3.533 \(\int \frac{A+B x}{x^{7/2} (a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=114 \[ \frac{8 \sqrt{a+b x} (6 A b-5 a B)}{15 a^3 x^{3/2}}-\frac{2 (6 A b-5 a B)}{5 a^2 x^{3/2} \sqrt{a+b x}}-\frac{16 b \sqrt{a+b x} (6 A b-5 a B)}{15 a^4 \sqrt{x}}-\frac{2 A}{5 a x^{5/2} \sqrt{a+b x}} \]

[Out]

(-2*A)/(5*a*x^(5/2)*Sqrt[a + b*x]) - (2*(6*A*b - 5*a*B))/(5*a^2*x^(3/2)*Sqrt[a + b*x]) + (8*(6*A*b - 5*a*B)*Sq
rt[a + b*x])/(15*a^3*x^(3/2)) - (16*b*(6*A*b - 5*a*B)*Sqrt[a + b*x])/(15*a^4*Sqrt[x])

________________________________________________________________________________________

Rubi [A]  time = 0.0369315, antiderivative size = 114, 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} \[ \frac{8 \sqrt{a+b x} (6 A b-5 a B)}{15 a^3 x^{3/2}}-\frac{2 (6 A b-5 a B)}{5 a^2 x^{3/2} \sqrt{a+b x}}-\frac{16 b \sqrt{a+b x} (6 A b-5 a B)}{15 a^4 \sqrt{x}}-\frac{2 A}{5 a x^{5/2} \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(7/2)*(a + b*x)^(3/2)),x]

[Out]

(-2*A)/(5*a*x^(5/2)*Sqrt[a + b*x]) - (2*(6*A*b - 5*a*B))/(5*a^2*x^(3/2)*Sqrt[a + b*x]) + (8*(6*A*b - 5*a*B)*Sq
rt[a + b*x])/(15*a^3*x^(3/2)) - (16*b*(6*A*b - 5*a*B)*Sqrt[a + b*x])/(15*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^{7/2} (a+b x)^{3/2}} \, dx &=-\frac{2 A}{5 a x^{5/2} \sqrt{a+b x}}+\frac{\left (2 \left (-3 A b+\frac{5 a B}{2}\right )\right ) \int \frac{1}{x^{5/2} (a+b x)^{3/2}} \, dx}{5 a}\\ &=-\frac{2 A}{5 a x^{5/2} \sqrt{a+b x}}-\frac{2 (6 A b-5 a B)}{5 a^2 x^{3/2} \sqrt{a+b x}}-\frac{(4 (6 A b-5 a B)) \int \frac{1}{x^{5/2} \sqrt{a+b x}} \, dx}{5 a^2}\\ &=-\frac{2 A}{5 a x^{5/2} \sqrt{a+b x}}-\frac{2 (6 A b-5 a B)}{5 a^2 x^{3/2} \sqrt{a+b x}}+\frac{8 (6 A b-5 a B) \sqrt{a+b x}}{15 a^3 x^{3/2}}+\frac{(8 b (6 A b-5 a B)) \int \frac{1}{x^{3/2} \sqrt{a+b x}} \, dx}{15 a^3}\\ &=-\frac{2 A}{5 a x^{5/2} \sqrt{a+b x}}-\frac{2 (6 A b-5 a B)}{5 a^2 x^{3/2} \sqrt{a+b x}}+\frac{8 (6 A b-5 a B) \sqrt{a+b x}}{15 a^3 x^{3/2}}-\frac{16 b (6 A b-5 a B) \sqrt{a+b x}}{15 a^4 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0225055, size = 75, normalized size = 0.66 \[ -\frac{2 \left (-2 a^2 b x (3 A+10 B x)+a^3 (3 A+5 B x)+8 a b^2 x^2 (3 A-5 B x)+48 A b^3 x^3\right )}{15 a^4 x^{5/2} \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(7/2)*(a + b*x)^(3/2)),x]

[Out]

(-2*(48*A*b^3*x^3 + 8*a*b^2*x^2*(3*A - 5*B*x) + a^3*(3*A + 5*B*x) - 2*a^2*b*x*(3*A + 10*B*x)))/(15*a^4*x^(5/2)
*Sqrt[a + b*x])

________________________________________________________________________________________

Maple [A]  time = 0.002, size = 77, normalized size = 0.7 \begin{align*} -{\frac{96\,A{b}^{3}{x}^{3}-80\,B{x}^{3}a{b}^{2}+48\,aA{b}^{2}{x}^{2}-40\,B{x}^{2}{a}^{2}b-12\,{a}^{2}Abx+10\,{a}^{3}Bx+6\,A{a}^{3}}{15\,{a}^{4}}{x}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(7/2)/(b*x+a)^(3/2),x)

[Out]

-2/15*(48*A*b^3*x^3-40*B*a*b^2*x^3+24*A*a*b^2*x^2-20*B*a^2*b*x^2-6*A*a^2*b*x+5*B*a^3*x+3*A*a^3)/x^(5/2)/(b*x+a
)^(1/2)/a^4

________________________________________________________________________________________

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^(7/2)/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.93645, size = 201, normalized size = 1.76 \begin{align*} -\frac{2 \,{\left (3 \, A a^{3} - 8 \,{\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x^{3} - 4 \,{\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} x^{2} +{\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{15 \,{\left (a^{4} b x^{4} + a^{5} x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

-2/15*(3*A*a^3 - 8*(5*B*a*b^2 - 6*A*b^3)*x^3 - 4*(5*B*a^2*b - 6*A*a*b^2)*x^2 + (5*B*a^3 - 6*A*a^2*b)*x)*sqrt(b
*x + a)*sqrt(x)/(a^4*b*x^4 + a^5*x^3)

________________________________________________________________________________________

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**(7/2)/(b*x+a)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 2.42516, size = 236, normalized size = 2.07 \begin{align*} -\frac{\sqrt{b x + a}{\left ({\left (b x + a\right )}{\left (\frac{{\left (25 \, B a^{6} b^{7} - 33 \, A a^{5} b^{8}\right )}{\left (b x + a\right )}}{a^{3} b^{9}} - \frac{5 \,{\left (11 \, B a^{7} b^{7} - 15 \, A a^{6} b^{8}\right )}}{a^{3} b^{9}}\right )} + \frac{15 \,{\left (2 \, B a^{8} b^{7} - 3 \, A a^{7} b^{8}\right )}}{a^{3} b^{9}}\right )}}{960 \,{\left ({\left (b x + a\right )} b - a b\right )}^{\frac{5}{2}}} + \frac{4 \,{\left (B a b^{\frac{7}{2}} - A b^{\frac{9}{2}}\right )}}{{\left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} a^{3}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

-1/960*sqrt(b*x + a)*((b*x + a)*((25*B*a^6*b^7 - 33*A*a^5*b^8)*(b*x + a)/(a^3*b^9) - 5*(11*B*a^7*b^7 - 15*A*a^
6*b^8)/(a^3*b^9)) + 15*(2*B*a^8*b^7 - 3*A*a^7*b^8)/(a^3*b^9))/((b*x + a)*b - a*b)^(5/2) + 4*(B*a*b^(7/2) - A*b
^(9/2))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)*a^3*abs(b))