3.2258 \(\int \frac{A+B x}{(a+b x)^{5/2} (d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=139 \[ -\frac{2 (B d-A e)}{e (a+b x)^{3/2} \sqrt{d+e x} (b d-a e)}-\frac{4 \sqrt{d+e x} (a B e-4 A b e+3 b B d)}{3 \sqrt{a+b x} (b d-a e)^3}+\frac{2 \sqrt{d+e x} (a B e-4 A b e+3 b B d)}{3 e (a+b x)^{3/2} (b d-a e)^2} \]

[Out]

(-2*(B*d - A*e))/(e*(b*d - a*e)*(a + b*x)^(3/2)*Sqrt[d + e*x]) + (2*(3*b*B*d - 4*A*b*e + a*B*e)*Sqrt[d + e*x])
/(3*e*(b*d - a*e)^2*(a + b*x)^(3/2)) - (4*(3*b*B*d - 4*A*b*e + a*B*e)*Sqrt[d + e*x])/(3*(b*d - a*e)^3*Sqrt[a +
b*x])

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Rubi [A]  time = 0.0866917, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {78, 45, 37} \[ -\frac{2 (B d-A e)}{e (a+b x)^{3/2} \sqrt{d+e x} (b d-a e)}-\frac{4 \sqrt{d+e x} (a B e-4 A b e+3 b B d)}{3 \sqrt{a+b x} (b d-a e)^3}+\frac{2 \sqrt{d+e x} (a B e-4 A b e+3 b B d)}{3 e (a+b x)^{3/2} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((a + b*x)^(5/2)*(d + e*x)^(3/2)),x]

[Out]

(-2*(B*d - A*e))/(e*(b*d - a*e)*(a + b*x)^(3/2)*Sqrt[d + e*x]) + (2*(3*b*B*d - 4*A*b*e + a*B*e)*Sqrt[d + e*x])
/(3*e*(b*d - a*e)^2*(a + b*x)^(3/2)) - (4*(3*b*B*d - 4*A*b*e + a*B*e)*Sqrt[d + e*x])/(3*(b*d - a*e)^3*Sqrt[a +
b*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}{(a+b x)^{5/2} (d+e x)^{3/2}} \, dx &=-\frac{2 (B d-A e)}{e (b d-a e) (a+b x)^{3/2} \sqrt{d+e x}}-\frac{(3 b B d-4 A b e+a B e) \int \frac{1}{(a+b x)^{5/2} \sqrt{d+e x}} \, dx}{e (b d-a e)}\\ &=-\frac{2 (B d-A e)}{e (b d-a e) (a+b x)^{3/2} \sqrt{d+e x}}+\frac{2 (3 b B d-4 A b e+a B e) \sqrt{d+e x}}{3 e (b d-a e)^2 (a+b x)^{3/2}}+\frac{(2 (3 b B d-4 A b e+a B e)) \int \frac{1}{(a+b x)^{3/2} \sqrt{d+e x}} \, dx}{3 (b d-a e)^2}\\ &=-\frac{2 (B d-A e)}{e (b d-a e) (a+b x)^{3/2} \sqrt{d+e x}}+\frac{2 (3 b B d-4 A b e+a B e) \sqrt{d+e x}}{3 e (b d-a e)^2 (a+b x)^{3/2}}-\frac{4 (3 b B d-4 A b e+a B e) \sqrt{d+e x}}{3 (b d-a e)^3 \sqrt{a+b x}}\\ \end{align*}

Mathematica [A]  time = 0.0699987, size = 125, normalized size = 0.9 \[ -\frac{2 \left (A \left (-3 a^2 e^2-6 a b e (d+2 e x)+b^2 \left (d^2-4 d e x-8 e^2 x^2\right )\right )+B \left (3 a^2 e (2 d+e x)+2 a b \left (d^2+5 d e x+e^2 x^2\right )+3 b^2 d x (d+2 e x)\right )\right )}{3 (a+b x)^{3/2} \sqrt{d+e x} (b d-a e)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((a + b*x)^(5/2)*(d + e*x)^(3/2)),x]

[Out]

(-2*(A*(-3*a^2*e^2 - 6*a*b*e*(d + 2*e*x) + b^2*(d^2 - 4*d*e*x - 8*e^2*x^2)) + B*(3*a^2*e*(2*d + e*x) + 3*b^2*d
*x*(d + 2*e*x) + 2*a*b*(d^2 + 5*d*e*x + e^2*x^2))))/(3*(b*d - a*e)^3*(a + b*x)^(3/2)*Sqrt[d + e*x])

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Maple [A]  time = 0.004, size = 177, normalized size = 1.3 \begin{align*} -{\frac{16\,A{b}^{2}{e}^{2}{x}^{2}-4\,Bab{e}^{2}{x}^{2}-12\,B{b}^{2}de{x}^{2}+24\,Aab{e}^{2}x+8\,A{b}^{2}dex-6\,B{a}^{2}{e}^{2}x-20\,Babdex-6\,B{b}^{2}{d}^{2}x+6\,A{a}^{2}{e}^{2}+12\,Aabde-2\,A{b}^{2}{d}^{2}-12\,B{a}^{2}de-4\,Bab{d}^{2}}{3\,{a}^{3}{e}^{3}-9\,{a}^{2}bd{e}^{2}+9\,a{b}^{2}{d}^{2}e-3\,{b}^{3}{d}^{3}} \left ( bx+a \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ex+d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(b*x+a)^(5/2)/(e*x+d)^(3/2),x)

[Out]

-2/3*(8*A*b^2*e^2*x^2-2*B*a*b*e^2*x^2-6*B*b^2*d*e*x^2+12*A*a*b*e^2*x+4*A*b^2*d*e*x-3*B*a^2*e^2*x-10*B*a*b*d*e*
x-3*B*b^2*d^2*x+3*A*a^2*e^2+6*A*a*b*d*e-A*b^2*d^2-6*B*a^2*d*e-2*B*a*b*d^2)/(b*x+a)^(3/2)/(e*x+d)^(1/2)/(a^3*e^
3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)

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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)/(b*x+a)^(5/2)/(e*x+d)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 14.1161, size = 682, normalized size = 4.91 \begin{align*} \frac{2 \,{\left (3 \, A a^{2} e^{2} -{\left (2 \, B a b + A b^{2}\right )} d^{2} - 6 \,{\left (B a^{2} - A a b\right )} d e - 2 \,{\left (3 \, B b^{2} d e +{\left (B a b - 4 \, A b^{2}\right )} e^{2}\right )} x^{2} -{\left (3 \, B b^{2} d^{2} + 2 \,{\left (5 \, B a b - 2 \, A b^{2}\right )} d e + 3 \,{\left (B a^{2} - 4 \, A a b\right )} e^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{3 \,{\left (a^{2} b^{3} d^{4} - 3 \, a^{3} b^{2} d^{3} e + 3 \, a^{4} b d^{2} e^{2} - a^{5} d e^{3} +{\left (b^{5} d^{3} e - 3 \, a b^{4} d^{2} e^{2} + 3 \, a^{2} b^{3} d e^{3} - a^{3} b^{2} e^{4}\right )} x^{3} +{\left (b^{5} d^{4} - a b^{4} d^{3} e - 3 \, a^{2} b^{3} d^{2} e^{2} + 5 \, a^{3} b^{2} d e^{3} - 2 \, a^{4} b e^{4}\right )} x^{2} +{\left (2 \, a b^{4} d^{4} - 5 \, a^{2} b^{3} d^{3} e + 3 \, a^{3} b^{2} d^{2} e^{2} + a^{4} b d e^{3} - a^{5} e^{4}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^(5/2)/(e*x+d)^(3/2),x, algorithm="fricas")

[Out]

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

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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)/(b*x+a)**(5/2)/(e*x+d)**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 2.82576, size = 763, normalized size = 5.49 \begin{align*} -\frac{2 \,{\left (B b^{2} d e - A b^{2} e^{2}\right )} \sqrt{b x + a}}{{\left (b^{3} d^{3}{\left | b \right |} - 3 \, a b^{2} d^{2}{\left | b \right |} e + 3 \, a^{2} b d{\left | b \right |} e^{2} - a^{3}{\left | b \right |} e^{3}\right )} \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e}} - \frac{4 \,{\left (3 \, B b^{\frac{13}{2}} d^{3} e^{\frac{1}{2}} - 4 \, B a b^{\frac{11}{2}} d^{2} e^{\frac{3}{2}} - 5 \, A b^{\frac{13}{2}} d^{2} e^{\frac{3}{2}} - 6 \,{\left (\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} - \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e}\right )}^{2} B b^{\frac{9}{2}} d^{2} e^{\frac{1}{2}} - B a^{2} b^{\frac{9}{2}} d e^{\frac{5}{2}} + 10 \, A a b^{\frac{11}{2}} d e^{\frac{5}{2}} + 12 \,{\left (\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} - \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e}\right )}^{2} A b^{\frac{9}{2}} d e^{\frac{3}{2}} + 3 \,{\left (\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} - \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e}\right )}^{4} B b^{\frac{5}{2}} d e^{\frac{1}{2}} + 2 \, B a^{3} b^{\frac{7}{2}} e^{\frac{7}{2}} - 5 \, A a^{2} b^{\frac{9}{2}} e^{\frac{7}{2}} + 6 \,{\left (\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} - \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e}\right )}^{2} B a^{2} b^{\frac{5}{2}} e^{\frac{5}{2}} - 12 \,{\left (\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} - \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e}\right )}^{2} A a b^{\frac{7}{2}} e^{\frac{5}{2}} - 3 \,{\left (\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} - \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e}\right )}^{4} A b^{\frac{5}{2}} e^{\frac{3}{2}}\right )}}{3 \,{\left (b^{2} d^{2}{\left | b \right |} - 2 \, a b d{\left | b \right |} e + a^{2}{\left | b \right |} e^{2}\right )}{\left (b^{2} d - a b e -{\left (\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} - \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e}\right )}^{2}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^(5/2)/(e*x+d)^(3/2),x, algorithm="giac")

[Out]

-2*(B*b^2*d*e - A*b^2*e^2)*sqrt(b*x + a)/((b^3*d^3*abs(b) - 3*a*b^2*d^2*abs(b)*e + 3*a^2*b*d*abs(b)*e^2 - a^3*
abs(b)*e^3)*sqrt(b^2*d + (b*x + a)*b*e - a*b*e)) - 4/3*(3*B*b^(13/2)*d^3*e^(1/2) - 4*B*a*b^(11/2)*d^2*e^(3/2)
- 5*A*b^(13/2)*d^2*e^(3/2) - 6*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*b^(9/
2)*d^2*e^(1/2) - B*a^2*b^(9/2)*d*e^(5/2) + 10*A*a*b^(11/2)*d*e^(5/2) + 12*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqr
t(b^2*d + (b*x + a)*b*e - a*b*e))^2*A*b^(9/2)*d*e^(3/2) + 3*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x
+ a)*b*e - a*b*e))^4*B*b^(5/2)*d*e^(1/2) + 2*B*a^3*b^(7/2)*e^(7/2) - 5*A*a^2*b^(9/2)*e^(7/2) + 6*(sqrt(b*x +
a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*a^2*b^(5/2)*e^(5/2) - 12*(sqrt(b*x + a)*sqrt(b)*
e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*A*a*b^(7/2)*e^(5/2) - 3*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt
(b^2*d + (b*x + a)*b*e - a*b*e))^4*A*b^(5/2)*e^(3/2))/((b^2*d^2*abs(b) - 2*a*b*d*abs(b)*e + a^2*abs(b)*e^2)*(b
^2*d - a*b*e - (sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2)^3)