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

Optimal. Leaf size=167 \[ \frac {5 a^2 (6 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{9/2}}-\frac {5 a \sqrt {x} \sqrt {a+b x} (6 A b-7 a B)}{8 b^4}+\frac {5 x^{3/2} \sqrt {a+b x} (6 A b-7 a B)}{12 b^3}-\frac {x^{5/2} \sqrt {a+b x} (6 A b-7 a B)}{3 a b^2}+\frac {2 x^{7/2} (A b-a B)}{a b \sqrt {a+b x}} \]

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Rubi [A]  time = 0.07, antiderivative size = 167, 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 (6 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{9/2}}-\frac {x^{5/2} \sqrt {a+b x} (6 A b-7 a B)}{3 a b^2}+\frac {5 x^{3/2} \sqrt {a+b x} (6 A b-7 a B)}{12 b^3}-\frac {5 a \sqrt {x} \sqrt {a+b x} (6 A b-7 a B)}{8 b^4}+\frac {2 x^{7/2} (A b-a B)}{a b \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b - a*B)*x^(7/2))/(a*b*Sqrt[a + b*x]) - (5*a*(6*A*b - 7*a*B)*Sqrt[x]*Sqrt[a + b*x])/(8*b^4) + (5*(6*A*b
- 7*a*B)*x^(3/2)*Sqrt[a + b*x])/(12*b^3) - ((6*A*b - 7*a*B)*x^(5/2)*Sqrt[a + b*x])/(3*a*b^2) + (5*a^2*(6*A*b -
7*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(8*b^(9/2))

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 {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx &=\frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}-\frac {\left (2 \left (3 A b-\frac {7 a B}{2}\right )\right ) \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx}{a b}\\ &=\frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}-\frac {(6 A b-7 a B) x^{5/2} \sqrt {a+b x}}{3 a b^2}+\frac {(5 (6 A b-7 a B)) \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{6 b^2}\\ &=\frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}+\frac {5 (6 A b-7 a B) x^{3/2} \sqrt {a+b x}}{12 b^3}-\frac {(6 A b-7 a B) x^{5/2} \sqrt {a+b x}}{3 a b^2}-\frac {(5 a (6 A b-7 a B)) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{8 b^3}\\ &=\frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}-\frac {5 a (6 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{8 b^4}+\frac {5 (6 A b-7 a B) x^{3/2} \sqrt {a+b x}}{12 b^3}-\frac {(6 A b-7 a B) x^{5/2} \sqrt {a+b x}}{3 a b^2}+\frac {\left (5 a^2 (6 A b-7 a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{16 b^4}\\ &=\frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}-\frac {5 a (6 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{8 b^4}+\frac {5 (6 A b-7 a B) x^{3/2} \sqrt {a+b x}}{12 b^3}-\frac {(6 A b-7 a B) x^{5/2} \sqrt {a+b x}}{3 a b^2}+\frac {\left (5 a^2 (6 A b-7 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{8 b^4}\\ &=\frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}-\frac {5 a (6 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{8 b^4}+\frac {5 (6 A b-7 a B) x^{3/2} \sqrt {a+b x}}{12 b^3}-\frac {(6 A b-7 a B) x^{5/2} \sqrt {a+b x}}{3 a b^2}+\frac {\left (5 a^2 (6 A b-7 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^4}\\ &=\frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}-\frac {5 a (6 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{8 b^4}+\frac {5 (6 A b-7 a B) x^{3/2} \sqrt {a+b x}}{12 b^3}-\frac {(6 A b-7 a B) x^{5/2} \sqrt {a+b x}}{3 a b^2}+\frac {5 a^2 (6 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 140, normalized size = 0.84 \begin {gather*} \frac {\frac {(a+b x) (7 a B-6 A b) \left (b x \sqrt {\frac {b x}{a}+1} \left (15 a^2-10 a b x+8 b^2 x^2\right )-15 a^{5/2} \sqrt {b} \sqrt {x} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{3 \sqrt {\frac {b x}{a}+1}}+16 b^4 x^4 (A b-a B)}{8 a b^5 \sqrt {x} \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(16*b^4*(A*b - a*B)*x^4 + ((-6*A*b + 7*a*B)*(a + b*x)*(b*x*Sqrt[1 + (b*x)/a]*(15*a^2 - 10*a*b*x + 8*b^2*x^2) -
15*a^(5/2)*Sqrt[b]*Sqrt[x]*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(3*Sqrt[1 + (b*x)/a]))/(8*a*b^5*Sqrt[x]*Sqrt[
a + b*x])

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IntegrateAlgebraic [A]  time = 0.27, size = 145, normalized size = 0.87 \begin {gather*} \frac {5 \left (7 a^3 B-6 a^2 A b\right ) \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{8 b^{9/2}}+\frac {105 a^3 B \sqrt {x}-90 a^2 A b \sqrt {x}+35 a^2 b B x^{3/2}-30 a A b^2 x^{3/2}-14 a b^2 B x^{5/2}+12 A b^3 x^{5/2}+8 b^3 B x^{7/2}}{24 b^4 \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-90*a^2*A*b*Sqrt[x] + 105*a^3*B*Sqrt[x] - 30*a*A*b^2*x^(3/2) + 35*a^2*b*B*x^(3/2) + 12*A*b^3*x^(5/2) - 14*a*b
^2*B*x^(5/2) + 8*b^3*B*x^(7/2))/(24*b^4*Sqrt[a + b*x]) + (5*(-6*a^2*A*b + 7*a^3*B)*Log[-(Sqrt[b]*Sqrt[x]) + Sq
rt[a + b*x]])/(8*b^(9/2))

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fricas [A]  time = 1.68, size = 306, normalized size = 1.83 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B a^{4} - 6 \, A a^{3} b + {\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (8 \, B b^{4} x^{3} + 105 \, B a^{3} b - 90 \, A a^{2} b^{2} - 2 \, {\left (7 \, B a b^{3} - 6 \, A b^{4}\right )} x^{2} + 5 \, {\left (7 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{48 \, {\left (b^{6} x + a b^{5}\right )}}, \frac {15 \, {\left (7 \, B a^{4} - 6 \, A a^{3} b + {\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (8 \, B b^{4} x^{3} + 105 \, B a^{3} b - 90 \, A a^{2} b^{2} - 2 \, {\left (7 \, B a b^{3} - 6 \, A b^{4}\right )} x^{2} + 5 \, {\left (7 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{24 \, {\left (b^{6} x + a b^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(15*(7*B*a^4 - 6*A*a^3*b + (7*B*a^3*b - 6*A*a^2*b^2)*x)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqr
t(x) + a) - 2*(8*B*b^4*x^3 + 105*B*a^3*b - 90*A*a^2*b^2 - 2*(7*B*a*b^3 - 6*A*b^4)*x^2 + 5*(7*B*a^2*b^2 - 6*A*a
*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^6*x + a*b^5), 1/24*(15*(7*B*a^4 - 6*A*a^3*b + (7*B*a^3*b - 6*A*a^2*b^2)*x)*
sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (8*B*b^4*x^3 + 105*B*a^3*b - 90*A*a^2*b^2 - 2*(7*B*a*b^3
- 6*A*b^4)*x^2 + 5*(7*B*a^2*b^2 - 6*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^6*x + a*b^5)]

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giac [A]  time = 107.34, size = 217, normalized size = 1.30 \begin {gather*} \frac {1}{24} \, \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} B {\left | b \right |}}{b^{6}} - \frac {19 \, B a b^{17} {\left | b \right |} - 6 \, A b^{18} {\left | b \right |}}{b^{23}}\right )} + \frac {3 \, {\left (29 \, B a^{2} b^{17} {\left | b \right |} - 18 \, A a b^{18} {\left | b \right |}\right )}}{b^{23}}\right )} + \frac {5 \, {\left (7 \, B a^{3} \sqrt {b} {\left | b \right |} - 6 \, A a^{2} b^{\frac {3}{2}} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{16 \, b^{6}} + \frac {4 \, {\left (B a^{4} \sqrt {b} {\left | b \right |} - A a^{3} b^{\frac {3}{2}} {\left | b \right |}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)*B*abs(b)/b^6 - (19*B*a*b^17*abs(b) - 6*A*
b^18*abs(b))/b^23) + 3*(29*B*a^2*b^17*abs(b) - 18*A*a*b^18*abs(b))/b^23) + 5/16*(7*B*a^3*sqrt(b)*abs(b) - 6*A*
a^2*b^(3/2)*abs(b))*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/b^6 + 4*(B*a^4*sqrt(b)*abs(b) - A
*a^3*b^(3/2)*abs(b))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)*b^5)

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maple [B]  time = 0.02, size = 288, normalized size = 1.72 \begin {gather*} \frac {\left (16 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {7}{2}} x^{3}+90 A \,a^{2} b^{2} x \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-105 B \,a^{3} b 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 {7}{2}} x^{2}-28 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {5}{2}} x^{2}+90 A \,a^{3} b \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-105 B \,a^{4} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-60 \sqrt {\left (b x +a \right ) x}\, A a \,b^{\frac {5}{2}} x +70 \sqrt {\left (b x +a \right ) x}\, B \,a^{2} b^{\frac {3}{2}} x -180 \sqrt {\left (b x +a \right ) x}\, A \,a^{2} b^{\frac {3}{2}}+210 \sqrt {\left (b x +a \right ) x}\, B \,a^{3} \sqrt {b}\right ) \sqrt {x}}{48 \sqrt {\left (b x +a \right ) x}\, \sqrt {b x +a}\, b^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*((b*x+a)*x)^(1/2)*B*b^(7/2)*x^3+24*((b*x+a)*x)^(1/2)*A*b^(7/2)*x^2-28*((b*x+a)*x)^(1/2)*B*a*b^(5/2)*x
^2+90*A*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x*a^2*b^2-60*((b*x+a)*x)^(1/2)*A*a*b^(5/2)*x-105
*B*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x*a^3*b+70*((b*x+a)*x)^(1/2)*B*a^2*b^(3/2)*x+90*A*a^3
*b*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))-180*((b*x+a)*x)^(1/2)*A*a^2*b^(3/2)-105*B*a^4*ln(1/2*
(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))+210*((b*x+a)*x)^(1/2)*B*a^3*b^(1/2))/b^(9/2)*x^(1/2)/((b*x+a)*x
)^(1/2)/(b*x+a)^(1/2)

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maxima [A]  time = 0.92, size = 212, normalized size = 1.27 \begin {gather*} \frac {B x^{4}}{3 \, \sqrt {b x^{2} + a x} b} - \frac {7 \, B a x^{3}}{12 \, \sqrt {b x^{2} + a x} b^{2}} + \frac {A x^{3}}{2 \, \sqrt {b x^{2} + a x} b} + \frac {35 \, B a^{2} x^{2}}{24 \, \sqrt {b x^{2} + a x} b^{3}} - \frac {5 \, A a x^{2}}{4 \, \sqrt {b x^{2} + a x} b^{2}} + \frac {35 \, B a^{3} x}{8 \, \sqrt {b x^{2} + a x} b^{4}} - \frac {15 \, A a^{2} x}{4 \, \sqrt {b x^{2} + a x} b^{3}} - \frac {35 \, B a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {9}{2}}} + \frac {15 \, A a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*B*x^4/(sqrt(b*x^2 + a*x)*b) - 7/12*B*a*x^3/(sqrt(b*x^2 + a*x)*b^2) + 1/2*A*x^3/(sqrt(b*x^2 + a*x)*b) + 35/
24*B*a^2*x^2/(sqrt(b*x^2 + a*x)*b^3) - 5/4*A*a*x^2/(sqrt(b*x^2 + a*x)*b^2) + 35/8*B*a^3*x/(sqrt(b*x^2 + a*x)*b
^4) - 15/4*A*a^2*x/(sqrt(b*x^2 + a*x)*b^3) - 35/16*B*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(9/2)
+ 15/8*A*a^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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