3.513 \(\int \frac{(a+b x)^{5/2} (A+B x)}{x^{19/2}} \, dx\)

Optimal. Leaf size=183 \[ \frac{256 b^4 (a+b x)^{7/2} (10 A b-17 a B)}{765765 a^6 x^{7/2}}-\frac{128 b^3 (a+b x)^{7/2} (10 A b-17 a B)}{109395 a^5 x^{9/2}}+\frac{32 b^2 (a+b x)^{7/2} (10 A b-17 a B)}{12155 a^4 x^{11/2}}-\frac{16 b (a+b x)^{7/2} (10 A b-17 a B)}{3315 a^3 x^{13/2}}+\frac{2 (a+b x)^{7/2} (10 A b-17 a B)}{255 a^2 x^{15/2}}-\frac{2 A (a+b x)^{7/2}}{17 a x^{17/2}} \]

[Out]

(-2*A*(a + b*x)^(7/2))/(17*a*x^(17/2)) + (2*(10*A*b - 17*a*B)*(a + b*x)^(7/2))/(255*a^2*x^(15/2)) - (16*b*(10*
A*b - 17*a*B)*(a + b*x)^(7/2))/(3315*a^3*x^(13/2)) + (32*b^2*(10*A*b - 17*a*B)*(a + b*x)^(7/2))/(12155*a^4*x^(
11/2)) - (128*b^3*(10*A*b - 17*a*B)*(a + b*x)^(7/2))/(109395*a^5*x^(9/2)) + (256*b^4*(10*A*b - 17*a*B)*(a + b*
x)^(7/2))/(765765*a^6*x^(7/2))

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Rubi [A]  time = 0.0707995, 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} \[ \frac{256 b^4 (a+b x)^{7/2} (10 A b-17 a B)}{765765 a^6 x^{7/2}}-\frac{128 b^3 (a+b x)^{7/2} (10 A b-17 a B)}{109395 a^5 x^{9/2}}+\frac{32 b^2 (a+b x)^{7/2} (10 A b-17 a B)}{12155 a^4 x^{11/2}}-\frac{16 b (a+b x)^{7/2} (10 A b-17 a B)}{3315 a^3 x^{13/2}}+\frac{2 (a+b x)^{7/2} (10 A b-17 a B)}{255 a^2 x^{15/2}}-\frac{2 A (a+b x)^{7/2}}{17 a x^{17/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x^(19/2),x]

[Out]

(-2*A*(a + b*x)^(7/2))/(17*a*x^(17/2)) + (2*(10*A*b - 17*a*B)*(a + b*x)^(7/2))/(255*a^2*x^(15/2)) - (16*b*(10*
A*b - 17*a*B)*(a + b*x)^(7/2))/(3315*a^3*x^(13/2)) + (32*b^2*(10*A*b - 17*a*B)*(a + b*x)^(7/2))/(12155*a^4*x^(
11/2)) - (128*b^3*(10*A*b - 17*a*B)*(a + b*x)^(7/2))/(109395*a^5*x^(9/2)) + (256*b^4*(10*A*b - 17*a*B)*(a + b*
x)^(7/2))/(765765*a^6*x^(7/2))

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)^{5/2} (A+B x)}{x^{19/2}} \, dx &=-\frac{2 A (a+b x)^{7/2}}{17 a x^{17/2}}+\frac{\left (2 \left (-5 A b+\frac{17 a B}{2}\right )\right ) \int \frac{(a+b x)^{5/2}}{x^{17/2}} \, dx}{17 a}\\ &=-\frac{2 A (a+b x)^{7/2}}{17 a x^{17/2}}+\frac{2 (10 A b-17 a B) (a+b x)^{7/2}}{255 a^2 x^{15/2}}+\frac{(8 b (10 A b-17 a B)) \int \frac{(a+b x)^{5/2}}{x^{15/2}} \, dx}{255 a^2}\\ &=-\frac{2 A (a+b x)^{7/2}}{17 a x^{17/2}}+\frac{2 (10 A b-17 a B) (a+b x)^{7/2}}{255 a^2 x^{15/2}}-\frac{16 b (10 A b-17 a B) (a+b x)^{7/2}}{3315 a^3 x^{13/2}}-\frac{\left (16 b^2 (10 A b-17 a B)\right ) \int \frac{(a+b x)^{5/2}}{x^{13/2}} \, dx}{1105 a^3}\\ &=-\frac{2 A (a+b x)^{7/2}}{17 a x^{17/2}}+\frac{2 (10 A b-17 a B) (a+b x)^{7/2}}{255 a^2 x^{15/2}}-\frac{16 b (10 A b-17 a B) (a+b x)^{7/2}}{3315 a^3 x^{13/2}}+\frac{32 b^2 (10 A b-17 a B) (a+b x)^{7/2}}{12155 a^4 x^{11/2}}+\frac{\left (64 b^3 (10 A b-17 a B)\right ) \int \frac{(a+b x)^{5/2}}{x^{11/2}} \, dx}{12155 a^4}\\ &=-\frac{2 A (a+b x)^{7/2}}{17 a x^{17/2}}+\frac{2 (10 A b-17 a B) (a+b x)^{7/2}}{255 a^2 x^{15/2}}-\frac{16 b (10 A b-17 a B) (a+b x)^{7/2}}{3315 a^3 x^{13/2}}+\frac{32 b^2 (10 A b-17 a B) (a+b x)^{7/2}}{12155 a^4 x^{11/2}}-\frac{128 b^3 (10 A b-17 a B) (a+b x)^{7/2}}{109395 a^5 x^{9/2}}-\frac{\left (128 b^4 (10 A b-17 a B)\right ) \int \frac{(a+b x)^{5/2}}{x^{9/2}} \, dx}{109395 a^5}\\ &=-\frac{2 A (a+b x)^{7/2}}{17 a x^{17/2}}+\frac{2 (10 A b-17 a B) (a+b x)^{7/2}}{255 a^2 x^{15/2}}-\frac{16 b (10 A b-17 a B) (a+b x)^{7/2}}{3315 a^3 x^{13/2}}+\frac{32 b^2 (10 A b-17 a B) (a+b x)^{7/2}}{12155 a^4 x^{11/2}}-\frac{128 b^3 (10 A b-17 a B) (a+b x)^{7/2}}{109395 a^5 x^{9/2}}+\frac{256 b^4 (10 A b-17 a B) (a+b x)^{7/2}}{765765 a^6 x^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0448331, size = 114, normalized size = 0.62 \[ -\frac{2 (a+b x)^{7/2} \left (336 a^3 b^2 x^2 (55 A+51 B x)-224 a^2 b^3 x^3 (45 A+34 B x)-462 a^4 b x (65 A+68 B x)+3003 a^5 (15 A+17 B x)+128 a b^4 x^4 (35 A+17 B x)-1280 A b^5 x^5\right )}{765765 a^6 x^{17/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^(19/2),x]

[Out]

(-2*(a + b*x)^(7/2)*(-1280*A*b^5*x^5 + 3003*a^5*(15*A + 17*B*x) + 128*a*b^4*x^4*(35*A + 17*B*x) - 224*a^2*b^3*
x^3*(45*A + 34*B*x) + 336*a^3*b^2*x^2*(55*A + 51*B*x) - 462*a^4*b*x*(65*A + 68*B*x)))/(765765*a^6*x^(17/2))

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Maple [A]  time = 0.004, size = 125, normalized size = 0.7 \begin{align*} -{\frac{-2560\,A{b}^{5}{x}^{5}+4352\,B{x}^{5}a{b}^{4}+8960\,aA{b}^{4}{x}^{4}-15232\,B{x}^{4}{a}^{2}{b}^{3}-20160\,{a}^{2}A{b}^{3}{x}^{3}+34272\,B{x}^{3}{a}^{3}{b}^{2}+36960\,{a}^{3}A{b}^{2}{x}^{2}-62832\,B{x}^{2}{a}^{4}b-60060\,{a}^{4}Abx+102102\,{a}^{5}Bx+90090\,A{a}^{5}}{765765\,{a}^{6}} \left ( bx+a \right ) ^{{\frac{7}{2}}}{x}^{-{\frac{17}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^(19/2),x)

[Out]

-2/765765*(b*x+a)^(7/2)*(-1280*A*b^5*x^5+2176*B*a*b^4*x^5+4480*A*a*b^4*x^4-7616*B*a^2*b^3*x^4-10080*A*a^2*b^3*
x^3+17136*B*a^3*b^2*x^3+18480*A*a^3*b^2*x^2-31416*B*a^4*b*x^2-30030*A*a^4*b*x+51051*B*a^5*x+45045*A*a^5)/x^(17
/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)^(5/2)*(B*x+A)/x^(19/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.63717, size = 479, normalized size = 2.62 \begin{align*} -\frac{2 \,{\left (45045 \, A a^{8} + 128 \,{\left (17 \, B a b^{7} - 10 \, A b^{8}\right )} x^{8} - 64 \,{\left (17 \, B a^{2} b^{6} - 10 \, A a b^{7}\right )} x^{7} + 48 \,{\left (17 \, B a^{3} b^{5} - 10 \, A a^{2} b^{6}\right )} x^{6} - 40 \,{\left (17 \, B a^{4} b^{4} - 10 \, A a^{3} b^{5}\right )} x^{5} + 35 \,{\left (17 \, B a^{5} b^{3} - 10 \, A a^{4} b^{4}\right )} x^{4} + 63 \,{\left (1207 \, B a^{6} b^{2} + 5 \, A a^{5} b^{3}\right )} x^{3} + 231 \,{\left (527 \, B a^{7} b + 275 \, A a^{6} b^{2}\right )} x^{2} + 3003 \,{\left (17 \, B a^{8} + 35 \, A a^{7} b\right )} x\right )} \sqrt{b x + a}}{765765 \, a^{6} x^{\frac{17}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(19/2),x, algorithm="fricas")

[Out]

-2/765765*(45045*A*a^8 + 128*(17*B*a*b^7 - 10*A*b^8)*x^8 - 64*(17*B*a^2*b^6 - 10*A*a*b^7)*x^7 + 48*(17*B*a^3*b
^5 - 10*A*a^2*b^6)*x^6 - 40*(17*B*a^4*b^4 - 10*A*a^3*b^5)*x^5 + 35*(17*B*a^5*b^3 - 10*A*a^4*b^4)*x^4 + 63*(120
7*B*a^6*b^2 + 5*A*a^5*b^3)*x^3 + 231*(527*B*a^7*b + 275*A*a^6*b^2)*x^2 + 3003*(17*B*a^8 + 35*A*a^7*b)*x)*sqrt(
b*x + a)/(a^6*x^(17/2))

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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)**(5/2)*(B*x+A)/x**(19/2),x)

[Out]

Timed out

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Giac [A]  time = 1.50431, size = 305, normalized size = 1.67 \begin{align*} \frac{{\left ({\left (8 \,{\left (2 \,{\left (b x + a\right )}{\left (4 \,{\left (b x + a\right )}{\left (\frac{2 \,{\left (17 \, B a^{3} b^{16} - 10 \, A a^{2} b^{17}\right )}{\left (b x + a\right )}}{a^{9} b^{27}} - \frac{17 \,{\left (17 \, B a^{4} b^{16} - 10 \, A a^{3} b^{17}\right )}}{a^{9} b^{27}}\right )} + \frac{255 \,{\left (17 \, B a^{5} b^{16} - 10 \, A a^{4} b^{17}\right )}}{a^{9} b^{27}}\right )} - \frac{1105 \,{\left (17 \, B a^{6} b^{16} - 10 \, A a^{5} b^{17}\right )}}{a^{9} b^{27}}\right )}{\left (b x + a\right )} + \frac{12155 \,{\left (17 \, B a^{7} b^{16} - 10 \, A a^{6} b^{17}\right )}}{a^{9} b^{27}}\right )}{\left (b x + a\right )} - \frac{109395 \,{\left (B a^{8} b^{16} - A a^{7} b^{17}\right )}}{a^{9} b^{27}}\right )}{\left (b x + a\right )}^{\frac{7}{2}} b}{200740700160 \,{\left ({\left (b x + a\right )} b - a b\right )}^{\frac{17}{2}}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(19/2),x, algorithm="giac")

[Out]

1/200740700160*((8*(2*(b*x + a)*(4*(b*x + a)*(2*(17*B*a^3*b^16 - 10*A*a^2*b^17)*(b*x + a)/(a^9*b^27) - 17*(17*
B*a^4*b^16 - 10*A*a^3*b^17)/(a^9*b^27)) + 255*(17*B*a^5*b^16 - 10*A*a^4*b^17)/(a^9*b^27)) - 1105*(17*B*a^6*b^1
6 - 10*A*a^5*b^17)/(a^9*b^27))*(b*x + a) + 12155*(17*B*a^7*b^16 - 10*A*a^6*b^17)/(a^9*b^27))*(b*x + a) - 10939
5*(B*a^8*b^16 - A*a^7*b^17)/(a^9*b^27))*(b*x + a)^(7/2)*b/(((b*x + a)*b - a*b)^(17/2)*abs(b))