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giac [A]  time = 1.34, size = 192, normalized size = 1.12 \[ \frac {\frac {15 \, {\left (2 \, B a b^{5} - A b^{6}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {30 \, {\left (b x + a\right )}^{\frac {9}{2}} B a b^{5} - 140 \, {\left (b x + a\right )}^{\frac {7}{2}} B a^{2} b^{5} + 140 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{4} b^{5} - 30 \, \sqrt {b x + a} B a^{5} b^{5} - 15 \, {\left (b x + a\right )}^{\frac {9}{2}} A b^{6} + 70 \, {\left (b x + a\right )}^{\frac {7}{2}} A a b^{6} - 128 \, {\left (b x + a\right )}^{\frac {5}{2}} A a^{2} b^{6} - 70 \, {\left (b x + a\right )}^{\frac {3}{2}} A a^{3} b^{6} + 15 \, \sqrt {b x + a} A a^{4} b^{6}}{a^{3} b^{5} x^{5}}}{640 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^6,x, algorithm="giac")

[Out]

1/640*(15*(2*B*a*b^5 - A*b^6)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + (30*(b*x + a)^(9/2)*B*a*b^5 - 14
0*(b*x + a)^(7/2)*B*a^2*b^5 + 140*(b*x + a)^(3/2)*B*a^4*b^5 - 30*sqrt(b*x + a)*B*a^5*b^5 - 15*(b*x + a)^(9/2)*
A*b^6 + 70*(b*x + a)^(7/2)*A*a*b^6 - 128*(b*x + a)^(5/2)*A*a^2*b^6 - 70*(b*x + a)^(3/2)*A*a^3*b^6 + 15*sqrt(b*
x + a)*A*a^4*b^6)/(a^3*b^5*x^5))/b

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maple [A]  time = 0.02, size = 129, normalized size = 0.75 \[ 2 \left (\frac {3 \left (A b -2 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{256 a^{\frac {7}{2}}}+\frac {-\frac {\left (b x +a \right )^{\frac {5}{2}} A b}{10 a}+\frac {3 \left (A b -2 B a \right ) \sqrt {b x +a}\, a}{256}+\frac {7 \left (A b -2 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{128 a^{2}}-\frac {3 \left (A b -2 B a \right ) \left (b x +a \right )^{\frac {9}{2}}}{256 a^{3}}+\left (-\frac {7 A b}{128}+\frac {7 B a}{64}\right ) \left (b x +a \right )^{\frac {3}{2}}}{b^{5} x^{5}}\right ) b^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(B*x+A)/x^6,x)

[Out]

2*b^4*((-3/256*(A*b-2*B*a)/a^3*(b*x+a)^(9/2)+7/128*(A*b-2*B*a)/a^2*(b*x+a)^(7/2)-1/10*A*b/a*(b*x+a)^(5/2)+(-7/
128*A*b+7/64*B*a)*(b*x+a)^(3/2)+3/256*a*(A*b-2*B*a)*(b*x+a)^(1/2))/x^5/b^5+3/256*(A*b-2*B*a)/a^(7/2)*arctanh((
b*x+a)^(1/2)/a^(1/2)))

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maxima [A]  time = 2.02, size = 224, normalized size = 1.30 \[ -\frac {1}{1280} \, b^{5} {\left (\frac {2 \, {\left (128 \, {\left (b x + a\right )}^{\frac {5}{2}} A a^{2} b - 15 \, {\left (2 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {9}{2}} + 70 \, {\left (2 \, B a^{2} - A a b\right )} {\left (b x + a\right )}^{\frac {7}{2}} - 70 \, {\left (2 \, B a^{4} - A a^{3} b\right )} {\left (b x + a\right )}^{\frac {3}{2}} + 15 \, {\left (2 \, B a^{5} - A a^{4} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{5} a^{3} b - 5 \, {\left (b x + a\right )}^{4} a^{4} b + 10 \, {\left (b x + a\right )}^{3} a^{5} b - 10 \, {\left (b x + a\right )}^{2} a^{6} b + 5 \, {\left (b x + a\right )} a^{7} b - a^{8} b} - \frac {15 \, {\left (2 \, B a - A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}} b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^6,x, algorithm="maxima")

[Out]

-1/1280*b^5*(2*(128*(b*x + a)^(5/2)*A*a^2*b - 15*(2*B*a - A*b)*(b*x + a)^(9/2) + 70*(2*B*a^2 - A*a*b)*(b*x + a
)^(7/2) - 70*(2*B*a^4 - A*a^3*b)*(b*x + a)^(3/2) + 15*(2*B*a^5 - A*a^4*b)*sqrt(b*x + a))/((b*x + a)^5*a^3*b -
5*(b*x + a)^4*a^4*b + 10*(b*x + a)^3*a^5*b - 10*(b*x + a)^2*a^6*b + 5*(b*x + a)*a^7*b - a^8*b) - 15*(2*B*a - A
*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(7/2)*b))

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mupad [B]  time = 0.44, size = 203, normalized size = 1.18 \[ \frac {\left (\frac {7\,A\,b^5}{64}-\frac {7\,B\,a\,b^4}{32}\right )\,{\left (a+b\,x\right )}^{3/2}+\left (\frac {3\,B\,a^2\,b^4}{64}-\frac {3\,A\,a\,b^5}{128}\right )\,\sqrt {a+b\,x}-\frac {7\,\left (A\,b^5-2\,B\,a\,b^4\right )\,{\left (a+b\,x\right )}^{7/2}}{64\,a^2}+\frac {3\,\left (A\,b^5-2\,B\,a\,b^4\right )\,{\left (a+b\,x\right )}^{9/2}}{128\,a^3}+\frac {A\,b^5\,{\left (a+b\,x\right )}^{5/2}}{5\,a}}{5\,a\,{\left (a+b\,x\right )}^4-5\,a^4\,\left (a+b\,x\right )-{\left (a+b\,x\right )}^5-10\,a^2\,{\left (a+b\,x\right )}^3+10\,a^3\,{\left (a+b\,x\right )}^2+a^5}+\frac {3\,b^4\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (A\,b-2\,B\,a\right )}{128\,a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(3/2))/x^6,x)

[Out]

(((7*A*b^5)/64 - (7*B*a*b^4)/32)*(a + b*x)^(3/2) + ((3*B*a^2*b^4)/64 - (3*A*a*b^5)/128)*(a + b*x)^(1/2) - (7*(
A*b^5 - 2*B*a*b^4)*(a + b*x)^(7/2))/(64*a^2) + (3*(A*b^5 - 2*B*a*b^4)*(a + b*x)^(9/2))/(128*a^3) + (A*b^5*(a +
b*x)^(5/2))/(5*a))/(5*a*(a + b*x)^4 - 5*a^4*(a + b*x) - (a + b*x)^5 - 10*a^2*(a + b*x)^3 + 10*a^3*(a + b*x)^2
+ a^5) + (3*b^4*atanh((a + b*x)^(1/2)/a^(1/2))*(A*b - 2*B*a))/(128*a^(7/2))

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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**6,x)

[Out]

Timed out

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