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

Optimal. Leaf size=123 \[ \frac {3 (A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{7/2}}-\frac {3 \sqrt {x} (A b-5 a B)}{4 a b^3}+\frac {x^{3/2} (A b-5 a B)}{4 a b^2 (a+b x)}+\frac {x^{5/2} (A b-a B)}{2 a b (a+b x)^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {78, 47, 50, 63, 205} \begin {gather*} \frac {x^{3/2} (A b-5 a B)}{4 a b^2 (a+b x)}-\frac {3 \sqrt {x} (A b-5 a B)}{4 a b^3}+\frac {3 (A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{7/2}}+\frac {x^{5/2} (A b-a B)}{2 a b (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(A*b - 5*a*B)*Sqrt[x])/(4*a*b^3) + ((A*b - a*B)*x^(5/2))/(2*a*b*(a + b*x)^2) + ((A*b - 5*a*B)*x^(3/2))/(4*
a*b^2*(a + b*x)) + (3*(A*b - 5*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*Sqrt[a]*b^(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 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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

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Mathematica [C]  time = 0.03, size = 61, normalized size = 0.50 \begin {gather*} \frac {x^{5/2} \left (\frac {5 a^2 (A b-a B)}{(a+b x)^2}+(5 a B-A b) \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};-\frac {b x}{a}\right )\right )}{10 a^3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(5/2)*((5*a^2*(A*b - a*B))/(a + b*x)^2 + (-(A*b) + 5*a*B)*Hypergeometric2F1[2, 5/2, 7/2, -((b*x)/a)]))/(10*
a^3*b)

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IntegrateAlgebraic [A]  time = 0.16, size = 94, normalized size = 0.76 \begin {gather*} \frac {\sqrt {x} \left (15 a^2 B-3 a A b+25 a b B x-5 A b^2 x+8 b^2 B x^2\right )}{4 b^3 (a+b x)^2}-\frac {3 (5 a B-A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(-3*a*A*b + 15*a^2*B - 5*A*b^2*x + 25*a*b*B*x + 8*b^2*B*x^2))/(4*b^3*(a + b*x)^2) - (3*(-(A*b) + 5*a*
B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*Sqrt[a]*b^(7/2))

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fricas [A]  time = 1.17, size = 319, normalized size = 2.59 \begin {gather*} \left [\frac {3 \, {\left (5 \, B a^{3} - A a^{2} b + {\left (5 \, B a b^{2} - A b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (8 \, B a b^{3} x^{2} + 15 \, B a^{3} b - 3 \, A a^{2} b^{2} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{8 \, {\left (a b^{6} x^{2} + 2 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}, \frac {3 \, {\left (5 \, B a^{3} - A a^{2} b + {\left (5 \, B a b^{2} - A b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (8 \, B a b^{3} x^{2} + 15 \, B a^{3} b - 3 \, A a^{2} b^{2} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{4 \, {\left (a b^{6} x^{2} + 2 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*(5*B*a^3 - A*a^2*b + (5*B*a*b^2 - A*b^3)*x^2 + 2*(5*B*a^2*b - A*a*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*
sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(8*B*a*b^3*x^2 + 15*B*a^3*b - 3*A*a^2*b^2 + 5*(5*B*a^2*b^2 - A*a*b^3)*x)*sq
rt(x))/(a*b^6*x^2 + 2*a^2*b^5*x + a^3*b^4), 1/4*(3*(5*B*a^3 - A*a^2*b + (5*B*a*b^2 - A*b^3)*x^2 + 2*(5*B*a^2*b
- A*a*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (8*B*a*b^3*x^2 + 15*B*a^3*b - 3*A*a^2*b^2 + 5*(5*B*a^
2*b^2 - A*a*b^3)*x)*sqrt(x))/(a*b^6*x^2 + 2*a^2*b^5*x + a^3*b^4)]

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giac [A]  time = 1.23, size = 87, normalized size = 0.71 \begin {gather*} \frac {2 \, B \sqrt {x}}{b^{3}} - \frac {3 \, {\left (5 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} + \frac {9 \, B a b x^{\frac {3}{2}} - 5 \, A b^{2} x^{\frac {3}{2}} + 7 \, B a^{2} \sqrt {x} - 3 \, A a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B*sqrt(x)/b^3 - 3/4*(5*B*a - A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/4*(9*B*a*b*x^(3/2) - 5*A*b
^2*x^(3/2) + 7*B*a^2*sqrt(x) - 3*A*a*b*sqrt(x))/((b*x + a)^2*b^3)

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maple [A]  time = 0.02, size = 125, normalized size = 1.02 \begin {gather*} -\frac {5 A \,x^{\frac {3}{2}}}{4 \left (b x +a \right )^{2} b}+\frac {9 B a \,x^{\frac {3}{2}}}{4 \left (b x +a \right )^{2} b^{2}}-\frac {3 A a \sqrt {x}}{4 \left (b x +a \right )^{2} b^{2}}+\frac {7 B \,a^{2} \sqrt {x}}{4 \left (b x +a \right )^{2} b^{3}}+\frac {3 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, b^{2}}-\frac {15 B a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, b^{3}}+\frac {2 B \sqrt {x}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B/b^3*x^(1/2)-5/4/b/(b*x+a)^2*x^(3/2)*A+9/4/b^2/(b*x+a)^2*x^(3/2)*B*a-3/4/b^2/(b*x+a)^2*A*x^(1/2)*a+7/4/b^3/
(b*x+a)^2*B*x^(1/2)*a^2+3/4/b^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*A-15/4/b^3/(a*b)^(1/2)*arctan(1/(a
*b)^(1/2)*b*x^(1/2))*B*a

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maxima [A]  time = 1.89, size = 99, normalized size = 0.80 \begin {gather*} \frac {{\left (9 \, B a b - 5 \, A b^{2}\right )} x^{\frac {3}{2}} + {\left (7 \, B a^{2} - 3 \, A a b\right )} \sqrt {x}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} + \frac {2 \, B \sqrt {x}}{b^{3}} - \frac {3 \, {\left (5 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((9*B*a*b - 5*A*b^2)*x^(3/2) + (7*B*a^2 - 3*A*a*b)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3) + 2*B*sqrt(x)/
b^3 - 3/4*(5*B*a - A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3)

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mupad [B]  time = 0.43, size = 96, normalized size = 0.78 \begin {gather*} \frac {\sqrt {x}\,\left (\frac {7\,B\,a^2}{4}-\frac {3\,A\,a\,b}{4}\right )-x^{3/2}\,\left (\frac {5\,A\,b^2}{4}-\frac {9\,B\,a\,b}{4}\right )}{a^2\,b^3+2\,a\,b^4\,x+b^5\,x^2}+\frac {2\,B\,\sqrt {x}}{b^3}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-5\,B\,a\right )}{4\,\sqrt {a}\,b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^(1/2)*((7*B*a^2)/4 - (3*A*a*b)/4) - x^(3/2)*((5*A*b^2)/4 - (9*B*a*b)/4))/(a^2*b^3 + b^5*x^2 + 2*a*b^4*x) +
(2*B*x^(1/2))/b^3 + (3*atan((b^(1/2)*x^(1/2))/a^(1/2))*(A*b - 5*B*a))/(4*a^(1/2)*b^(7/2))

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sympy [A]  time = 35.50, size = 1586, normalized size = 12.89

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/sqrt(x) + 2*B*sqrt(x)), Eq(a, 0) & Eq(b, 0)), ((-2*A/sqrt(x) + 2*B*sqrt(x))/b**3, Eq(a, 0
)), ((2*A*x**(5/2)/5 + 2*B*x**(7/2)/7)/a**3, Eq(b, 0)), (-6*I*A*a**(3/2)*b**2*sqrt(x)*sqrt(1/b)/(8*I*a**(5/2)*
b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) - 10*I*A*sqrt(a)*b**3*x**(3
/2)*sqrt(1/b)/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b))
+ 3*A*a**2*b*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b
) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) - 3*A*a**2*b*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(5/2)*b**4*sqrt(1
/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) + 6*A*a*b**2*x*log(-I*sqrt(a)*sqrt(1/b
) + sqrt(x))/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b))
- 6*A*a*b**2*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b
) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) + 3*A*b**3*x**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(5/2)*b**4*sq
rt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) - 3*A*b**3*x**2*log(I*sqrt(a)*sqrt
(1/b) + sqrt(x))/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/
b)) + 30*I*B*a**(5/2)*b*sqrt(x)*sqrt(1/b)/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*
sqrt(a)*b**6*x**2*sqrt(1/b)) + 50*I*B*a**(3/2)*b**2*x**(3/2)*sqrt(1/b)/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**
(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) + 16*I*B*sqrt(a)*b**3*x**(5/2)*sqrt(1/b)/(8*I*a**(5/
2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) - 15*B*a**3*log(-I*sqrt(
a)*sqrt(1/b) + sqrt(x))/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*
sqrt(1/b)) + 15*B*a**3*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*
sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) - 30*B*a**2*b*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(5/2)
*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) + 30*B*a**2*b*x*log(I*sqrt
(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2
*sqrt(1/b)) - 15*B*a*b**2*x**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2
)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) + 15*B*a*b**2*x**2*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8
*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)), True))

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