3.510 \(\int \frac{(a+b x)^{5/2} (A+B x)}{x^{13/2}} \, dx\)

Optimal. Leaf size=84 \[ -\frac{4 b (a+b x)^{7/2} (4 A b-11 a B)}{693 a^3 x^{7/2}}+\frac{2 (a+b x)^{7/2} (4 A b-11 a B)}{99 a^2 x^{9/2}}-\frac{2 A (a+b x)^{7/2}}{11 a x^{11/2}} \]

[Out]

(-2*A*(a + b*x)^(7/2))/(11*a*x^(11/2)) + (2*(4*A*b - 11*a*B)*(a + b*x)^(7/2))/(99*a^2*x^(9/2)) - (4*b*(4*A*b -
11*a*B)*(a + b*x)^(7/2))/(693*a^3*x^(7/2))

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Rubi [A]  time = 0.0268538, 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} \[ -\frac{4 b (a+b x)^{7/2} (4 A b-11 a B)}{693 a^3 x^{7/2}}+\frac{2 (a+b x)^{7/2} (4 A b-11 a B)}{99 a^2 x^{9/2}}-\frac{2 A (a+b x)^{7/2}}{11 a x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x^(13/2),x]

[Out]

(-2*A*(a + b*x)^(7/2))/(11*a*x^(11/2)) + (2*(4*A*b - 11*a*B)*(a + b*x)^(7/2))/(99*a^2*x^(9/2)) - (4*b*(4*A*b -
11*a*B)*(a + b*x)^(7/2))/(693*a^3*x^(7/2))

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)^{5/2} (A+B x)}{x^{13/2}} \, dx &=-\frac{2 A (a+b x)^{7/2}}{11 a x^{11/2}}+\frac{\left (2 \left (-2 A b+\frac{11 a B}{2}\right )\right ) \int \frac{(a+b x)^{5/2}}{x^{11/2}} \, dx}{11 a}\\ &=-\frac{2 A (a+b x)^{7/2}}{11 a x^{11/2}}+\frac{2 (4 A b-11 a B) (a+b x)^{7/2}}{99 a^2 x^{9/2}}+\frac{(2 b (4 A b-11 a B)) \int \frac{(a+b x)^{5/2}}{x^{9/2}} \, dx}{99 a^2}\\ &=-\frac{2 A (a+b x)^{7/2}}{11 a x^{11/2}}+\frac{2 (4 A b-11 a B) (a+b x)^{7/2}}{99 a^2 x^{9/2}}-\frac{4 b (4 A b-11 a B) (a+b x)^{7/2}}{693 a^3 x^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0280224, size = 57, normalized size = 0.68 \[ -\frac{2 (a+b x)^{7/2} \left (7 a^2 (9 A+11 B x)-2 a b x (14 A+11 B x)+8 A b^2 x^2\right )}{693 a^3 x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^(13/2),x]

[Out]

(-2*(a + b*x)^(7/2)*(8*A*b^2*x^2 + 7*a^2*(9*A + 11*B*x) - 2*a*b*x*(14*A + 11*B*x)))/(693*a^3*x^(11/2))

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Maple [A]  time = 0.003, size = 53, normalized size = 0.6 \begin{align*} -{\frac{16\,A{b}^{2}{x}^{2}-44\,B{x}^{2}ab-56\,aAbx+154\,{a}^{2}Bx+126\,A{a}^{2}}{693\,{a}^{3}} \left ( bx+a \right ) ^{{\frac{7}{2}}}{x}^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^(13/2),x)

[Out]

-2/693*(b*x+a)^(7/2)*(8*A*b^2*x^2-22*B*a*b*x^2-28*A*a*b*x+77*B*a^2*x+63*A*a^2)/x^(11/2)/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)^(5/2)*(B*x+A)/x^(13/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.53569, size = 288, normalized size = 3.43 \begin{align*} -\frac{2 \,{\left (63 \, A a^{5} - 2 \,{\left (11 \, B a b^{4} - 4 \, A b^{5}\right )} x^{5} +{\left (11 \, B a^{2} b^{3} - 4 \, A a b^{4}\right )} x^{4} + 3 \,{\left (55 \, B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} +{\left (209 \, B a^{4} b + 113 \, A a^{3} b^{2}\right )} x^{2} + 7 \,{\left (11 \, B a^{5} + 23 \, A a^{4} b\right )} x\right )} \sqrt{b x + a}}{693 \, a^{3} x^{\frac{11}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(13/2),x, algorithm="fricas")

[Out]

-2/693*(63*A*a^5 - 2*(11*B*a*b^4 - 4*A*b^5)*x^5 + (11*B*a^2*b^3 - 4*A*a*b^4)*x^4 + 3*(55*B*a^3*b^2 + A*a^2*b^3
)*x^3 + (209*B*a^4*b + 113*A*a^3*b^2)*x^2 + 7*(11*B*a^5 + 23*A*a^4*b)*x)*sqrt(b*x + a)/(a^3*x^(11/2))

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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)**(5/2)*(B*x+A)/x**(13/2),x)

[Out]

Timed out

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Giac [A]  time = 1.38203, size = 163, normalized size = 1.94 \begin{align*} -\frac{{\left (b x + a\right )}^{\frac{7}{2}}{\left ({\left (b x + a\right )}{\left (\frac{2 \,{\left (11 \, B a^{3} b^{10} - 4 \, A a^{2} b^{11}\right )}{\left (b x + a\right )}}{a^{6} b^{18}} - \frac{11 \,{\left (11 \, B a^{4} b^{10} - 4 \, A a^{3} b^{11}\right )}}{a^{6} b^{18}}\right )} + \frac{99 \,{\left (B a^{5} b^{10} - A a^{4} b^{11}\right )}}{a^{6} b^{18}}\right )} b}{2838528 \,{\left ({\left (b x + a\right )} b - a b\right )}^{\frac{11}{2}}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(13/2),x, algorithm="giac")

[Out]

-1/2838528*(b*x + a)^(7/2)*((b*x + a)*(2*(11*B*a^3*b^10 - 4*A*a^2*b^11)*(b*x + a)/(a^6*b^18) - 11*(11*B*a^4*b^
10 - 4*A*a^3*b^11)/(a^6*b^18)) + 99*(B*a^5*b^10 - A*a^4*b^11)/(a^6*b^18))*b/(((b*x + a)*b - a*b)^(11/2)*abs(b)
)