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

Optimal. Leaf size=117 \[ -\frac {2 (a+b x)^{3/2} (3 a B+2 A b)}{3 a \sqrt {x}}+\frac {b \sqrt {x} \sqrt {a+b x} (3 a B+2 A b)}{a}+\sqrt {b} (3 a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 A (a+b x)^{5/2}}{3 a x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {78, 47, 50, 63, 217, 206} \[ -\frac {2 (a+b x)^{3/2} (3 a B+2 A b)}{3 a \sqrt {x}}+\frac {b \sqrt {x} \sqrt {a+b x} (3 a B+2 A b)}{a}+\sqrt {b} (3 a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 A (a+b x)^{5/2}}{3 a x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 206

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

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

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Mathematica [C]  time = 0.09, size = 73, normalized size = 0.62 \[ \frac {2 \sqrt {a+b x} \left (-\frac {x (3 a B+2 A b) \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};-\frac {b x}{a}\right )}{\sqrt {\frac {b x}{a}+1}}-\frac {A (a+b x)^2}{a}\right )}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x]*(-((A*(a + b*x)^2)/a) - ((2*A*b + 3*a*B)*x*Hypergeometric2F1[-3/2, -1/2, 1/2, -((b*x)/a)])/Sq
rt[1 + (b*x)/a]))/(3*x^(3/2))

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fricas [A]  time = 0.57, size = 163, normalized size = 1.39 \[ \left [\frac {3 \, {\left (3 \, B a + 2 \, A b\right )} \sqrt {b} x^{2} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (3 \, B b x^{2} - 2 \, A a - 2 \, {\left (3 \, B a + 4 \, A b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{6 \, x^{2}}, -\frac {3 \, {\left (3 \, B a + 2 \, A b\right )} \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (3 \, B b x^{2} - 2 \, A a - 2 \, {\left (3 \, B a + 4 \, A b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{3 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(3*B*a + 2*A*b)*sqrt(b)*x^2*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(3*B*b*x^2 - 2*A*a -
2*(3*B*a + 4*A*b)*x)*sqrt(b*x + a)*sqrt(x))/x^2, -1/3*(3*(3*B*a + 2*A*b)*sqrt(-b)*x^2*arctan(sqrt(b*x + a)*sqr
t(-b)/(b*sqrt(x))) - (3*B*b*x^2 - 2*A*a - 2*(3*B*a + 4*A*b)*x)*sqrt(b*x + a)*sqrt(x))/x^2]

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giac [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^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.02, size = 162, normalized size = 1.38 \[ \frac {\sqrt {b x +a}\, \left (6 A \,b^{2} x^{2} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+9 B a b \,x^{2} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+6 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {3}{2}} x^{2}-16 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {3}{2}} x -12 \sqrt {\left (b x +a \right ) x}\, B a \sqrt {b}\, x -4 \sqrt {\left (b x +a \right ) x}\, A a \sqrt {b}\right )}{6 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}\, x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(b*x+a)^(1/2)/x^(3/2)*(6*A*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x^2*b^2+9*B*a*b*x^2*ln(1/
2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))+6*((b*x+a)*x)^(1/2)*B*b^(3/2)*x^2-16*((b*x+a)*x)^(1/2)*A*b^(3
/2)*x-12*((b*x+a)*x)^(1/2)*B*a*b^(1/2)*x-4*((b*x+a)*x)^(1/2)*A*a*b^(1/2))/((b*x+a)*x)^(1/2)/b^(1/2)

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maxima [A]  time = 0.97, size = 146, normalized size = 1.25 \[ \frac {3}{2} \, B a \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + A b^{\frac {3}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {3 \, \sqrt {b x^{2} + a x} B a}{x} - \frac {7 \, \sqrt {b x^{2} + a x} A b}{3 \, x} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{x^{2}} - \frac {\sqrt {b x^{2} + a x} A a}{3 \, x^{2}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*B*a*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + A*b^(3/2)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*s
qrt(b)) - 3*sqrt(b*x^2 + a*x)*B*a/x - 7/3*sqrt(b*x^2 + a*x)*A*b/x + (b*x^2 + a*x)^(3/2)*B/x^2 - 1/3*sqrt(b*x^2
+ a*x)*A*a/x^2 - 1/3*(b*x^2 + a*x)^(3/2)*A/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{3/2}}{x^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 45.81, size = 168, normalized size = 1.44 \[ A \left (- \frac {2 a \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {8 b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3} - b^{\frac {3}{2}} \log {\left (\frac {a}{b x} \right )} + 2 b^{\frac {3}{2}} \log {\left (\sqrt {\frac {a}{b x} + 1} + 1 \right )}\right ) + B \left (- \frac {2 a^{\frac {3}{2}}}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} - \frac {\sqrt {a} b \sqrt {x}}{\sqrt {1 + \frac {b x}{a}}} + 3 a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} + \frac {b^{2} x^{\frac {3}{2}}}{\sqrt {a} \sqrt {1 + \frac {b x}{a}}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-2*a*sqrt(b)*sqrt(a/(b*x) + 1)/(3*x) - 8*b**(3/2)*sqrt(a/(b*x) + 1)/3 - b**(3/2)*log(a/(b*x)) + 2*b**(3/2)*
log(sqrt(a/(b*x) + 1) + 1)) + B*(-2*a**(3/2)/(sqrt(x)*sqrt(1 + b*x/a)) - sqrt(a)*b*sqrt(x)/sqrt(1 + b*x/a) + 3
*a*sqrt(b)*asinh(sqrt(b)*sqrt(x)/sqrt(a)) + b**2*x**(3/2)/(sqrt(a)*sqrt(1 + b*x/a)))

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