3.522 \(\int \frac{A+B x}{x^{9/2} \sqrt{a+b x}} \, dx\)

Optimal. Leaf size=117 \[ \frac{16 b^2 \sqrt{a+b x} (6 A b-7 a B)}{105 a^4 \sqrt{x}}-\frac{8 b \sqrt{a+b x} (6 A b-7 a B)}{105 a^3 x^{3/2}}+\frac{2 \sqrt{a+b x} (6 A b-7 a B)}{35 a^2 x^{5/2}}-\frac{2 A \sqrt{a+b x}}{7 a x^{7/2}} \]

[Out]

(-2*A*Sqrt[a + b*x])/(7*a*x^(7/2)) + (2*(6*A*b - 7*a*B)*Sqrt[a + b*x])/(35*a^2*x^(5/2)) - (8*b*(6*A*b - 7*a*B)
*Sqrt[a + b*x])/(105*a^3*x^(3/2)) + (16*b^2*(6*A*b - 7*a*B)*Sqrt[a + b*x])/(105*a^4*Sqrt[x])

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Rubi [A]  time = 0.039461, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {78, 45, 37} \[ \frac{16 b^2 \sqrt{a+b x} (6 A b-7 a B)}{105 a^4 \sqrt{x}}-\frac{8 b \sqrt{a+b x} (6 A b-7 a B)}{105 a^3 x^{3/2}}+\frac{2 \sqrt{a+b x} (6 A b-7 a B)}{35 a^2 x^{5/2}}-\frac{2 A \sqrt{a+b x}}{7 a x^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(9/2)*Sqrt[a + b*x]),x]

[Out]

(-2*A*Sqrt[a + b*x])/(7*a*x^(7/2)) + (2*(6*A*b - 7*a*B)*Sqrt[a + b*x])/(35*a^2*x^(5/2)) - (8*b*(6*A*b - 7*a*B)
*Sqrt[a + b*x])/(105*a^3*x^(3/2)) + (16*b^2*(6*A*b - 7*a*B)*Sqrt[a + b*x])/(105*a^4*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^{9/2} \sqrt{a+b x}} \, dx &=-\frac{2 A \sqrt{a+b x}}{7 a x^{7/2}}+\frac{\left (2 \left (-3 A b+\frac{7 a B}{2}\right )\right ) \int \frac{1}{x^{7/2} \sqrt{a+b x}} \, dx}{7 a}\\ &=-\frac{2 A \sqrt{a+b x}}{7 a x^{7/2}}+\frac{2 (6 A b-7 a B) \sqrt{a+b x}}{35 a^2 x^{5/2}}+\frac{(4 b (6 A b-7 a B)) \int \frac{1}{x^{5/2} \sqrt{a+b x}} \, dx}{35 a^2}\\ &=-\frac{2 A \sqrt{a+b x}}{7 a x^{7/2}}+\frac{2 (6 A b-7 a B) \sqrt{a+b x}}{35 a^2 x^{5/2}}-\frac{8 b (6 A b-7 a B) \sqrt{a+b x}}{105 a^3 x^{3/2}}-\frac{\left (8 b^2 (6 A b-7 a B)\right ) \int \frac{1}{x^{3/2} \sqrt{a+b x}} \, dx}{105 a^3}\\ &=-\frac{2 A \sqrt{a+b x}}{7 a x^{7/2}}+\frac{2 (6 A b-7 a B) \sqrt{a+b x}}{35 a^2 x^{5/2}}-\frac{8 b (6 A b-7 a B) \sqrt{a+b x}}{105 a^3 x^{3/2}}+\frac{16 b^2 (6 A b-7 a B) \sqrt{a+b x}}{105 a^4 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0223751, size = 76, normalized size = 0.65 \[ -\frac{2 \sqrt{a+b x} \left (-2 a^2 b x (9 A+14 B x)+3 a^3 (5 A+7 B x)+8 a b^2 x^2 (3 A+7 B x)-48 A b^3 x^3\right )}{105 a^4 x^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(9/2)*Sqrt[a + b*x]),x]

[Out]

(-2*Sqrt[a + b*x]*(-48*A*b^3*x^3 + 8*a*b^2*x^2*(3*A + 7*B*x) + 3*a^3*(5*A + 7*B*x) - 2*a^2*b*x*(9*A + 14*B*x))
)/(105*a^4*x^(7/2))

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Maple [A]  time = 0.004, size = 77, normalized size = 0.7 \begin{align*} -{\frac{-96\,A{b}^{3}{x}^{3}+112\,B{x}^{3}a{b}^{2}+48\,aA{b}^{2}{x}^{2}-56\,B{x}^{2}{a}^{2}b-36\,{a}^{2}Abx+42\,{a}^{3}Bx+30\,A{a}^{3}}{105\,{a}^{4}}\sqrt{bx+a}{x}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(9/2)/(b*x+a)^(1/2),x)

[Out]

-2/105*(b*x+a)^(1/2)*(-48*A*b^3*x^3+56*B*a*b^2*x^3+24*A*a*b^2*x^2-28*B*a^2*b*x^2-18*A*a^2*b*x+21*B*a^3*x+15*A*
a^3)/x^(7/2)/a^4

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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^(9/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.6929, size = 185, normalized size = 1.58 \begin{align*} -\frac{2 \,{\left (15 \, A a^{3} + 8 \,{\left (7 \, B a b^{2} - 6 \, A b^{3}\right )} x^{3} - 4 \,{\left (7 \, B a^{2} b - 6 \, A a b^{2}\right )} x^{2} + 3 \,{\left (7 \, B a^{3} - 6 \, A a^{2} b\right )} x\right )} \sqrt{b x + a}}{105 \, a^{4} x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(9/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

-2/105*(15*A*a^3 + 8*(7*B*a*b^2 - 6*A*b^3)*x^3 - 4*(7*B*a^2*b - 6*A*a*b^2)*x^2 + 3*(7*B*a^3 - 6*A*a^2*b)*x)*sq
rt(b*x + a)/(a^4*x^(7/2))

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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**(9/2)/(b*x+a)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.3993, size = 201, normalized size = 1.72 \begin{align*} \frac{{\left ({\left (b x + a\right )}{\left (4 \,{\left (b x + a\right )}{\left (\frac{2 \,{\left (7 \, B a b^{6} - 6 \, A b^{7}\right )}{\left (b x + a\right )}}{a^{4} b^{12}} - \frac{7 \,{\left (7 \, B a^{2} b^{6} - 6 \, A a b^{7}\right )}}{a^{4} b^{12}}\right )} + \frac{35 \,{\left (7 \, B a^{3} b^{6} - 6 \, A a^{2} b^{7}\right )}}{a^{4} b^{12}}\right )} - \frac{105 \,{\left (B a^{4} b^{6} - A a^{3} b^{7}\right )}}{a^{4} b^{12}}\right )} \sqrt{b x + a} b}{80640 \,{\left ({\left (b x + a\right )} b - a b\right )}^{\frac{7}{2}}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(9/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/80640*((b*x + a)*(4*(b*x + a)*(2*(7*B*a*b^6 - 6*A*b^7)*(b*x + a)/(a^4*b^12) - 7*(7*B*a^2*b^6 - 6*A*a*b^7)/(a
^4*b^12)) + 35*(7*B*a^3*b^6 - 6*A*a^2*b^7)/(a^4*b^12)) - 105*(B*a^4*b^6 - A*a^3*b^7)/(a^4*b^12))*sqrt(b*x + a)
*b/(((b*x + a)*b - a*b)^(7/2)*abs(b))