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

Optimal. Leaf size=144 \[ \frac {5 a^2 (a B+6 A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 \sqrt {b}}+\frac {\sqrt {x} (a+b x)^{5/2} (a B+6 A b)}{3 a}+\frac {5}{12} \sqrt {x} (a+b x)^{3/2} (a B+6 A b)+\frac {5}{8} a \sqrt {x} \sqrt {a+b x} (a B+6 A b)-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}} \]

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Rubi [A]  time = 0.06, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {78, 50, 63, 217, 206} \begin {gather*} \frac {5 a^2 (a B+6 A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 \sqrt {b}}+\frac {\sqrt {x} (a+b x)^{5/2} (a B+6 A b)}{3 a}+\frac {5}{12} \sqrt {x} (a+b x)^{3/2} (a B+6 A b)+\frac {5}{8} a \sqrt {x} \sqrt {a+b x} (a B+6 A b)-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*(6*A*b + a*B)*Sqrt[x]*Sqrt[a + b*x])/8 + (5*(6*A*b + a*B)*Sqrt[x]*(a + b*x)^(3/2))/12 + ((6*A*b + a*B)*Sq
rt[x]*(a + b*x)^(5/2))/(3*a) - (2*A*(a + b*x)^(7/2))/(a*Sqrt[x]) + (5*a^2*(6*A*b + a*B)*ArcTanh[(Sqrt[b]*Sqrt[
x])/Sqrt[a + b*x]])/(8*Sqrt[b])

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)^{5/2} (A+B x)}{x^{3/2}} \, dx &=-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {\left (2 \left (3 A b+\frac {a B}{2}\right )\right ) \int \frac {(a+b x)^{5/2}}{\sqrt {x}} \, dx}{a}\\ &=\frac {(6 A b+a B) \sqrt {x} (a+b x)^{5/2}}{3 a}-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {1}{6} (5 (6 A b+a B)) \int \frac {(a+b x)^{3/2}}{\sqrt {x}} \, dx\\ &=\frac {5}{12} (6 A b+a B) \sqrt {x} (a+b x)^{3/2}+\frac {(6 A b+a B) \sqrt {x} (a+b x)^{5/2}}{3 a}-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {1}{8} (5 a (6 A b+a B)) \int \frac {\sqrt {a+b x}}{\sqrt {x}} \, dx\\ &=\frac {5}{8} a (6 A b+a B) \sqrt {x} \sqrt {a+b x}+\frac {5}{12} (6 A b+a B) \sqrt {x} (a+b x)^{3/2}+\frac {(6 A b+a B) \sqrt {x} (a+b x)^{5/2}}{3 a}-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {1}{16} \left (5 a^2 (6 A b+a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx\\ &=\frac {5}{8} a (6 A b+a B) \sqrt {x} \sqrt {a+b x}+\frac {5}{12} (6 A b+a B) \sqrt {x} (a+b x)^{3/2}+\frac {(6 A b+a B) \sqrt {x} (a+b x)^{5/2}}{3 a}-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {1}{8} \left (5 a^2 (6 A b+a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )\\ &=\frac {5}{8} a (6 A b+a B) \sqrt {x} \sqrt {a+b x}+\frac {5}{12} (6 A b+a B) \sqrt {x} (a+b x)^{3/2}+\frac {(6 A b+a B) \sqrt {x} (a+b x)^{5/2}}{3 a}-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {1}{8} \left (5 a^2 (6 A b+a B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )\\ &=\frac {5}{8} a (6 A b+a B) \sqrt {x} \sqrt {a+b x}+\frac {5}{12} (6 A b+a B) \sqrt {x} (a+b x)^{3/2}+\frac {(6 A b+a B) \sqrt {x} (a+b x)^{5/2}}{3 a}-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {5 a^2 (6 A b+a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 111, normalized size = 0.77 \begin {gather*} \frac {1}{24} \sqrt {a+b x} \left (\frac {15 a^{3/2} (a B+6 A b) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {b} \sqrt {\frac {b x}{a}+1}}+\frac {a^2 (33 B x-48 A)+2 a b x (27 A+13 B x)+4 b^2 x^2 (3 A+2 B x)}{\sqrt {x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*((4*b^2*x^2*(3*A + 2*B*x) + 2*a*b*x*(27*A + 13*B*x) + a^2*(-48*A + 33*B*x))/Sqrt[x] + (15*a^(3/
2)*(6*A*b + a*B)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(Sqrt[b]*Sqrt[1 + (b*x)/a])))/24

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IntegrateAlgebraic [A]  time = 0.31, size = 110, normalized size = 0.76 \begin {gather*} \frac {\sqrt {a+b x} \left (-48 a^2 A+33 a^2 B x+54 a A b x+26 a b B x^2+12 A b^2 x^2+8 b^2 B x^3\right )}{24 \sqrt {x}}-\frac {5 \left (a^3 B+6 a^2 A b\right ) \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{8 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(-48*a^2*A + 54*a*A*b*x + 33*a^2*B*x + 12*A*b^2*x^2 + 26*a*b*B*x^2 + 8*b^2*B*x^3))/(24*Sqrt[x])
- (5*(6*a^2*A*b + a^3*B)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/(8*Sqrt[b])

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fricas [A]  time = 1.71, size = 233, normalized size = 1.62 \begin {gather*} \left [\frac {15 \, {\left (B a^{3} + 6 \, A a^{2} b\right )} \sqrt {b} x \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (8 \, B b^{3} x^{3} - 48 \, A a^{2} b + 2 \, {\left (13 \, B a b^{2} + 6 \, A b^{3}\right )} x^{2} + 3 \, {\left (11 \, B a^{2} b + 18 \, A a b^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b x}, -\frac {15 \, {\left (B a^{3} + 6 \, A a^{2} b\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (8 \, B b^{3} x^{3} - 48 \, A a^{2} b + 2 \, {\left (13 \, B a b^{2} + 6 \, A b^{3}\right )} x^{2} + 3 \, {\left (11 \, B a^{2} b + 18 \, A a b^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*(B*a^3 + 6*A*a^2*b)*sqrt(b)*x*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(8*B*b^3*x^3 - 48
*A*a^2*b + 2*(13*B*a*b^2 + 6*A*b^3)*x^2 + 3*(11*B*a^2*b + 18*A*a*b^2)*x)*sqrt(b*x + a)*sqrt(x))/(b*x), -1/24*(
15*(B*a^3 + 6*A*a^2*b)*sqrt(-b)*x*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (8*B*b^3*x^3 - 48*A*a^2*b + 2*(
13*B*a*b^2 + 6*A*b^3)*x^2 + 3*(11*B*a^2*b + 18*A*a*b^2)*x)*sqrt(b*x + a)*sqrt(x))/(b*x)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.02, size = 202, normalized size = 1.40 \begin {gather*} \frac {\sqrt {b x +a}\, \left (16 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {5}{2}} x^{3}+90 A \,a^{2} b x \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+15 B \,a^{3} x \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+24 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {5}{2}} x^{2}+52 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {3}{2}} x^{2}+108 \sqrt {\left (b x +a \right ) x}\, A a \,b^{\frac {3}{2}} x +66 \sqrt {\left (b x +a \right ) x}\, B \,a^{2} \sqrt {b}\, x -96 \sqrt {\left (b x +a \right ) x}\, A \,a^{2} \sqrt {b}\right )}{48 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}\, \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(b*x+a)^(1/2)*(16*B*b^(5/2)*((b*x+a)*x)^(1/2)*x^3+24*((b*x+a)*x)^(1/2)*A*b^(5/2)*x^2+52*((b*x+a)*x)^(1/2)
*B*a*b^(3/2)*x^2+90*A*b*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x*a^2+108*((b*x+a)*x)^(1/2)*A*a*
b^(3/2)*x+15*B*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x*a^3+66*((b*x+a)*x)^(1/2)*B*a^2*b^(1/2)*
x-96*((b*x+a)*x)^(1/2)*A*a^2*b^(1/2))/x^(1/2)/((b*x+a)*x)^(1/2)/b^(1/2)

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maxima [B]  time = 0.90, size = 487, normalized size = 3.38 \begin {gather*} \frac {B b^{3} x^{4}}{3 \, \sqrt {b x^{2} + a x}} - \frac {7 \, B a b^{2} x^{3}}{12 \, \sqrt {b x^{2} + a x}} + \frac {35 \, B a^{2} b x^{2}}{24 \, \sqrt {b x^{2} + a x}} + \frac {51 \, B a^{3} x}{8 \, \sqrt {b x^{2} + a x}} + \frac {4 \, A a^{2} b x}{\sqrt {b x^{2} + a x}} - \frac {35 \, B a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, \sqrt {b}} - \frac {2 \, A a^{3}}{\sqrt {b x^{2} + a x}} + \frac {{\left (4 \, B a b^{3} + A b^{4}\right )} x^{3}}{2 \, \sqrt {b x^{2} + a x} b} - \frac {5 \, {\left (4 \, B a b^{3} + A b^{4}\right )} a x^{2}}{4 \, \sqrt {b x^{2} + a x} b^{2}} + \frac {2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{2}}{\sqrt {b x^{2} + a x} b} - \frac {15 \, {\left (4 \, B a b^{3} + A b^{4}\right )} a^{2} x}{4 \, \sqrt {b x^{2} + a x} b^{3}} + \frac {6 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} a x}{\sqrt {b x^{2} + a x} b^{2}} - \frac {4 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x}{\sqrt {b x^{2} + a x} b} + \frac {15 \, {\left (4 \, B a b^{3} + A b^{4}\right )} a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {7}{2}}} - \frac {3 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} a \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{b^{\frac {5}{2}}} + \frac {2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*B*b^3*x^4/sqrt(b*x^2 + a*x) - 7/12*B*a*b^2*x^3/sqrt(b*x^2 + a*x) + 35/24*B*a^2*b*x^2/sqrt(b*x^2 + a*x) + 5
1/8*B*a^3*x/sqrt(b*x^2 + a*x) + 4*A*a^2*b*x/sqrt(b*x^2 + a*x) - 35/16*B*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x
)*sqrt(b))/sqrt(b) - 2*A*a^3/sqrt(b*x^2 + a*x) + 1/2*(4*B*a*b^3 + A*b^4)*x^3/(sqrt(b*x^2 + a*x)*b) - 5/4*(4*B*
a*b^3 + A*b^4)*a*x^2/(sqrt(b*x^2 + a*x)*b^2) + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*x^2/(sqrt(b*x^2 + a*x)*b) - 15/4*(4
*B*a*b^3 + A*b^4)*a^2*x/(sqrt(b*x^2 + a*x)*b^3) + 6*(3*B*a^2*b^2 + 2*A*a*b^3)*a*x/(sqrt(b*x^2 + a*x)*b^2) - 4*
(2*B*a^3*b + 3*A*a^2*b^2)*x/(sqrt(b*x^2 + a*x)*b) + 15/8*(4*B*a*b^3 + A*b^4)*a^2*log(2*b*x + a + 2*sqrt(b*x^2
+ a*x)*sqrt(b))/b^(7/2) - 3*(3*B*a^2*b^2 + 2*A*a*b^3)*a*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/2) +
2*(2*B*a^3*b + 3*A*a^2*b^2)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 64.94, size = 233, normalized size = 1.62 \begin {gather*} A \left (- \frac {2 a^{\frac {5}{2}}}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} + \frac {a^{\frac {3}{2}} b \sqrt {x}}{4 \sqrt {1 + \frac {b x}{a}}} + \frac {11 \sqrt {a} b^{2} x^{\frac {3}{2}}}{4 \sqrt {1 + \frac {b x}{a}}} + \frac {15 a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4} + \frac {b^{3} x^{\frac {5}{2}}}{2 \sqrt {a} \sqrt {1 + \frac {b x}{a}}}\right ) + B \left (\frac {11 a^{\frac {5}{2}} \sqrt {x} \sqrt {1 + \frac {b x}{a}}}{8} + \frac {13 a^{\frac {3}{2}} b x^{\frac {3}{2}} \sqrt {1 + \frac {b x}{a}}}{12} + \frac {\sqrt {a} b^{2} x^{\frac {5}{2}} \sqrt {1 + \frac {b x}{a}}}{3} + \frac {5 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 \sqrt {b}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-2*a**(5/2)/(sqrt(x)*sqrt(1 + b*x/a)) + a**(3/2)*b*sqrt(x)/(4*sqrt(1 + b*x/a)) + 11*sqrt(a)*b**2*x**(3/2)/(
4*sqrt(1 + b*x/a)) + 15*a**2*sqrt(b)*asinh(sqrt(b)*sqrt(x)/sqrt(a))/4 + b**3*x**(5/2)/(2*sqrt(a)*sqrt(1 + b*x/
a))) + B*(11*a**(5/2)*sqrt(x)*sqrt(1 + b*x/a)/8 + 13*a**(3/2)*b*x**(3/2)*sqrt(1 + b*x/a)/12 + sqrt(a)*b**2*x**
(5/2)*sqrt(1 + b*x/a)/3 + 5*a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*sqrt(b)))

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