3.544 \(\int \frac{A+B x}{x^{7/2} (a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=144 \[ \frac{16 \sqrt{a+b x} (8 A b-5 a B)}{15 a^4 x^{3/2}}-\frac{4 (8 A b-5 a B)}{5 a^3 x^{3/2} \sqrt{a+b x}}-\frac{2 (8 A b-5 a B)}{15 a^2 x^{3/2} (a+b x)^{3/2}}-\frac{32 b \sqrt{a+b x} (8 A b-5 a B)}{15 a^5 \sqrt{x}}-\frac{2 A}{5 a x^{5/2} (a+b x)^{3/2}} \]

[Out]

(-2*A)/(5*a*x^(5/2)*(a + b*x)^(3/2)) - (2*(8*A*b - 5*a*B))/(15*a^2*x^(3/2)*(a + b*x)^(3/2)) - (4*(8*A*b - 5*a*
B))/(5*a^3*x^(3/2)*Sqrt[a + b*x]) + (16*(8*A*b - 5*a*B)*Sqrt[a + b*x])/(15*a^4*x^(3/2)) - (32*b*(8*A*b - 5*a*B
)*Sqrt[a + b*x])/(15*a^5*Sqrt[x])

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Rubi [A]  time = 0.0493079, antiderivative size = 144, 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} \[ \frac{16 \sqrt{a+b x} (8 A b-5 a B)}{15 a^4 x^{3/2}}-\frac{4 (8 A b-5 a B)}{5 a^3 x^{3/2} \sqrt{a+b x}}-\frac{2 (8 A b-5 a B)}{15 a^2 x^{3/2} (a+b x)^{3/2}}-\frac{32 b \sqrt{a+b x} (8 A b-5 a B)}{15 a^5 \sqrt{x}}-\frac{2 A}{5 a x^{5/2} (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(7/2)*(a + b*x)^(5/2)),x]

[Out]

(-2*A)/(5*a*x^(5/2)*(a + b*x)^(3/2)) - (2*(8*A*b - 5*a*B))/(15*a^2*x^(3/2)*(a + b*x)^(3/2)) - (4*(8*A*b - 5*a*
B))/(5*a^3*x^(3/2)*Sqrt[a + b*x]) + (16*(8*A*b - 5*a*B)*Sqrt[a + b*x])/(15*a^4*x^(3/2)) - (32*b*(8*A*b - 5*a*B
)*Sqrt[a + b*x])/(15*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^{7/2} (a+b x)^{5/2}} \, dx &=-\frac{2 A}{5 a x^{5/2} (a+b x)^{3/2}}+\frac{\left (2 \left (-4 A b+\frac{5 a B}{2}\right )\right ) \int \frac{1}{x^{5/2} (a+b x)^{5/2}} \, dx}{5 a}\\ &=-\frac{2 A}{5 a x^{5/2} (a+b x)^{3/2}}-\frac{2 (8 A b-5 a B)}{15 a^2 x^{3/2} (a+b x)^{3/2}}-\frac{(2 (8 A b-5 a B)) \int \frac{1}{x^{5/2} (a+b x)^{3/2}} \, dx}{5 a^2}\\ &=-\frac{2 A}{5 a x^{5/2} (a+b x)^{3/2}}-\frac{2 (8 A b-5 a B)}{15 a^2 x^{3/2} (a+b x)^{3/2}}-\frac{4 (8 A b-5 a B)}{5 a^3 x^{3/2} \sqrt{a+b x}}-\frac{(8 (8 A b-5 a B)) \int \frac{1}{x^{5/2} \sqrt{a+b x}} \, dx}{5 a^3}\\ &=-\frac{2 A}{5 a x^{5/2} (a+b x)^{3/2}}-\frac{2 (8 A b-5 a B)}{15 a^2 x^{3/2} (a+b x)^{3/2}}-\frac{4 (8 A b-5 a B)}{5 a^3 x^{3/2} \sqrt{a+b x}}+\frac{16 (8 A b-5 a B) \sqrt{a+b x}}{15 a^4 x^{3/2}}+\frac{(16 b (8 A b-5 a B)) \int \frac{1}{x^{3/2} \sqrt{a+b x}} \, dx}{15 a^4}\\ &=-\frac{2 A}{5 a x^{5/2} (a+b x)^{3/2}}-\frac{2 (8 A b-5 a B)}{15 a^2 x^{3/2} (a+b x)^{3/2}}-\frac{4 (8 A b-5 a B)}{5 a^3 x^{3/2} \sqrt{a+b x}}+\frac{16 (8 A b-5 a B) \sqrt{a+b x}}{15 a^4 x^{3/2}}-\frac{32 b (8 A b-5 a B) \sqrt{a+b x}}{15 a^5 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0303987, size = 94, normalized size = 0.65 \[ -\frac{2 \left (24 a^2 b^2 x^2 (2 A-5 B x)-2 a^3 b x (4 A+15 B x)+a^4 (3 A+5 B x)+16 a b^3 x^3 (12 A-5 B x)+128 A b^4 x^4\right )}{15 a^5 x^{5/2} (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(7/2)*(a + b*x)^(5/2)),x]

[Out]

(-2*(128*A*b^4*x^4 + 24*a^2*b^2*x^2*(2*A - 5*B*x) + 16*a*b^3*x^3*(12*A - 5*B*x) + a^4*(3*A + 5*B*x) - 2*a^3*b*
x*(4*A + 15*B*x)))/(15*a^5*x^(5/2)*(a + b*x)^(3/2))

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Maple [A]  time = 0.006, size = 101, normalized size = 0.7 \begin{align*} -{\frac{256\,A{b}^{4}{x}^{4}-160\,Ba{b}^{3}{x}^{4}+384\,Aa{b}^{3}{x}^{3}-240\,B{a}^{2}{b}^{2}{x}^{3}+96\,A{a}^{2}{b}^{2}{x}^{2}-60\,B{a}^{3}b{x}^{2}-16\,A{a}^{3}bx+10\,B{a}^{4}x+6\,A{a}^{4}}{15\,{a}^{5}}{x}^{-{\frac{5}{2}}} \left ( bx+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(7/2)/(b*x+a)^(5/2),x)

[Out]

-2/15*(128*A*b^4*x^4-80*B*a*b^3*x^4+192*A*a*b^3*x^3-120*B*a^2*b^2*x^3+48*A*a^2*b^2*x^2-30*B*a^3*b*x^2-8*A*a^3*
b*x+5*B*a^4*x+3*A*a^4)/x^(5/2)/(b*x+a)^(3/2)/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^(7/2)/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.69795, size = 274, normalized size = 1.9 \begin{align*} -\frac{2 \,{\left (3 \, A a^{4} - 16 \,{\left (5 \, B a b^{3} - 8 \, A b^{4}\right )} x^{4} - 24 \,{\left (5 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x^{3} - 6 \,{\left (5 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x^{2} +{\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{15 \,{\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} 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)^(5/2),x, algorithm="fricas")

[Out]

-2/15*(3*A*a^4 - 16*(5*B*a*b^3 - 8*A*b^4)*x^4 - 24*(5*B*a^2*b^2 - 8*A*a*b^3)*x^3 - 6*(5*B*a^3*b - 8*A*a^2*b^2)
*x^2 + (5*B*a^4 - 8*A*a^3*b)*x)*sqrt(b*x + a)*sqrt(x)/(a^5*b^2*x^5 + 2*a^6*b*x^4 + a^7*x^3)

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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**(7/2)/(b*x+a)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.57463, size = 444, normalized size = 3.08 \begin{align*} -\frac{\sqrt{b x + a}{\left ({\left (b x + a\right )}{\left (\frac{{\left (40 \, B a^{8} b^{7} - 73 \, A a^{7} b^{8}\right )}{\left (b x + a\right )}}{a^{3} b^{9}} - \frac{5 \,{\left (17 \, B a^{9} b^{7} - 32 \, A a^{8} b^{8}\right )}}{a^{3} b^{9}}\right )} + \frac{45 \,{\left (B a^{10} b^{7} - 2 \, A a^{9} 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 (6 \, B a{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac{7}{2}} + 18 \, B a^{2}{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac{9}{2}} - 9 \, A{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac{9}{2}} + 8 \, B a^{3} b^{\frac{11}{2}} - 24 \, A a{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac{11}{2}} - 11 \, A a^{2} b^{\frac{13}{2}}\right )}}{3 \,{\left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3} a^{4}{\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)^(5/2),x, algorithm="giac")

[Out]

-1/960*sqrt(b*x + a)*((b*x + a)*((40*B*a^8*b^7 - 73*A*a^7*b^8)*(b*x + a)/(a^3*b^9) - 5*(17*B*a^9*b^7 - 32*A*a^
8*b^8)/(a^3*b^9)) + 45*(B*a^10*b^7 - 2*A*a^9*b^8)/(a^3*b^9))/((b*x + a)*b - a*b)^(5/2) + 4/3*(6*B*a*(sqrt(b*x
+ a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*b^(7/2) + 18*B*a^2*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))
^2*b^(9/2) - 9*A*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*b^(9/2) + 8*B*a^3*b^(11/2) - 24*A*a*(sqrt
(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(11/2) - 11*A*a^2*b^(13/2))/(((sqrt(b*x + a)*sqrt(b) - sqrt((
b*x + a)*b - a*b))^2 + a*b)^3*a^4*abs(b))