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

Optimal. Leaf size=183 \[ \frac {256 b^4 (a+b x)^{3/2} (10 A b-13 a B)}{45045 a^6 x^{3/2}}-\frac {128 b^3 (a+b x)^{3/2} (10 A b-13 a B)}{15015 a^5 x^{5/2}}+\frac {32 b^2 (a+b x)^{3/2} (10 A b-13 a B)}{3003 a^4 x^{7/2}}-\frac {16 b (a+b x)^{3/2} (10 A b-13 a B)}{1287 a^3 x^{9/2}}+\frac {2 (a+b x)^{3/2} (10 A b-13 a B)}{143 a^2 x^{11/2}}-\frac {2 A (a+b x)^{3/2}}{13 a x^{13/2}} \]

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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} \begin {gather*} \frac {256 b^4 (a+b x)^{3/2} (10 A b-13 a B)}{45045 a^6 x^{3/2}}-\frac {128 b^3 (a+b x)^{3/2} (10 A b-13 a B)}{15015 a^5 x^{5/2}}+\frac {32 b^2 (a+b x)^{3/2} (10 A b-13 a B)}{3003 a^4 x^{7/2}}-\frac {16 b (a+b x)^{3/2} (10 A b-13 a B)}{1287 a^3 x^{9/2}}+\frac {2 (a+b x)^{3/2} (10 A b-13 a B)}{143 a^2 x^{11/2}}-\frac {2 A (a+b x)^{3/2}}{13 a x^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*(a + b*x)^(3/2))/(13*a*x^(13/2)) + (2*(10*A*b - 13*a*B)*(a + b*x)^(3/2))/(143*a^2*x^(11/2)) - (16*b*(10*
A*b - 13*a*B)*(a + b*x)^(3/2))/(1287*a^3*x^(9/2)) + (32*b^2*(10*A*b - 13*a*B)*(a + b*x)^(3/2))/(3003*a^4*x^(7/
2)) - (128*b^3*(10*A*b - 13*a*B)*(a + b*x)^(3/2))/(15015*a^5*x^(5/2)) + (256*b^4*(10*A*b - 13*a*B)*(a + b*x)^(
3/2))/(45045*a^6*x^(3/2))

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 {\sqrt {a+b x} (A+B x)}{x^{15/2}} \, dx &=-\frac {2 A (a+b x)^{3/2}}{13 a x^{13/2}}+\frac {\left (2 \left (-5 A b+\frac {13 a B}{2}\right )\right ) \int \frac {\sqrt {a+b x}}{x^{13/2}} \, dx}{13 a}\\ &=-\frac {2 A (a+b x)^{3/2}}{13 a x^{13/2}}+\frac {2 (10 A b-13 a B) (a+b x)^{3/2}}{143 a^2 x^{11/2}}+\frac {(8 b (10 A b-13 a B)) \int \frac {\sqrt {a+b x}}{x^{11/2}} \, dx}{143 a^2}\\ &=-\frac {2 A (a+b x)^{3/2}}{13 a x^{13/2}}+\frac {2 (10 A b-13 a B) (a+b x)^{3/2}}{143 a^2 x^{11/2}}-\frac {16 b (10 A b-13 a B) (a+b x)^{3/2}}{1287 a^3 x^{9/2}}-\frac {\left (16 b^2 (10 A b-13 a B)\right ) \int \frac {\sqrt {a+b x}}{x^{9/2}} \, dx}{429 a^3}\\ &=-\frac {2 A (a+b x)^{3/2}}{13 a x^{13/2}}+\frac {2 (10 A b-13 a B) (a+b x)^{3/2}}{143 a^2 x^{11/2}}-\frac {16 b (10 A b-13 a B) (a+b x)^{3/2}}{1287 a^3 x^{9/2}}+\frac {32 b^2 (10 A b-13 a B) (a+b x)^{3/2}}{3003 a^4 x^{7/2}}+\frac {\left (64 b^3 (10 A b-13 a B)\right ) \int \frac {\sqrt {a+b x}}{x^{7/2}} \, dx}{3003 a^4}\\ &=-\frac {2 A (a+b x)^{3/2}}{13 a x^{13/2}}+\frac {2 (10 A b-13 a B) (a+b x)^{3/2}}{143 a^2 x^{11/2}}-\frac {16 b (10 A b-13 a B) (a+b x)^{3/2}}{1287 a^3 x^{9/2}}+\frac {32 b^2 (10 A b-13 a B) (a+b x)^{3/2}}{3003 a^4 x^{7/2}}-\frac {128 b^3 (10 A b-13 a B) (a+b x)^{3/2}}{15015 a^5 x^{5/2}}-\frac {\left (128 b^4 (10 A b-13 a B)\right ) \int \frac {\sqrt {a+b x}}{x^{5/2}} \, dx}{15015 a^5}\\ &=-\frac {2 A (a+b x)^{3/2}}{13 a x^{13/2}}+\frac {2 (10 A b-13 a B) (a+b x)^{3/2}}{143 a^2 x^{11/2}}-\frac {16 b (10 A b-13 a B) (a+b x)^{3/2}}{1287 a^3 x^{9/2}}+\frac {32 b^2 (10 A b-13 a B) (a+b x)^{3/2}}{3003 a^4 x^{7/2}}-\frac {128 b^3 (10 A b-13 a B) (a+b x)^{3/2}}{15015 a^5 x^{5/2}}+\frac {256 b^4 (10 A b-13 a B) (a+b x)^{3/2}}{45045 a^6 x^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 114, normalized size = 0.62 \begin {gather*} -\frac {2 (a+b x)^{3/2} \left (315 a^5 (11 A+13 B x)-70 a^4 b x (45 A+52 B x)+80 a^3 b^2 x^2 (35 A+39 B x)-96 a^2 b^3 x^3 (25 A+26 B x)+128 a b^4 x^4 (15 A+13 B x)-1280 A b^5 x^5\right )}{45045 a^6 x^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*x)^(3/2)*(-1280*A*b^5*x^5 + 315*a^5*(11*A + 13*B*x) + 128*a*b^4*x^4*(15*A + 13*B*x) - 96*a^2*b^3*x^
3*(25*A + 26*B*x) + 80*a^3*b^2*x^2*(35*A + 39*B*x) - 70*a^4*b*x*(45*A + 52*B*x)))/(45045*a^6*x^(13/2))

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IntegrateAlgebraic [A]  time = 0.28, size = 154, normalized size = 0.84 \begin {gather*} \frac {2 \sqrt {a+b x} \left (-3465 a^6 A-4095 a^6 B x-315 a^5 A b x-455 a^5 b B x^2+350 a^4 A b^2 x^2+520 a^4 b^2 B x^3-400 a^3 A b^3 x^3-624 a^3 b^3 B x^4+480 a^2 A b^4 x^4+832 a^2 b^4 B x^5-640 a A b^5 x^5-1664 a b^5 B x^6+1280 A b^6 x^6\right )}{45045 a^6 x^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x]*(-3465*a^6*A - 315*a^5*A*b*x - 4095*a^6*B*x + 350*a^4*A*b^2*x^2 - 455*a^5*b*B*x^2 - 400*a^3*A
*b^3*x^3 + 520*a^4*b^2*B*x^3 + 480*a^2*A*b^4*x^4 - 624*a^3*b^3*B*x^4 - 640*a*A*b^5*x^5 + 832*a^2*b^4*B*x^5 + 1
280*A*b^6*x^6 - 1664*a*b^5*B*x^6))/(45045*a^6*x^(13/2))

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fricas [A]  time = 0.90, size = 149, normalized size = 0.81 \begin {gather*} -\frac {2 \, {\left (3465 \, A a^{6} + 128 \, {\left (13 \, B a b^{5} - 10 \, A b^{6}\right )} x^{6} - 64 \, {\left (13 \, B a^{2} b^{4} - 10 \, A a b^{5}\right )} x^{5} + 48 \, {\left (13 \, B a^{3} b^{3} - 10 \, A a^{2} b^{4}\right )} x^{4} - 40 \, {\left (13 \, B a^{4} b^{2} - 10 \, A a^{3} b^{3}\right )} x^{3} + 35 \, {\left (13 \, B a^{5} b - 10 \, A a^{4} b^{2}\right )} x^{2} + 315 \, {\left (13 \, B a^{6} + A a^{5} b\right )} x\right )} \sqrt {b x + a}}{45045 \, a^{6} x^{\frac {13}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^(15/2),x, algorithm="fricas")

[Out]

-2/45045*(3465*A*a^6 + 128*(13*B*a*b^5 - 10*A*b^6)*x^6 - 64*(13*B*a^2*b^4 - 10*A*a*b^5)*x^5 + 48*(13*B*a^3*b^3
- 10*A*a^2*b^4)*x^4 - 40*(13*B*a^4*b^2 - 10*A*a^3*b^3)*x^3 + 35*(13*B*a^5*b - 10*A*a^4*b^2)*x^2 + 315*(13*B*a
^6 + A*a^5*b)*x)*sqrt(b*x + a)/(a^6*x^(13/2))

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giac [A]  time = 1.22, size = 201, normalized size = 1.10 \begin {gather*} -\frac {2 \, {\left ({\left (8 \, {\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (13 \, B a b^{12} - 10 \, A b^{13}\right )} {\left (b x + a\right )}}{a^{6}} - \frac {13 \, {\left (13 \, B a^{2} b^{12} - 10 \, A a b^{13}\right )}}{a^{6}}\right )} + \frac {143 \, {\left (13 \, B a^{3} b^{12} - 10 \, A a^{2} b^{13}\right )}}{a^{6}}\right )} - \frac {429 \, {\left (13 \, B a^{4} b^{12} - 10 \, A a^{3} b^{13}\right )}}{a^{6}}\right )} {\left (b x + a\right )} + \frac {3003 \, {\left (13 \, B a^{5} b^{12} - 10 \, A a^{4} b^{13}\right )}}{a^{6}}\right )} {\left (b x + a\right )} - \frac {15015 \, {\left (B a^{6} b^{12} - A a^{5} b^{13}\right )}}{a^{6}}\right )} {\left (b x + a\right )}^{\frac {3}{2}} b}{45045 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {13}{2}} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^(15/2),x, algorithm="giac")

[Out]

-2/45045*((8*(2*(b*x + a)*(4*(b*x + a)*(2*(13*B*a*b^12 - 10*A*b^13)*(b*x + a)/a^6 - 13*(13*B*a^2*b^12 - 10*A*a
*b^13)/a^6) + 143*(13*B*a^3*b^12 - 10*A*a^2*b^13)/a^6) - 429*(13*B*a^4*b^12 - 10*A*a^3*b^13)/a^6)*(b*x + a) +
3003*(13*B*a^5*b^12 - 10*A*a^4*b^13)/a^6)*(b*x + a) - 15015*(B*a^6*b^12 - A*a^5*b^13)/a^6)*(b*x + a)^(3/2)*b/(
((b*x + a)*b - a*b)^(13/2)*abs(b))

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maple [A]  time = 0.01, size = 125, normalized size = 0.68 \begin {gather*} -\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (-1280 A \,b^{5} x^{5}+1664 B a \,b^{4} x^{5}+1920 A a \,b^{4} x^{4}-2496 B \,a^{2} b^{3} x^{4}-2400 A \,a^{2} b^{3} x^{3}+3120 B \,a^{3} b^{2} x^{3}+2800 A \,a^{3} b^{2} x^{2}-3640 B \,a^{4} b \,x^{2}-3150 A \,a^{4} b x +4095 B \,a^{5} x +3465 A \,a^{5}\right )}{45045 a^{6} x^{\frac {13}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b*x+a)^(1/2)/x^(15/2),x)

[Out]

-2/45045*(b*x+a)^(3/2)*(-1280*A*b^5*x^5+1664*B*a*b^4*x^5+1920*A*a*b^4*x^4-2496*B*a^2*b^3*x^4-2400*A*a^2*b^3*x^
3+3120*B*a^3*b^2*x^3+2800*A*a^3*b^2*x^2-3640*B*a^4*b*x^2-3150*A*a^4*b*x+4095*B*a^5*x+3465*A*a^5)/x^(13/2)/a^6

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maxima [A]  time = 0.91, size = 284, normalized size = 1.55 \begin {gather*} -\frac {256 \, \sqrt {b x^{2} + a x} B b^{5}}{3465 \, a^{5} x} + \frac {512 \, \sqrt {b x^{2} + a x} A b^{6}}{9009 \, a^{6} x} + \frac {128 \, \sqrt {b x^{2} + a x} B b^{4}}{3465 \, a^{4} x^{2}} - \frac {256 \, \sqrt {b x^{2} + a x} A b^{5}}{9009 \, a^{5} x^{2}} - \frac {32 \, \sqrt {b x^{2} + a x} B b^{3}}{1155 \, a^{3} x^{3}} + \frac {64 \, \sqrt {b x^{2} + a x} A b^{4}}{3003 \, a^{4} x^{3}} + \frac {16 \, \sqrt {b x^{2} + a x} B b^{2}}{693 \, a^{2} x^{4}} - \frac {160 \, \sqrt {b x^{2} + a x} A b^{3}}{9009 \, a^{3} x^{4}} - \frac {2 \, \sqrt {b x^{2} + a x} B b}{99 \, a x^{5}} + \frac {20 \, \sqrt {b x^{2} + a x} A b^{2}}{1287 \, a^{2} x^{5}} - \frac {2 \, \sqrt {b x^{2} + a x} B}{11 \, x^{6}} - \frac {2 \, \sqrt {b x^{2} + a x} A b}{143 \, a x^{6}} - \frac {2 \, \sqrt {b x^{2} + a x} A}{13 \, x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^(15/2),x, algorithm="maxima")

[Out]

-256/3465*sqrt(b*x^2 + a*x)*B*b^5/(a^5*x) + 512/9009*sqrt(b*x^2 + a*x)*A*b^6/(a^6*x) + 128/3465*sqrt(b*x^2 + a
*x)*B*b^4/(a^4*x^2) - 256/9009*sqrt(b*x^2 + a*x)*A*b^5/(a^5*x^2) - 32/1155*sqrt(b*x^2 + a*x)*B*b^3/(a^3*x^3) +
64/3003*sqrt(b*x^2 + a*x)*A*b^4/(a^4*x^3) + 16/693*sqrt(b*x^2 + a*x)*B*b^2/(a^2*x^4) - 160/9009*sqrt(b*x^2 +
a*x)*A*b^3/(a^3*x^4) - 2/99*sqrt(b*x^2 + a*x)*B*b/(a*x^5) + 20/1287*sqrt(b*x^2 + a*x)*A*b^2/(a^2*x^5) - 2/11*s
qrt(b*x^2 + a*x)*B/x^6 - 2/143*sqrt(b*x^2 + a*x)*A*b/(a*x^6) - 2/13*sqrt(b*x^2 + a*x)*A/x^7

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mupad [B]  time = 0.80, size = 128, normalized size = 0.70 \begin {gather*} -\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{13}+\frac {2\,x\,\left (A\,b+13\,B\,a\right )}{143\,a}+\frac {16\,b^2\,x^3\,\left (10\,A\,b-13\,B\,a\right )}{9009\,a^3}-\frac {32\,b^3\,x^4\,\left (10\,A\,b-13\,B\,a\right )}{15015\,a^4}+\frac {128\,b^4\,x^5\,\left (10\,A\,b-13\,B\,a\right )}{45045\,a^5}-\frac {256\,b^5\,x^6\,\left (10\,A\,b-13\,B\,a\right )}{45045\,a^6}-\frac {2\,b\,x^2\,\left (10\,A\,b-13\,B\,a\right )}{1287\,a^2}\right )}{x^{13/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(1/2))/x^(15/2),x)

[Out]

-((a + b*x)^(1/2)*((2*A)/13 + (2*x*(A*b + 13*B*a))/(143*a) + (16*b^2*x^3*(10*A*b - 13*B*a))/(9009*a^3) - (32*b
^3*x^4*(10*A*b - 13*B*a))/(15015*a^4) + (128*b^4*x^5*(10*A*b - 13*B*a))/(45045*a^5) - (256*b^5*x^6*(10*A*b - 1
3*B*a))/(45045*a^6) - (2*b*x^2*(10*A*b - 13*B*a))/(1287*a^2)))/x^(13/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)**(1/2)/x**(15/2),x)

[Out]

Timed out

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