3.511 \(\int \frac{(a+b x)^{5/2} (A+B x)}{x^{15/2}} \, dx\)

Optimal. Leaf size=117 \[ \frac{16 b^2 (a+b x)^{7/2} (6 A b-13 a B)}{9009 a^4 x^{7/2}}-\frac{8 b (a+b x)^{7/2} (6 A b-13 a B)}{1287 a^3 x^{9/2}}+\frac{2 (a+b x)^{7/2} (6 A b-13 a B)}{143 a^2 x^{11/2}}-\frac{2 A (a+b x)^{7/2}}{13 a x^{13/2}} \]

[Out]

(-2*A*(a + b*x)^(7/2))/(13*a*x^(13/2)) + (2*(6*A*b - 13*a*B)*(a + b*x)^(7/2))/(143*a^2*x^(11/2)) - (8*b*(6*A*b
- 13*a*B)*(a + b*x)^(7/2))/(1287*a^3*x^(9/2)) + (16*b^2*(6*A*b - 13*a*B)*(a + b*x)^(7/2))/(9009*a^4*x^(7/2))

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Rubi [A]  time = 0.0400097, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {78, 45, 37} \[ \frac{16 b^2 (a+b x)^{7/2} (6 A b-13 a B)}{9009 a^4 x^{7/2}}-\frac{8 b (a+b x)^{7/2} (6 A b-13 a B)}{1287 a^3 x^{9/2}}+\frac{2 (a+b x)^{7/2} (6 A b-13 a B)}{143 a^2 x^{11/2}}-\frac{2 A (a+b x)^{7/2}}{13 a x^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x^(15/2),x]

[Out]

(-2*A*(a + b*x)^(7/2))/(13*a*x^(13/2)) + (2*(6*A*b - 13*a*B)*(a + b*x)^(7/2))/(143*a^2*x^(11/2)) - (8*b*(6*A*b
- 13*a*B)*(a + b*x)^(7/2))/(1287*a^3*x^(9/2)) + (16*b^2*(6*A*b - 13*a*B)*(a + b*x)^(7/2))/(9009*a^4*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^{15/2}} \, dx &=-\frac{2 A (a+b x)^{7/2}}{13 a x^{13/2}}+\frac{\left (2 \left (-3 A b+\frac{13 a B}{2}\right )\right ) \int \frac{(a+b x)^{5/2}}{x^{13/2}} \, dx}{13 a}\\ &=-\frac{2 A (a+b x)^{7/2}}{13 a x^{13/2}}+\frac{2 (6 A b-13 a B) (a+b x)^{7/2}}{143 a^2 x^{11/2}}+\frac{(4 b (6 A b-13 a B)) \int \frac{(a+b x)^{5/2}}{x^{11/2}} \, dx}{143 a^2}\\ &=-\frac{2 A (a+b x)^{7/2}}{13 a x^{13/2}}+\frac{2 (6 A b-13 a B) (a+b x)^{7/2}}{143 a^2 x^{11/2}}-\frac{8 b (6 A b-13 a B) (a+b x)^{7/2}}{1287 a^3 x^{9/2}}-\frac{\left (8 b^2 (6 A b-13 a B)\right ) \int \frac{(a+b x)^{5/2}}{x^{9/2}} \, dx}{1287 a^3}\\ &=-\frac{2 A (a+b x)^{7/2}}{13 a x^{13/2}}+\frac{2 (6 A b-13 a B) (a+b x)^{7/2}}{143 a^2 x^{11/2}}-\frac{8 b (6 A b-13 a B) (a+b x)^{7/2}}{1287 a^3 x^{9/2}}+\frac{16 b^2 (6 A b-13 a B) (a+b x)^{7/2}}{9009 a^4 x^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0346603, size = 76, normalized size = 0.65 \[ -\frac{2 (a+b x)^{7/2} \left (-14 a^2 b x (27 A+26 B x)+63 a^3 (11 A+13 B x)+8 a b^2 x^2 (21 A+13 B x)-48 A b^3 x^3\right )}{9009 a^4 x^{13/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^(15/2),x]

[Out]

(-2*(a + b*x)^(7/2)*(-48*A*b^3*x^3 + 63*a^3*(11*A + 13*B*x) + 8*a*b^2*x^2*(21*A + 13*B*x) - 14*a^2*b*x*(27*A +
26*B*x)))/(9009*a^4*x^(13/2))

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Maple [A]  time = 0.003, size = 77, normalized size = 0.7 \begin{align*} -{\frac{-96\,A{b}^{3}{x}^{3}+208\,B{x}^{3}a{b}^{2}+336\,aA{b}^{2}{x}^{2}-728\,B{x}^{2}{a}^{2}b-756\,{a}^{2}Abx+1638\,{a}^{3}Bx+1386\,A{a}^{3}}{9009\,{a}^{4}} \left ( bx+a \right ) ^{{\frac{7}{2}}}{x}^{-{\frac{13}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^(15/2),x)

[Out]

-2/9009*(b*x+a)^(7/2)*(-48*A*b^3*x^3+104*B*a*b^2*x^3+168*A*a*b^2*x^2-364*B*a^2*b*x^2-378*A*a^2*b*x+819*B*a^3*x
+693*A*a^3)/x^(13/2)/a^4

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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^(15/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.64339, size = 351, normalized size = 3. \begin{align*} -\frac{2 \,{\left (693 \, A a^{6} + 8 \,{\left (13 \, B a b^{5} - 6 \, A b^{6}\right )} x^{6} - 4 \,{\left (13 \, B a^{2} b^{4} - 6 \, A a b^{5}\right )} x^{5} + 3 \,{\left (13 \, B a^{3} b^{3} - 6 \, A a^{2} b^{4}\right )} x^{4} +{\left (1469 \, B a^{4} b^{2} + 15 \, A a^{3} b^{3}\right )} x^{3} + 7 \,{\left (299 \, B a^{5} b + 159 \, A a^{4} b^{2}\right )} x^{2} + 63 \,{\left (13 \, B a^{6} + 27 \, A a^{5} b\right )} x\right )} \sqrt{b x + a}}{9009 \, a^{4} x^{\frac{13}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(15/2),x, algorithm="fricas")

[Out]

-2/9009*(693*A*a^6 + 8*(13*B*a*b^5 - 6*A*b^6)*x^6 - 4*(13*B*a^2*b^4 - 6*A*a*b^5)*x^5 + 3*(13*B*a^3*b^3 - 6*A*a
^2*b^4)*x^4 + (1469*B*a^4*b^2 + 15*A*a^3*b^3)*x^3 + 7*(299*B*a^5*b + 159*A*a^4*b^2)*x^2 + 63*(13*B*a^6 + 27*A*
a^5*b)*x)*sqrt(b*x + a)/(a^4*x^(13/2))

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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**(15/2),x)

[Out]

Timed out

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Giac [A]  time = 2.66771, size = 211, normalized size = 1.8 \begin{align*} \frac{{\left ({\left (b x + a\right )}{\left (4 \,{\left (b x + a\right )}{\left (\frac{2 \,{\left (13 \, B a^{3} b^{12} - 6 \, A a^{2} b^{13}\right )}{\left (b x + a\right )}}{a^{7} b^{21}} - \frac{13 \,{\left (13 \, B a^{4} b^{12} - 6 \, A a^{3} b^{13}\right )}}{a^{7} b^{21}}\right )} + \frac{143 \,{\left (13 \, B a^{5} b^{12} - 6 \, A a^{4} b^{13}\right )}}{a^{7} b^{21}}\right )} - \frac{1287 \,{\left (B a^{6} b^{12} - A a^{5} b^{13}\right )}}{a^{7} b^{21}}\right )}{\left (b x + a\right )}^{\frac{7}{2}} b}{6642155520 \,{\left ({\left (b x + a\right )} b - a b\right )}^{\frac{13}{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^(15/2),x, algorithm="giac")

[Out]

1/6642155520*((b*x + a)*(4*(b*x + a)*(2*(13*B*a^3*b^12 - 6*A*a^2*b^13)*(b*x + a)/(a^7*b^21) - 13*(13*B*a^4*b^1
2 - 6*A*a^3*b^13)/(a^7*b^21)) + 143*(13*B*a^5*b^12 - 6*A*a^4*b^13)/(a^7*b^21)) - 1287*(B*a^6*b^12 - A*a^5*b^13
)/(a^7*b^21))*(b*x + a)^(7/2)*b/(((b*x + a)*b - a*b)^(13/2)*abs(b))