3.523 \(\int \frac{A+B x}{x^{11/2} \sqrt{a+b x}} \, dx\)

Optimal. Leaf size=150 \[ \frac{16 b^2 \sqrt{a+b x} (8 A b-9 a B)}{315 a^4 x^{3/2}}-\frac{32 b^3 \sqrt{a+b x} (8 A b-9 a B)}{315 a^5 \sqrt{x}}-\frac{4 b \sqrt{a+b x} (8 A b-9 a B)}{105 a^3 x^{5/2}}+\frac{2 \sqrt{a+b x} (8 A b-9 a B)}{63 a^2 x^{7/2}}-\frac{2 A \sqrt{a+b x}}{9 a x^{9/2}} \]

[Out]

(-2*A*Sqrt[a + b*x])/(9*a*x^(9/2)) + (2*(8*A*b - 9*a*B)*Sqrt[a + b*x])/(63*a^2*x^(7/2)) - (4*b*(8*A*b - 9*a*B)
*Sqrt[a + b*x])/(105*a^3*x^(5/2)) + (16*b^2*(8*A*b - 9*a*B)*Sqrt[a + b*x])/(315*a^4*x^(3/2)) - (32*b^3*(8*A*b
- 9*a*B)*Sqrt[a + b*x])/(315*a^5*Sqrt[x])

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Rubi [A]  time = 0.0573826, 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} \[ \frac{16 b^2 \sqrt{a+b x} (8 A b-9 a B)}{315 a^4 x^{3/2}}-\frac{32 b^3 \sqrt{a+b x} (8 A b-9 a B)}{315 a^5 \sqrt{x}}-\frac{4 b \sqrt{a+b x} (8 A b-9 a B)}{105 a^3 x^{5/2}}+\frac{2 \sqrt{a+b x} (8 A b-9 a B)}{63 a^2 x^{7/2}}-\frac{2 A \sqrt{a+b x}}{9 a x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(11/2)*Sqrt[a + b*x]),x]

[Out]

(-2*A*Sqrt[a + b*x])/(9*a*x^(9/2)) + (2*(8*A*b - 9*a*B)*Sqrt[a + b*x])/(63*a^2*x^(7/2)) - (4*b*(8*A*b - 9*a*B)
*Sqrt[a + b*x])/(105*a^3*x^(5/2)) + (16*b^2*(8*A*b - 9*a*B)*Sqrt[a + b*x])/(315*a^4*x^(3/2)) - (32*b^3*(8*A*b
- 9*a*B)*Sqrt[a + b*x])/(315*a^5*Sqrt[x])

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}{x^{11/2} \sqrt{a+b x}} \, dx &=-\frac{2 A \sqrt{a+b x}}{9 a x^{9/2}}+\frac{\left (2 \left (-4 A b+\frac{9 a B}{2}\right )\right ) \int \frac{1}{x^{9/2} \sqrt{a+b x}} \, dx}{9 a}\\ &=-\frac{2 A \sqrt{a+b x}}{9 a x^{9/2}}+\frac{2 (8 A b-9 a B) \sqrt{a+b x}}{63 a^2 x^{7/2}}+\frac{(2 b (8 A b-9 a B)) \int \frac{1}{x^{7/2} \sqrt{a+b x}} \, dx}{21 a^2}\\ &=-\frac{2 A \sqrt{a+b x}}{9 a x^{9/2}}+\frac{2 (8 A b-9 a B) \sqrt{a+b x}}{63 a^2 x^{7/2}}-\frac{4 b (8 A b-9 a B) \sqrt{a+b x}}{105 a^3 x^{5/2}}-\frac{\left (8 b^2 (8 A b-9 a B)\right ) \int \frac{1}{x^{5/2} \sqrt{a+b x}} \, dx}{105 a^3}\\ &=-\frac{2 A \sqrt{a+b x}}{9 a x^{9/2}}+\frac{2 (8 A b-9 a B) \sqrt{a+b x}}{63 a^2 x^{7/2}}-\frac{4 b (8 A b-9 a B) \sqrt{a+b x}}{105 a^3 x^{5/2}}+\frac{16 b^2 (8 A b-9 a B) \sqrt{a+b x}}{315 a^4 x^{3/2}}+\frac{\left (16 b^3 (8 A b-9 a B)\right ) \int \frac{1}{x^{3/2} \sqrt{a+b x}} \, dx}{315 a^4}\\ &=-\frac{2 A \sqrt{a+b x}}{9 a x^{9/2}}+\frac{2 (8 A b-9 a B) \sqrt{a+b x}}{63 a^2 x^{7/2}}-\frac{4 b (8 A b-9 a B) \sqrt{a+b x}}{105 a^3 x^{5/2}}+\frac{16 b^2 (8 A b-9 a B) \sqrt{a+b x}}{315 a^4 x^{3/2}}-\frac{32 b^3 (8 A b-9 a B) \sqrt{a+b x}}{315 a^5 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.028888, size = 95, normalized size = 0.63 \[ -\frac{2 \sqrt{a+b x} \left (24 a^2 b^2 x^2 (2 A+3 B x)-2 a^3 b x (20 A+27 B x)+5 a^4 (7 A+9 B x)-16 a b^3 x^3 (4 A+9 B x)+128 A b^4 x^4\right )}{315 a^5 x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(11/2)*Sqrt[a + b*x]),x]

[Out]

(-2*Sqrt[a + b*x]*(128*A*b^4*x^4 + 24*a^2*b^2*x^2*(2*A + 3*B*x) - 16*a*b^3*x^3*(4*A + 9*B*x) + 5*a^4*(7*A + 9*
B*x) - 2*a^3*b*x*(20*A + 27*B*x)))/(315*a^5*x^(9/2))

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Maple [A]  time = 0.005, size = 101, normalized size = 0.7 \begin{align*} -{\frac{256\,A{b}^{4}{x}^{4}-288\,Ba{b}^{3}{x}^{4}-128\,Aa{b}^{3}{x}^{3}+144\,B{a}^{2}{b}^{2}{x}^{3}+96\,A{a}^{2}{b}^{2}{x}^{2}-108\,B{a}^{3}b{x}^{2}-80\,A{a}^{3}bx+90\,B{a}^{4}x+70\,A{a}^{4}}{315\,{a}^{5}}\sqrt{bx+a}{x}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(11/2)/(b*x+a)^(1/2),x)

[Out]

-2/315*(b*x+a)^(1/2)*(128*A*b^4*x^4-144*B*a*b^3*x^4-64*A*a*b^3*x^3+72*B*a^2*b^2*x^3+48*A*a^2*b^2*x^2-54*B*a^3*
b*x^2-40*A*a^3*b*x+45*B*a^4*x+35*A*a^4)/x^(9/2)/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)/x^(11/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.69322, size = 235, normalized size = 1.57 \begin{align*} -\frac{2 \,{\left (35 \, A a^{4} - 16 \,{\left (9 \, B a b^{3} - 8 \, A b^{4}\right )} x^{4} + 8 \,{\left (9 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x^{3} - 6 \,{\left (9 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x^{2} + 5 \,{\left (9 \, B a^{4} - 8 \, A a^{3} b\right )} x\right )} \sqrt{b x + a}}{315 \, a^{5} x^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(11/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

-2/315*(35*A*a^4 - 16*(9*B*a*b^3 - 8*A*b^4)*x^4 + 8*(9*B*a^2*b^2 - 8*A*a*b^3)*x^3 - 6*(9*B*a^3*b - 8*A*a^2*b^2
)*x^2 + 5*(9*B*a^4 - 8*A*a^3*b)*x)*sqrt(b*x + a)/(a^5*x^(9/2))

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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)/x**(11/2)/(b*x+a)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.2511, size = 248, normalized size = 1.65 \begin{align*} -\frac{{\left ({\left (2 \,{\left (b x + a\right )}{\left (4 \,{\left (b x + a\right )}{\left (\frac{2 \,{\left (9 \, B a b^{8} - 8 \, A b^{9}\right )}{\left (b x + a\right )}}{a^{5} b^{15}} - \frac{9 \,{\left (9 \, B a^{2} b^{8} - 8 \, A a b^{9}\right )}}{a^{5} b^{15}}\right )} + \frac{63 \,{\left (9 \, B a^{3} b^{8} - 8 \, A a^{2} b^{9}\right )}}{a^{5} b^{15}}\right )} - \frac{105 \,{\left (9 \, B a^{4} b^{8} - 8 \, A a^{3} b^{9}\right )}}{a^{5} b^{15}}\right )}{\left (b x + a\right )} + \frac{315 \,{\left (B a^{5} b^{8} - A a^{4} b^{9}\right )}}{a^{5} b^{15}}\right )} \sqrt{b x + a} b}{322560 \,{\left ({\left (b x + a\right )} b - a b\right )}^{\frac{9}{2}}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(11/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

-1/322560*((2*(b*x + a)*(4*(b*x + a)*(2*(9*B*a*b^8 - 8*A*b^9)*(b*x + a)/(a^5*b^15) - 9*(9*B*a^2*b^8 - 8*A*a*b^
9)/(a^5*b^15)) + 63*(9*B*a^3*b^8 - 8*A*a^2*b^9)/(a^5*b^15)) - 105*(9*B*a^4*b^8 - 8*A*a^3*b^9)/(a^5*b^15))*(b*x
+ a) + 315*(B*a^5*b^8 - A*a^4*b^9)/(a^5*b^15))*sqrt(b*x + a)*b/(((b*x + a)*b - a*b)^(9/2)*abs(b))