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]

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

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Rubi [A]  time = 0.07, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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 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]

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 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]))))

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*}

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Mathematica [A]  time = 0.06, size = 111, normalized size = 0.61 \[ -\frac {2 (a+b x)^{5/2} \left (231 a^5 (13 A+15 B x)-210 a^4 b x (11 A+12 B x)+1680 a^3 b^2 x^2 (A+B x)-160 a^2 b^3 x^3 (7 A+6 B x)+128 a b^4 x^4 (5 A+3 B x)-256 A b^5 x^5\right )}{45045 a^6 x^{15/2}} \]

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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fricas [A]  time = 0.58, size = 173, normalized size = 0.95 \[ -\frac {2 \, {\left (3003 \, A a^{7} + 128 \, {\left (3 \, B a b^{6} - 2 \, A b^{7}\right )} x^{7} - 64 \, {\left (3 \, B a^{2} b^{5} - 2 \, A a b^{6}\right )} x^{6} + 48 \, {\left (3 \, B a^{3} b^{4} - 2 \, A a^{2} b^{5}\right )} x^{5} - 40 \, {\left (3 \, B a^{4} b^{3} - 2 \, A a^{3} b^{4}\right )} x^{4} + 35 \, {\left (3 \, B a^{5} b^{2} - 2 \, A a^{4} b^{3}\right )} x^{3} + 63 \, {\left (70 \, B a^{6} b + A a^{5} b^{2}\right )} x^{2} + 231 \, {\left (15 \, B a^{7} + 16 \, A a^{6} b\right )} x\right )} \sqrt {b x + a}}{45045 \, a^{6} x^{\frac {15}{2}}} \]

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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giac [A]  time = 1.86, size = 206, normalized size = 1.13 \[ -\frac {2 \, {\left ({\left (8 \, {\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (3 \, B a^{2} b^{14} - 2 \, A a b^{15}\right )} {\left (b x + a\right )}}{a^{7}} - \frac {15 \, {\left (3 \, B a^{3} b^{14} - 2 \, A a^{2} b^{15}\right )}}{a^{7}}\right )} + \frac {195 \, {\left (3 \, B a^{4} b^{14} - 2 \, A a^{3} b^{15}\right )}}{a^{7}}\right )} - \frac {715 \, {\left (3 \, B a^{5} b^{14} - 2 \, A a^{4} b^{15}\right )}}{a^{7}}\right )} {\left (b x + a\right )} + \frac {6435 \, {\left (3 \, B a^{6} b^{14} - 2 \, A a^{5} b^{15}\right )}}{a^{7}}\right )} {\left (b x + a\right )} - \frac {9009 \, {\left (B a^{7} b^{14} - A a^{6} b^{15}\right )}}{a^{7}}\right )} {\left (b x + a\right )}^{\frac {5}{2}} b}{45045 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {15}{2}} {\left | b \right |}} \]

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]

-2/45045*((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^7 - 15*(3*B*a^3*b^14 - 2*A*a
^2*b^15)/a^7) + 195*(3*B*a^4*b^14 - 2*A*a^3*b^15)/a^7) - 715*(3*B*a^5*b^14 - 2*A*a^4*b^15)/a^7)*(b*x + a) + 64
35*(3*B*a^6*b^14 - 2*A*a^5*b^15)/a^7)*(b*x + a) - 9009*(B*a^7*b^14 - A*a^6*b^15)/a^7)*(b*x + a)^(5/2)*b/(((b*x
+ a)*b - a*b)^(15/2)*abs(b))

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maple [A]  time = 0.01, size = 125, normalized size = 0.68 \[ -\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (-256 A \,b^{5} x^{5}+384 B a \,b^{4} x^{5}+640 A a \,b^{4} x^{4}-960 B \,a^{2} b^{3} x^{4}-1120 A \,a^{2} b^{3} x^{3}+1680 B \,a^{3} b^{2} x^{3}+1680 A \,a^{3} b^{2} x^{2}-2520 B \,a^{4} b \,x^{2}-2310 A \,a^{4} b x +3465 B \,a^{5} x +3003 A \,a^{5}\right )}{45045 a^{6} x^{\frac {15}{2}}} \]

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 [B]  time = 0.99, size = 360, normalized size = 1.97 \[ -\frac {256 \, \sqrt {b x^{2} + a x} B b^{6}}{15015 \, a^{5} x} + \frac {512 \, \sqrt {b x^{2} + a x} A b^{7}}{45045 \, a^{6} x} + \frac {128 \, \sqrt {b x^{2} + a x} B b^{5}}{15015 \, a^{4} x^{2}} - \frac {256 \, \sqrt {b x^{2} + a x} A b^{6}}{45045 \, a^{5} x^{2}} - \frac {32 \, \sqrt {b x^{2} + a x} B b^{4}}{5005 \, a^{3} x^{3}} + \frac {64 \, \sqrt {b x^{2} + a x} A b^{5}}{15015 \, a^{4} x^{3}} + \frac {16 \, \sqrt {b x^{2} + a x} B b^{3}}{3003 \, a^{2} x^{4}} - \frac {32 \, \sqrt {b x^{2} + a x} A b^{4}}{9009 \, a^{3} x^{4}} - \frac {2 \, \sqrt {b x^{2} + a x} B b^{2}}{429 \, a x^{5}} + \frac {4 \, \sqrt {b x^{2} + a x} A b^{3}}{1287 \, a^{2} x^{5}} + \frac {3 \, \sqrt {b x^{2} + a x} B b}{715 \, x^{6}} - \frac {2 \, \sqrt {b x^{2} + a x} A b^{2}}{715 \, a x^{6}} + \frac {3 \, \sqrt {b x^{2} + a x} B a}{65 \, x^{7}} + \frac {\sqrt {b x^{2} + a x} A b}{390 \, x^{7}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{5 \, x^{8}} + \frac {\sqrt {b x^{2} + a x} A a}{30 \, x^{8}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{6 \, x^{9}} \]

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]

-256/15015*sqrt(b*x^2 + a*x)*B*b^6/(a^5*x) + 512/45045*sqrt(b*x^2 + a*x)*A*b^7/(a^6*x) + 128/15015*sqrt(b*x^2
+ a*x)*B*b^5/(a^4*x^2) - 256/45045*sqrt(b*x^2 + a*x)*A*b^6/(a^5*x^2) - 32/5005*sqrt(b*x^2 + a*x)*B*b^4/(a^3*x^
3) + 64/15015*sqrt(b*x^2 + a*x)*A*b^5/(a^4*x^3) + 16/3003*sqrt(b*x^2 + a*x)*B*b^3/(a^2*x^4) - 32/9009*sqrt(b*x
^2 + a*x)*A*b^4/(a^3*x^4) - 2/429*sqrt(b*x^2 + a*x)*B*b^2/(a*x^5) + 4/1287*sqrt(b*x^2 + a*x)*A*b^3/(a^2*x^5) +
3/715*sqrt(b*x^2 + a*x)*B*b/x^6 - 2/715*sqrt(b*x^2 + a*x)*A*b^2/(a*x^6) + 3/65*sqrt(b*x^2 + a*x)*B*a/x^7 + 1/
390*sqrt(b*x^2 + a*x)*A*b/x^7 - 1/5*(b*x^2 + a*x)^(3/2)*B/x^8 + 1/30*sqrt(b*x^2 + a*x)*A*a/x^8 - 1/6*(b*x^2 +
a*x)^(3/2)*A/x^9

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mupad [B]  time = 1.00, size = 147, normalized size = 0.80 \[ -\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A\,a}{15}+x\,\left (\frac {32\,A\,b}{195}+\frac {2\,B\,a}{13}\right )-\frac {x^7\,\left (512\,A\,b^7-768\,B\,a\,b^6\right )}{45045\,a^6}-\frac {2\,b^2\,x^3\,\left (2\,A\,b-3\,B\,a\right )}{1287\,a^2}+\frac {16\,b^3\,x^4\,\left (2\,A\,b-3\,B\,a\right )}{9009\,a^3}-\frac {32\,b^4\,x^5\,\left (2\,A\,b-3\,B\,a\right )}{15015\,a^4}+\frac {128\,b^5\,x^6\,\left (2\,A\,b-3\,B\,a\right )}{45045\,a^5}+\frac {2\,b\,x^2\,\left (A\,b+70\,B\,a\right )}{715\,a}\right )}{x^{15/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a + b*x)^(1/2)*((2*A*a)/15 + x*((32*A*b)/195 + (2*B*a)/13) - (x^7*(512*A*b^7 - 768*B*a*b^6))/(45045*a^6) -
(2*b^2*x^3*(2*A*b - 3*B*a))/(1287*a^2) + (16*b^3*x^4*(2*A*b - 3*B*a))/(9009*a^3) - (32*b^4*x^5*(2*A*b - 3*B*a)
)/(15015*a^4) + (128*b^5*x^6*(2*A*b - 3*B*a))/(45045*a^5) + (2*b*x^2*(A*b + 70*B*a))/(715*a)))/x^(15/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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