3.501 \(\int \frac{(a+b x)^{3/2} (A+B x)}{x^{17/2}} \, dx\)

Optimal. Leaf size=183 \[ \frac{256 b^4 (a+b x)^{5/2} (2 A b-3 a B)}{45045 a^6 x^{5/2}}-\frac{128 b^3 (a+b x)^{5/2} (2 A b-3 a B)}{9009 a^5 x^{7/2}}+\frac{32 b^2 (a+b x)^{5/2} (2 A b-3 a B)}{1287 a^4 x^{9/2}}-\frac{16 b (a+b x)^{5/2} (2 A b-3 a B)}{429 a^3 x^{11/2}}+\frac{2 (a+b x)^{5/2} (2 A b-3 a B)}{39 a^2 x^{13/2}}-\frac{2 A (a+b x)^{5/2}}{15 a x^{15/2}} \]

[Out]

(-2*A*(a + b*x)^(5/2))/(15*a*x^(15/2)) + (2*(2*A*b - 3*a*B)*(a + b*x)^(5/2))/(39*a^2*x^(13/2)) - (16*b*(2*A*b
- 3*a*B)*(a + b*x)^(5/2))/(429*a^3*x^(11/2)) + (32*b^2*(2*A*b - 3*a*B)*(a + b*x)^(5/2))/(1287*a^4*x^(9/2)) - (
128*b^3*(2*A*b - 3*a*B)*(a + b*x)^(5/2))/(9009*a^5*x^(7/2)) + (256*b^4*(2*A*b - 3*a*B)*(a + b*x)^(5/2))/(45045
*a^6*x^(5/2))

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Rubi [A]  time = 0.0675422, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {78, 45, 37} \[ \frac{256 b^4 (a+b x)^{5/2} (2 A b-3 a B)}{45045 a^6 x^{5/2}}-\frac{128 b^3 (a+b x)^{5/2} (2 A b-3 a B)}{9009 a^5 x^{7/2}}+\frac{32 b^2 (a+b x)^{5/2} (2 A b-3 a B)}{1287 a^4 x^{9/2}}-\frac{16 b (a+b x)^{5/2} (2 A b-3 a B)}{429 a^3 x^{11/2}}+\frac{2 (a+b x)^{5/2} (2 A b-3 a B)}{39 a^2 x^{13/2}}-\frac{2 A (a+b x)^{5/2}}{15 a x^{15/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(3/2)*(A + B*x))/x^(17/2),x]

[Out]

(-2*A*(a + b*x)^(5/2))/(15*a*x^(15/2)) + (2*(2*A*b - 3*a*B)*(a + b*x)^(5/2))/(39*a^2*x^(13/2)) - (16*b*(2*A*b
- 3*a*B)*(a + b*x)^(5/2))/(429*a^3*x^(11/2)) + (32*b^2*(2*A*b - 3*a*B)*(a + b*x)^(5/2))/(1287*a^4*x^(9/2)) - (
128*b^3*(2*A*b - 3*a*B)*(a + b*x)^(5/2))/(9009*a^5*x^(7/2)) + (256*b^4*(2*A*b - 3*a*B)*(a + b*x)^(5/2))/(45045
*a^6*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^{17/2}} \, dx &=-\frac{2 A (a+b x)^{5/2}}{15 a x^{15/2}}+\frac{\left (2 \left (-5 A b+\frac{15 a B}{2}\right )\right ) \int \frac{(a+b x)^{3/2}}{x^{15/2}} \, dx}{15 a}\\ &=-\frac{2 A (a+b x)^{5/2}}{15 a x^{15/2}}+\frac{2 (2 A b-3 a B) (a+b x)^{5/2}}{39 a^2 x^{13/2}}+\frac{(8 b (2 A b-3 a B)) \int \frac{(a+b x)^{3/2}}{x^{13/2}} \, dx}{39 a^2}\\ &=-\frac{2 A (a+b x)^{5/2}}{15 a x^{15/2}}+\frac{2 (2 A b-3 a B) (a+b x)^{5/2}}{39 a^2 x^{13/2}}-\frac{16 b (2 A b-3 a B) (a+b x)^{5/2}}{429 a^3 x^{11/2}}-\frac{\left (16 b^2 (2 A b-3 a B)\right ) \int \frac{(a+b x)^{3/2}}{x^{11/2}} \, dx}{143 a^3}\\ &=-\frac{2 A (a+b x)^{5/2}}{15 a x^{15/2}}+\frac{2 (2 A b-3 a B) (a+b x)^{5/2}}{39 a^2 x^{13/2}}-\frac{16 b (2 A b-3 a B) (a+b x)^{5/2}}{429 a^3 x^{11/2}}+\frac{32 b^2 (2 A b-3 a B) (a+b x)^{5/2}}{1287 a^4 x^{9/2}}+\frac{\left (64 b^3 (2 A b-3 a B)\right ) \int \frac{(a+b x)^{3/2}}{x^{9/2}} \, dx}{1287 a^4}\\ &=-\frac{2 A (a+b x)^{5/2}}{15 a x^{15/2}}+\frac{2 (2 A b-3 a B) (a+b x)^{5/2}}{39 a^2 x^{13/2}}-\frac{16 b (2 A b-3 a B) (a+b x)^{5/2}}{429 a^3 x^{11/2}}+\frac{32 b^2 (2 A b-3 a B) (a+b x)^{5/2}}{1287 a^4 x^{9/2}}-\frac{128 b^3 (2 A b-3 a B) (a+b x)^{5/2}}{9009 a^5 x^{7/2}}-\frac{\left (128 b^4 (2 A b-3 a B)\right ) \int \frac{(a+b x)^{3/2}}{x^{7/2}} \, dx}{9009 a^5}\\ &=-\frac{2 A (a+b x)^{5/2}}{15 a x^{15/2}}+\frac{2 (2 A b-3 a B) (a+b x)^{5/2}}{39 a^2 x^{13/2}}-\frac{16 b (2 A b-3 a B) (a+b x)^{5/2}}{429 a^3 x^{11/2}}+\frac{32 b^2 (2 A b-3 a B) (a+b x)^{5/2}}{1287 a^4 x^{9/2}}-\frac{128 b^3 (2 A b-3 a B) (a+b x)^{5/2}}{9009 a^5 x^{7/2}}+\frac{256 b^4 (2 A b-3 a B) (a+b x)^{5/2}}{45045 a^6 x^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0411764, size = 111, normalized size = 0.61 \[ -\frac{2 (a+b x)^{5/2} \left (1680 a^3 b^2 x^2 (A+B x)-160 a^2 b^3 x^3 (7 A+6 B x)-210 a^4 b x (11 A+12 B x)+231 a^5 (13 A+15 B x)+128 a b^4 x^4 (5 A+3 B x)-256 A b^5 x^5\right )}{45045 a^6 x^{15/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(3/2)*(A + B*x))/x^(17/2),x]

[Out]

(-2*(a + b*x)^(5/2)*(-256*A*b^5*x^5 + 1680*a^3*b^2*x^2*(A + B*x) + 128*a*b^4*x^4*(5*A + 3*B*x) - 160*a^2*b^3*x
^3*(7*A + 6*B*x) - 210*a^4*b*x*(11*A + 12*B*x) + 231*a^5*(13*A + 15*B*x)))/(45045*a^6*x^(15/2))

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Maple [A]  time = 0.004, size = 125, normalized size = 0.7 \begin{align*} -{\frac{-512\,A{b}^{5}{x}^{5}+768\,B{x}^{5}a{b}^{4}+1280\,aA{b}^{4}{x}^{4}-1920\,B{x}^{4}{a}^{2}{b}^{3}-2240\,{a}^{2}A{b}^{3}{x}^{3}+3360\,B{x}^{3}{a}^{3}{b}^{2}+3360\,{a}^{3}A{b}^{2}{x}^{2}-5040\,B{x}^{2}{a}^{4}b-4620\,{a}^{4}Abx+6930\,{a}^{5}Bx+6006\,A{a}^{5}}{45045\,{a}^{6}} \left ( bx+a \right ) ^{{\frac{5}{2}}}{x}^{-{\frac{15}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(B*x+A)/x^(17/2),x)

[Out]

-2/45045*(b*x+a)^(5/2)*(-256*A*b^5*x^5+384*B*a*b^4*x^5+640*A*a*b^4*x^4-960*B*a^2*b^3*x^4-1120*A*a^2*b^3*x^3+16
80*B*a^3*b^2*x^3+1680*A*a^3*b^2*x^2-2520*B*a^4*b*x^2-2310*A*a^4*b*x+3465*B*a^5*x+3003*A*a^5)/x^(15/2)/a^6

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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^(17/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.60287, size = 400, normalized size = 2.19 \begin{align*} -\frac{2 \,{\left (3003 \, A a^{7} + 128 \,{\left (3 \, B a b^{6} - 2 \, A b^{7}\right )} x^{7} - 64 \,{\left (3 \, B a^{2} b^{5} - 2 \, A a b^{6}\right )} x^{6} + 48 \,{\left (3 \, B a^{3} b^{4} - 2 \, A a^{2} b^{5}\right )} x^{5} - 40 \,{\left (3 \, B a^{4} b^{3} - 2 \, A a^{3} b^{4}\right )} x^{4} + 35 \,{\left (3 \, B a^{5} b^{2} - 2 \, A a^{4} b^{3}\right )} x^{3} + 63 \,{\left (70 \, B a^{6} b + A a^{5} b^{2}\right )} x^{2} + 231 \,{\left (15 \, B a^{7} + 16 \, A a^{6} b\right )} x\right )} \sqrt{b x + a}}{45045 \, a^{6} x^{\frac{15}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^(17/2),x, algorithm="fricas")

[Out]

-2/45045*(3003*A*a^7 + 128*(3*B*a*b^6 - 2*A*b^7)*x^7 - 64*(3*B*a^2*b^5 - 2*A*a*b^6)*x^6 + 48*(3*B*a^3*b^4 - 2*
A*a^2*b^5)*x^5 - 40*(3*B*a^4*b^3 - 2*A*a^3*b^4)*x^4 + 35*(3*B*a^5*b^2 - 2*A*a^4*b^3)*x^3 + 63*(70*B*a^6*b + A*
a^5*b^2)*x^2 + 231*(15*B*a^7 + 16*A*a^6*b)*x)*sqrt(b*x + a)/(a^6*x^(15/2))

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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**(17/2),x)

[Out]

Timed out

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Giac [A]  time = 1.29059, size = 302, normalized size = 1.65 \begin{align*} \frac{{\left ({\left (8 \,{\left (2 \,{\left (b x + a\right )}{\left (4 \,{\left (b x + a\right )}{\left (\frac{2 \,{\left (3 \, B a^{2} b^{14} - 2 \, A a b^{15}\right )}{\left (b x + a\right )}}{a^{8} b^{24}} - \frac{15 \,{\left (3 \, B a^{3} b^{14} - 2 \, A a^{2} b^{15}\right )}}{a^{8} b^{24}}\right )} + \frac{195 \,{\left (3 \, B a^{4} b^{14} - 2 \, A a^{3} b^{15}\right )}}{a^{8} b^{24}}\right )} - \frac{715 \,{\left (3 \, B a^{5} b^{14} - 2 \, A a^{4} b^{15}\right )}}{a^{8} b^{24}}\right )}{\left (b x + a\right )} + \frac{6435 \,{\left (3 \, B a^{6} b^{14} - 2 \, A a^{5} b^{15}\right )}}{a^{8} b^{24}}\right )}{\left (b x + a\right )} - \frac{9009 \,{\left (B a^{7} b^{14} - A a^{6} b^{15}\right )}}{a^{8} b^{24}}\right )}{\left (b x + a\right )}^{\frac{5}{2}} b}{2952069120 \,{\left ({\left (b x + a\right )} b - a b\right )}^{\frac{15}{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^(17/2),x, algorithm="giac")

[Out]

1/2952069120*((8*(2*(b*x + a)*(4*(b*x + a)*(2*(3*B*a^2*b^14 - 2*A*a*b^15)*(b*x + a)/(a^8*b^24) - 15*(3*B*a^3*b
^14 - 2*A*a^2*b^15)/(a^8*b^24)) + 195*(3*B*a^4*b^14 - 2*A*a^3*b^15)/(a^8*b^24)) - 715*(3*B*a^5*b^14 - 2*A*a^4*
b^15)/(a^8*b^24))*(b*x + a) + 6435*(3*B*a^6*b^14 - 2*A*a^5*b^15)/(a^8*b^24))*(b*x + a) - 9009*(B*a^7*b^14 - A*
a^6*b^15)/(a^8*b^24))*(b*x + a)^(5/2)*b/(((b*x + a)*b - a*b)^(15/2)*abs(b))