3.5.75 \(\int \frac {(a+b x)^{3/2} (A+B x)}{x^{15/2}} \, dx\)

Optimal. Leaf size=150 \[ -\frac {32 b^3 (a+b x)^{5/2} (8 A b-13 a B)}{15015 a^5 x^{5/2}}+\frac {16 b^2 (a+b x)^{5/2} (8 A b-13 a B)}{3003 a^4 x^{7/2}}-\frac {4 b (a+b x)^{5/2} (8 A b-13 a B)}{429 a^3 x^{9/2}}+\frac {2 (a+b x)^{5/2} (8 A b-13 a B)}{143 a^2 x^{11/2}}-\frac {2 A (a+b x)^{5/2}}{13 a x^{13/2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {32 b^3 (a+b x)^{5/2} (8 A b-13 a B)}{15015 a^5 x^{5/2}}+\frac {16 b^2 (a+b x)^{5/2} (8 A b-13 a B)}{3003 a^4 x^{7/2}}-\frac {4 b (a+b x)^{5/2} (8 A b-13 a B)}{429 a^3 x^{9/2}}+\frac {2 (a+b x)^{5/2} (8 A b-13 a B)}{143 a^2 x^{11/2}}-\frac {2 A (a+b x)^{5/2}}{13 a x^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(3/2)*(A + B*x))/x^(15/2),x]

[Out]

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

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Mathematica [A]  time = 0.05, size = 95, normalized size = 0.63 \begin {gather*} -\frac {2 (a+b x)^{5/2} \left (105 a^4 (11 A+13 B x)-70 a^3 b x (12 A+13 B x)+40 a^2 b^2 x^2 (14 A+13 B x)-16 a b^3 x^3 (20 A+13 B x)+128 A b^4 x^4\right )}{15015 a^5 x^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(3/2)*(A + B*x))/x^(15/2),x]

[Out]

(-2*(a + b*x)^(5/2)*(128*A*b^4*x^4 + 105*a^4*(11*A + 13*B*x) - 70*a^3*b*x*(12*A + 13*B*x) + 40*a^2*b^2*x^2*(14
*A + 13*B*x) - 16*a*b^3*x^3*(20*A + 13*B*x)))/(15015*a^5*x^(13/2))

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IntegrateAlgebraic [A]  time = 0.40, size = 154, normalized size = 1.03 \begin {gather*} \frac {2 \sqrt {a+b x} \left (-1155 a^6 A-1365 a^6 B x-1470 a^5 A b x-1820 a^5 b B x^2-35 a^4 A b^2 x^2-65 a^4 b^2 B x^3+40 a^3 A b^3 x^3+78 a^3 b^3 B x^4-48 a^2 A b^4 x^4-104 a^2 b^4 B x^5+64 a A b^5 x^5+208 a b^5 B x^6-128 A b^6 x^6\right )}{15015 a^5 x^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x)^(3/2)*(A + B*x))/x^(15/2),x]

[Out]

(2*Sqrt[a + b*x]*(-1155*a^6*A - 1470*a^5*A*b*x - 1365*a^6*B*x - 35*a^4*A*b^2*x^2 - 1820*a^5*b*B*x^2 + 40*a^3*A
*b^3*x^3 - 65*a^4*b^2*B*x^3 - 48*a^2*A*b^4*x^4 + 78*a^3*b^3*B*x^4 + 64*a*A*b^5*x^5 - 104*a^2*b^4*B*x^5 - 128*A
*b^6*x^6 + 208*a*b^5*B*x^6))/(15015*a^5*x^(13/2))

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fricas [A]  time = 1.52, size = 149, normalized size = 0.99 \begin {gather*} -\frac {2 \, {\left (1155 \, A a^{6} - 16 \, {\left (13 \, B a b^{5} - 8 \, A b^{6}\right )} x^{6} + 8 \, {\left (13 \, B a^{2} b^{4} - 8 \, A a b^{5}\right )} x^{5} - 6 \, {\left (13 \, B a^{3} b^{3} - 8 \, A a^{2} b^{4}\right )} x^{4} + 5 \, {\left (13 \, B a^{4} b^{2} - 8 \, A a^{3} b^{3}\right )} x^{3} + 35 \, {\left (52 \, B a^{5} b + A a^{4} b^{2}\right )} x^{2} + 105 \, {\left (13 \, B a^{6} + 14 \, A a^{5} b\right )} x\right )} \sqrt {b x + a}}{15015 \, a^{5} x^{\frac {13}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^(15/2),x, algorithm="fricas")

[Out]

-2/15015*(1155*A*a^6 - 16*(13*B*a*b^5 - 8*A*b^6)*x^6 + 8*(13*B*a^2*b^4 - 8*A*a*b^5)*x^5 - 6*(13*B*a^3*b^3 - 8*
A*a^2*b^4)*x^4 + 5*(13*B*a^4*b^2 - 8*A*a^3*b^3)*x^3 + 35*(52*B*a^5*b + A*a^4*b^2)*x^2 + 105*(13*B*a^6 + 14*A*a
^5*b)*x)*sqrt(b*x + a)/(a^5*x^(13/2))

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giac [A]  time = 2.19, size = 174, normalized size = 1.16 \begin {gather*} \frac {2 \, {\left ({\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (13 \, B a^{2} b^{12} - 8 \, A a b^{13}\right )} {\left (b x + a\right )}}{a^{6}} - \frac {13 \, {\left (13 \, B a^{3} b^{12} - 8 \, A a^{2} b^{13}\right )}}{a^{6}}\right )} + \frac {143 \, {\left (13 \, B a^{4} b^{12} - 8 \, A a^{3} b^{13}\right )}}{a^{6}}\right )} - \frac {429 \, {\left (13 \, B a^{5} b^{12} - 8 \, A a^{4} b^{13}\right )}}{a^{6}}\right )} {\left (b x + a\right )} + \frac {3003 \, {\left (B a^{6} b^{12} - A a^{5} b^{13}\right )}}{a^{6}}\right )} {\left (b x + a\right )}^{\frac {5}{2}} b}{15015 \, {\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)^(3/2)*(B*x+A)/x^(15/2),x, algorithm="giac")

[Out]

2/15015*((2*(b*x + a)*(4*(b*x + a)*(2*(13*B*a^2*b^12 - 8*A*a*b^13)*(b*x + a)/a^6 - 13*(13*B*a^3*b^12 - 8*A*a^2
*b^13)/a^6) + 143*(13*B*a^4*b^12 - 8*A*a^3*b^13)/a^6) - 429*(13*B*a^5*b^12 - 8*A*a^4*b^13)/a^6)*(b*x + a) + 30
03*(B*a^6*b^12 - A*a^5*b^13)/a^6)*(b*x + a)^(5/2)*b/(((b*x + a)*b - a*b)^(13/2)*abs(b))

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maple [A]  time = 0.01, size = 101, normalized size = 0.67 \begin {gather*} -\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (128 A \,b^{4} x^{4}-208 B a \,b^{3} x^{4}-320 A a \,b^{3} x^{3}+520 B \,a^{2} b^{2} x^{3}+560 A \,a^{2} b^{2} x^{2}-910 B \,a^{3} b \,x^{2}-840 A \,a^{3} b x +1365 B \,a^{4} x +1155 A \,a^{4}\right )}{15015 a^{5} x^{\frac {13}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(B*x+A)/x^(15/2),x)

[Out]

-2/15015*(b*x+a)^(5/2)*(128*A*b^4*x^4-208*B*a*b^3*x^4-320*A*a*b^3*x^3+520*B*a^2*b^2*x^3+560*A*a^2*b^2*x^2-910*
B*a^3*b*x^2-840*A*a^3*b*x+1365*B*a^4*x+1155*A*a^4)/x^(13/2)/a^5

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maxima [B]  time = 0.98, size = 314, normalized size = 2.09 \begin {gather*} \frac {32 \, \sqrt {b x^{2} + a x} B b^{5}}{1155 \, a^{4} x} - \frac {256 \, \sqrt {b x^{2} + a x} A b^{6}}{15015 \, a^{5} x} - \frac {16 \, \sqrt {b x^{2} + a x} B b^{4}}{1155 \, a^{3} x^{2}} + \frac {128 \, \sqrt {b x^{2} + a x} A b^{5}}{15015 \, a^{4} x^{2}} + \frac {4 \, \sqrt {b x^{2} + a x} B b^{3}}{385 \, a^{2} x^{3}} - \frac {32 \, \sqrt {b x^{2} + a x} A b^{4}}{5005 \, a^{3} x^{3}} - \frac {2 \, \sqrt {b x^{2} + a x} B b^{2}}{231 \, a x^{4}} + \frac {16 \, \sqrt {b x^{2} + a x} A b^{3}}{3003 \, a^{2} x^{4}} + \frac {\sqrt {b x^{2} + a x} B b}{132 \, x^{5}} - \frac {2 \, \sqrt {b x^{2} + a x} A b^{2}}{429 \, a x^{5}} + \frac {3 \, \sqrt {b x^{2} + a x} B a}{44 \, x^{6}} + \frac {3 \, \sqrt {b x^{2} + a x} A b}{715 \, x^{6}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{4 \, x^{7}} + \frac {3 \, \sqrt {b x^{2} + a x} A a}{65 \, x^{7}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{5 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^(15/2),x, algorithm="maxima")

[Out]

32/1155*sqrt(b*x^2 + a*x)*B*b^5/(a^4*x) - 256/15015*sqrt(b*x^2 + a*x)*A*b^6/(a^5*x) - 16/1155*sqrt(b*x^2 + a*x
)*B*b^4/(a^3*x^2) + 128/15015*sqrt(b*x^2 + a*x)*A*b^5/(a^4*x^2) + 4/385*sqrt(b*x^2 + a*x)*B*b^3/(a^2*x^3) - 32
/5005*sqrt(b*x^2 + a*x)*A*b^4/(a^3*x^3) - 2/231*sqrt(b*x^2 + a*x)*B*b^2/(a*x^4) + 16/3003*sqrt(b*x^2 + a*x)*A*
b^3/(a^2*x^4) + 1/132*sqrt(b*x^2 + a*x)*B*b/x^5 - 2/429*sqrt(b*x^2 + a*x)*A*b^2/(a*x^5) + 3/44*sqrt(b*x^2 + a*
x)*B*a/x^6 + 3/715*sqrt(b*x^2 + a*x)*A*b/x^6 - 1/4*(b*x^2 + a*x)^(3/2)*B/x^7 + 3/65*sqrt(b*x^2 + a*x)*A*a/x^7
- 1/5*(b*x^2 + a*x)^(3/2)*A/x^8

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mupad [B]  time = 0.84, size = 127, normalized size = 0.85 \begin {gather*} -\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A\,a}{13}+x\,\left (\frac {28\,A\,b}{143}+\frac {2\,B\,a}{11}\right )+\frac {x^6\,\left (256\,A\,b^6-416\,B\,a\,b^5\right )}{15015\,a^5}-\frac {2\,b^2\,x^3\,\left (8\,A\,b-13\,B\,a\right )}{3003\,a^2}+\frac {4\,b^3\,x^4\,\left (8\,A\,b-13\,B\,a\right )}{5005\,a^3}-\frac {16\,b^4\,x^5\,\left (8\,A\,b-13\,B\,a\right )}{15015\,a^4}+\frac {2\,b\,x^2\,\left (A\,b+52\,B\,a\right )}{429\,a}\right )}{x^{13/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(3/2))/x^(15/2),x)

[Out]

-((a + b*x)^(1/2)*((2*A*a)/13 + x*((28*A*b)/143 + (2*B*a)/11) + (x^6*(256*A*b^6 - 416*B*a*b^5))/(15015*a^5) -
(2*b^2*x^3*(8*A*b - 13*B*a))/(3003*a^2) + (4*b^3*x^4*(8*A*b - 13*B*a))/(5005*a^3) - (16*b^4*x^5*(8*A*b - 13*B*
a))/(15015*a^4) + (2*b*x^2*(A*b + 52*B*a))/(429*a)))/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)**(3/2)*(B*x+A)/x**(15/2),x)

[Out]

Timed out

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