3.498 \(\int \frac{(a+b x)^{3/2} (A+B x)}{x^{11/2}} \, dx\)

Optimal. Leaf size=84 \[ -\frac{4 b (a+b x)^{5/2} (4 A b-9 a B)}{315 a^3 x^{5/2}}+\frac{2 (a+b x)^{5/2} (4 A b-9 a B)}{63 a^2 x^{7/2}}-\frac{2 A (a+b x)^{5/2}}{9 a x^{9/2}} \]

[Out]

(-2*A*(a + b*x)^(5/2))/(9*a*x^(9/2)) + (2*(4*A*b - 9*a*B)*(a + b*x)^(5/2))/(63*a^2*x^(7/2)) - (4*b*(4*A*b - 9*
a*B)*(a + b*x)^(5/2))/(315*a^3*x^(5/2))

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Rubi [A]  time = 0.0272937, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {78, 45, 37} \[ -\frac{4 b (a+b x)^{5/2} (4 A b-9 a B)}{315 a^3 x^{5/2}}+\frac{2 (a+b x)^{5/2} (4 A b-9 a B)}{63 a^2 x^{7/2}}-\frac{2 A (a+b x)^{5/2}}{9 a x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(3/2)*(A + B*x))/x^(11/2),x]

[Out]

(-2*A*(a + b*x)^(5/2))/(9*a*x^(9/2)) + (2*(4*A*b - 9*a*B)*(a + b*x)^(5/2))/(63*a^2*x^(7/2)) - (4*b*(4*A*b - 9*
a*B)*(a + b*x)^(5/2))/(315*a^3*x^(5/2))

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 45

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{(a+b x)^{3/2} (A+B x)}{x^{11/2}} \, dx &=-\frac{2 A (a+b x)^{5/2}}{9 a x^{9/2}}+\frac{\left (2 \left (-2 A b+\frac{9 a B}{2}\right )\right ) \int \frac{(a+b x)^{3/2}}{x^{9/2}} \, dx}{9 a}\\ &=-\frac{2 A (a+b x)^{5/2}}{9 a x^{9/2}}+\frac{2 (4 A b-9 a B) (a+b x)^{5/2}}{63 a^2 x^{7/2}}+\frac{(2 b (4 A b-9 a B)) \int \frac{(a+b x)^{3/2}}{x^{7/2}} \, dx}{63 a^2}\\ &=-\frac{2 A (a+b x)^{5/2}}{9 a x^{9/2}}+\frac{2 (4 A b-9 a B) (a+b x)^{5/2}}{63 a^2 x^{7/2}}-\frac{4 b (4 A b-9 a B) (a+b x)^{5/2}}{315 a^3 x^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0241387, size = 57, normalized size = 0.68 \[ -\frac{2 (a+b x)^{5/2} \left (5 a^2 (7 A+9 B x)-2 a b x (10 A+9 B x)+8 A b^2 x^2\right )}{315 a^3 x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(3/2)*(A + B*x))/x^(11/2),x]

[Out]

(-2*(a + b*x)^(5/2)*(8*A*b^2*x^2 + 5*a^2*(7*A + 9*B*x) - 2*a*b*x*(10*A + 9*B*x)))/(315*a^3*x^(9/2))

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Maple [A]  time = 0.003, size = 53, normalized size = 0.6 \begin{align*} -{\frac{16\,A{b}^{2}{x}^{2}-36\,B{x}^{2}ab-40\,aAbx+90\,{a}^{2}Bx+70\,A{a}^{2}}{315\,{a}^{3}} \left ( bx+a \right ) ^{{\frac{5}{2}}}{x}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(B*x+A)/x^(11/2),x)

[Out]

-2/315*(b*x+a)^(5/2)*(8*A*b^2*x^2-18*B*a*b*x^2-20*A*a*b*x+45*B*a^2*x+35*A*a^2)/x^(9/2)/a^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^(11/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.56902, size = 231, normalized size = 2.75 \begin{align*} -\frac{2 \,{\left (35 \, A a^{4} - 2 \,{\left (9 \, B a b^{3} - 4 \, A b^{4}\right )} x^{4} +{\left (9 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{3} + 3 \,{\left (24 \, B a^{3} b + A a^{2} b^{2}\right )} x^{2} + 5 \,{\left (9 \, B a^{4} + 10 \, A a^{3} b\right )} x\right )} \sqrt{b x + a}}{315 \, a^{3} x^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^(11/2),x, algorithm="fricas")

[Out]

-2/315*(35*A*a^4 - 2*(9*B*a*b^3 - 4*A*b^4)*x^4 + (9*B*a^2*b^2 - 4*A*a*b^3)*x^3 + 3*(24*B*a^3*b + A*a^2*b^2)*x^
2 + 5*(9*B*a^4 + 10*A*a^3*b)*x)*sqrt(b*x + a)/(a^3*x^(9/2))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)*(B*x+A)/x**(11/2),x)

[Out]

Timed out

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Giac [A]  time = 1.53667, size = 161, normalized size = 1.92 \begin{align*} -\frac{{\left (b x + a\right )}^{\frac{5}{2}}{\left ({\left (b x + a\right )}{\left (\frac{2 \,{\left (9 \, B a^{2} b^{8} - 4 \, A a b^{9}\right )}{\left (b x + a\right )}}{a^{5} b^{15}} - \frac{9 \,{\left (9 \, B a^{3} b^{8} - 4 \, A a^{2} b^{9}\right )}}{a^{5} b^{15}}\right )} + \frac{63 \,{\left (B a^{4} b^{8} - A a^{3} b^{9}\right )}}{a^{5} b^{15}}\right )} b}{322560 \,{\left ({\left (b x + a\right )} b - a b\right )}^{\frac{9}{2}}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^(11/2),x, algorithm="giac")

[Out]

-1/322560*(b*x + a)^(5/2)*((b*x + a)*(2*(9*B*a^2*b^8 - 4*A*a*b^9)*(b*x + a)/(a^5*b^15) - 9*(9*B*a^3*b^8 - 4*A*
a^2*b^9)/(a^5*b^15)) + 63*(B*a^4*b^8 - A*a^3*b^9)/(a^5*b^15))*b/(((b*x + a)*b - a*b)^(9/2)*abs(b))