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

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

[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))

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Rubi [A]  time = 0.0563563, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {78, 45, 37} \[ -\frac{32 b^3 (a+b x)^{5/2} (8 A b-13 a B)}{15015 a^5 x^{5/2}}+\frac{16 b^2 (a+b x)^{5/2} (8 A b-13 a B)}{3003 a^4 x^{7/2}}-\frac{4 b (a+b x)^{5/2} (8 A b-13 a B)}{429 a^3 x^{9/2}}+\frac{2 (a+b x)^{5/2} (8 A b-13 a B)}{143 a^2 x^{11/2}}-\frac{2 A (a+b x)^{5/2}}{13 a x^{13/2}} \]

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 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^{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*}

Mathematica [A]  time = 0.0360798, size = 95, normalized size = 0.63 \[ -\frac{2 (a+b x)^{5/2} \left (40 a^2 b^2 x^2 (14 A+13 B x)-70 a^3 b x (12 A+13 B x)+105 a^4 (11 A+13 B x)-16 a b^3 x^3 (20 A+13 B x)+128 A b^4 x^4\right )}{15015 a^5 x^{13/2}} \]

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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Maple [A]  time = 0.006, size = 101, normalized size = 0.7 \begin{align*} -{\frac{256\,A{b}^{4}{x}^{4}-416\,Ba{b}^{3}{x}^{4}-640\,Aa{b}^{3}{x}^{3}+1040\,B{a}^{2}{b}^{2}{x}^{3}+1120\,A{a}^{2}{b}^{2}{x}^{2}-1820\,B{a}^{3}b{x}^{2}-1680\,A{a}^{3}bx+2730\,B{a}^{4}x+2310\,A{a}^{4}}{15015\,{a}^{5}} \left ( bx+a \right ) ^{{\frac{5}{2}}}{x}^{-{\frac{13}{2}}}} \end{align*}

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 [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^(15/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.60978, size = 350, normalized size = 2.33 \begin{align*} -\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{align*}

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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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**(15/2),x)

[Out]

Timed out

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Giac [A]  time = 2.53017, size = 255, normalized size = 1.7 \begin{align*} -\frac{{\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^{7} b^{21}} - \frac{13 \,{\left (13 \, B a^{3} b^{12} - 8 \, A a^{2} b^{13}\right )}}{a^{7} b^{21}}\right )} + \frac{143 \,{\left (13 \, B a^{4} b^{12} - 8 \, A a^{3} b^{13}\right )}}{a^{7} b^{21}}\right )} - \frac{429 \,{\left (13 \, B a^{5} b^{12} - 8 \, A a^{4} b^{13}\right )}}{a^{7} b^{21}}\right )}{\left (b x + a\right )} + \frac{3003 \,{\left (B a^{6} b^{12} - A a^{5} b^{13}\right )}}{a^{7} b^{21}}\right )}{\left (b x + a\right )}^{\frac{5}{2}} b}{11070259200 \,{\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)^(3/2)*(B*x+A)/x^(15/2),x, algorithm="giac")

[Out]

-1/11070259200*((2*(b*x + a)*(4*(b*x + a)*(2*(13*B*a^2*b^12 - 8*A*a*b^13)*(b*x + a)/(a^7*b^21) - 13*(13*B*a^3*
b^12 - 8*A*a^2*b^13)/(a^7*b^21)) + 143*(13*B*a^4*b^12 - 8*A*a^3*b^13)/(a^7*b^21)) - 429*(13*B*a^5*b^12 - 8*A*a
^4*b^13)/(a^7*b^21))*(b*x + a) + 3003*(B*a^6*b^12 - A*a^5*b^13)/(a^7*b^21))*(b*x + a)^(5/2)*b/(((b*x + a)*b -
a*b)^(13/2)*abs(b))