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

Optimal. Leaf size=152 \[ -\frac {2 (a+b x)^{5/2} (3 a B+4 A b)}{3 a \sqrt {x}}+\frac {5 b \sqrt {x} (a+b x)^{3/2} (3 a B+4 A b)}{6 a}+\frac {5}{4} b \sqrt {x} \sqrt {a+b x} (3 a B+4 A b)+\frac {5}{4} a \sqrt {b} (3 a B+4 A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 A (a+b x)^{7/2}}{3 a x^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \begin {gather*} -\frac {2 (a+b x)^{5/2} (3 a B+4 A b)}{3 a \sqrt {x}}+\frac {5 b \sqrt {x} (a+b x)^{3/2} (3 a B+4 A b)}{6 a}+\frac {5}{4} b \sqrt {x} \sqrt {a+b x} (3 a B+4 A b)+\frac {5}{4} a \sqrt {b} (3 a B+4 A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 A (a+b x)^{7/2}}{3 a x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b*(4*A*b + 3*a*B)*Sqrt[x]*Sqrt[a + b*x])/4 + (5*b*(4*A*b + 3*a*B)*Sqrt[x]*(a + b*x)^(3/2))/(6*a) - (2*(4*A*
b + 3*a*B)*(a + b*x)^(5/2))/(3*a*Sqrt[x]) - (2*A*(a + b*x)^(7/2))/(3*a*x^(3/2)) + (5*a*Sqrt[b]*(4*A*b + 3*a*B)
*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/4

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

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 76, normalized size = 0.50 \begin {gather*} \frac {2 \sqrt {a+b x} \left (-\frac {a^2 x (3 a B+4 A b) \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};-\frac {b x}{a}\right )}{\sqrt {\frac {b x}{a}+1}}-A (a+b x)^3\right )}{3 a x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.44, size = 113, normalized size = 0.74 \begin {gather*} \frac {\sqrt {a+b x} \left (-8 a^2 A-24 a^2 B x-56 a A b x+27 a b B x^2+12 A b^2 x^2+6 b^2 B x^3\right )}{12 x^{3/2}}-\frac {5}{4} \left (3 a^2 \sqrt {b} B+4 a A b^{3/2}\right ) \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(-8*a^2*A - 56*a*A*b*x - 24*a^2*B*x + 12*A*b^2*x^2 + 27*a*b*B*x^2 + 6*b^2*B*x^3))/(12*x^(3/2))
- (5*(4*a*A*b^(3/2) + 3*a^2*Sqrt[b]*B)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/4

________________________________________________________________________________________

fricas [A]  time = 1.44, size = 217, normalized size = 1.43 \begin {gather*} \left [\frac {15 \, {\left (3 \, B a^{2} + 4 \, A 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 (6 \, B b^{2} x^{3} - 8 \, A a^{2} + 3 \, {\left (9 \, B a b + 4 \, A b^{2}\right )} x^{2} - 8 \, {\left (3 \, B a^{2} + 7 \, A a b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{24 \, x^{2}}, -\frac {15 \, {\left (3 \, B a^{2} + 4 \, A a b\right )} \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (6 \, B b^{2} x^{3} - 8 \, A a^{2} + 3 \, {\left (9 \, B a b + 4 \, A b^{2}\right )} x^{2} - 8 \, {\left (3 \, B a^{2} + 7 \, A a b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{12 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(15*(3*B*a^2 + 4*A*a*b)*sqrt(b)*x^2*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(6*B*b^2*x^3 -
8*A*a^2 + 3*(9*B*a*b + 4*A*b^2)*x^2 - 8*(3*B*a^2 + 7*A*a*b)*x)*sqrt(b*x + a)*sqrt(x))/x^2, -1/12*(15*(3*B*a^2
+ 4*A*a*b)*sqrt(-b)*x^2*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (6*B*b^2*x^3 - 8*A*a^2 + 3*(9*B*a*b + 4*A
*b^2)*x^2 - 8*(3*B*a^2 + 7*A*a*b)*x)*sqrt(b*x + a)*sqrt(x))/x^2]

________________________________________________________________________________________

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^(5/2),x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.01, size = 207, normalized size = 1.36 \begin {gather*} \frac {\sqrt {b x +a}\, \left (60 A 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 )+45 B \,a^{2} b \,x^{2} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+12 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {5}{2}} x^{3}+24 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {5}{2}} x^{2}+54 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {3}{2}} x^{2}-112 \sqrt {\left (b x +a \right ) x}\, A a \,b^{\frac {3}{2}} x -48 \sqrt {\left (b x +a \right ) x}\, B \,a^{2} \sqrt {b}\, x -16 \sqrt {\left (b x +a \right ) x}\, A \,a^{2} \sqrt {b}\right )}{24 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}\, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b*x+a)^(1/2)/x^(3/2)*(12*((b*x+a)*x)^(1/2)*B*b^(5/2)*x^3+60*A*b^2*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^
(1/2))/b^(1/2))*x^2*a+24*((b*x+a)*x)^(1/2)*A*b^(5/2)*x^2+45*B*b*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b
^(1/2))*x^2*a^2+54*((b*x+a)*x)^(1/2)*B*a*b^(3/2)*x^2-112*((b*x+a)*x)^(1/2)*A*a*b^(3/2)*x-48*((b*x+a)*x)^(1/2)*
B*a^2*b^(1/2)*x-16*((b*x+a)*x)^(1/2)*A*a^2*b^(1/2))/((b*x+a)*x)^(1/2)/b^(1/2)

________________________________________________________________________________________

maxima [A]  time = 0.99, size = 191, normalized size = 1.26 \begin {gather*} \frac {15}{8} \, B a^{2} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + \frac {5}{2} \, A a b^{\frac {3}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {15 \, \sqrt {b x^{2} + a x} B a^{2}}{4 \, x} - \frac {35 \, \sqrt {b x^{2} + a x} A a b}{6 \, x} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{4 \, x^{2}} - \frac {5 \, \sqrt {b x^{2} + a x} A a^{2}}{6 \, x^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B}{2 \, x^{3}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{6 \, x^{3}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} A}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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^{5/2}} \,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^(5/2),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 68.98, size = 230, normalized size = 1.51 \begin {gather*} A \left (- \frac {2 a^{2} \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {14 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3} - \frac {5 a b^{\frac {3}{2}} \log {\left (\frac {a}{b x} \right )}}{2} + 5 a b^{\frac {3}{2}} \log {\left (\sqrt {\frac {a}{b x} + 1} + 1 \right )} + b^{\frac {5}{2}} x \sqrt {\frac {a}{b x} + 1}\right ) + B \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-2*a**2*sqrt(b)*sqrt(a/(b*x) + 1)/(3*x) - 14*a*b**(3/2)*sqrt(a/(b*x) + 1)/3 - 5*a*b**(3/2)*log(a/(b*x))/2 +
5*a*b**(3/2)*log(sqrt(a/(b*x) + 1) + 1) + b**(5/2)*x*sqrt(a/(b*x) + 1)) + B*(-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)))

________________________________________________________________________________________