3.2241 \(\int \frac {A+B x}{\sqrt {a+b x} (d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=94 \[ \frac {2 \sqrt {a+b x} (-3 a B e+2 A b e+b B d)}{3 e \sqrt {d+e x} (b d-a e)^2}-\frac {2 \sqrt {a+b x} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)} \]

[Out]

-2/3*(-A*e+B*d)*(b*x+a)^(1/2)/e/(-a*e+b*d)/(e*x+d)^(3/2)+2/3*(2*A*b*e-3*B*a*e+B*b*d)*(b*x+a)^(1/2)/e/(-a*e+b*d
)^2/(e*x+d)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {78, 37} \[ \frac {2 \sqrt {a+b x} (-3 a B e+2 A b e+b B d)}{3 e \sqrt {d+e x} (b d-a e)^2}-\frac {2 \sqrt {a+b x} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]))))

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 65, normalized size = 0.69 \[ \frac {2 \sqrt {a+b x} (A (-a e+3 b d+2 b e x)+B (-2 a d-3 a e x+b d x))}{3 (d+e x)^{3/2} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 2.63, size = 140, normalized size = 1.49 \[ -\frac {2 \, {\left (A a e + {\left (2 \, B a - 3 \, A b\right )} d - {\left (B b d - {\left (3 \, B a - 2 \, A b\right )} e\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{3 \, {\left (b^{2} d^{4} - 2 \, a b d^{3} e + a^{2} d^{2} e^{2} + {\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} x^{2} + 2 \, {\left (b^{2} d^{3} e - 2 \, a b d^{2} e^{2} + a^{2} d e^{3}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^(5/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.29, size = 179, normalized size = 1.90 \[ \frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (B b^{4} d {\left | b \right |} e - 3 \, B a b^{3} {\left | b \right |} e^{2} + 2 \, A b^{4} {\left | b \right |} e^{2}\right )} {\left (b x + a\right )}}{b^{4} d^{2} e - 2 \, a b^{3} d e^{2} + a^{2} b^{2} e^{3}} - \frac {3 \, {\left (B a b^{4} d {\left | b \right |} e - A b^{5} d {\left | b \right |} e - B a^{2} b^{3} {\left | b \right |} e^{2} + A a b^{4} {\left | b \right |} e^{2}\right )}}{b^{4} d^{2} e - 2 \, a b^{3} d e^{2} + a^{2} b^{2} e^{3}}\right )}}{3 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^(5/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 73, normalized size = 0.78 \[ -\frac {2 \sqrt {b x +a}\, \left (-2 A b e x +3 B a e x -B b d x +A a e -3 A b d +2 B a d \right )}{3 \left (e x +d \right )^{\frac {3}{2}} \left (a^{2} e^{2}-2 b e a d +b^{2} d^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(e*x+d)^(5/2)/(b*x+a)^(1/2),x)

[Out]

-2/3*(b*x+a)^(1/2)*(-2*A*b*e*x+3*B*a*e*x-B*b*d*x+A*a*e-3*A*b*d+2*B*a*d)/(e*x+d)^(3/2)/(a^2*e^2-2*a*b*d*e+b^2*d
^2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^(5/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [B]  time = 2.04, size = 169, normalized size = 1.80 \[ \frac {\sqrt {d+e\,x}\,\left (\frac {x\,\left (6\,A\,b^2\,d-6\,B\,a^2\,e+2\,A\,a\,b\,e-2\,B\,a\,b\,d\right )}{3\,e^2\,{\left (a\,e-b\,d\right )}^2}-\frac {2\,A\,a^2\,e+4\,B\,a^2\,d-6\,A\,a\,b\,d}{3\,e^2\,{\left (a\,e-b\,d\right )}^2}+\frac {x^2\,\left (4\,A\,b^2\,e+2\,B\,b^2\,d-6\,B\,a\,b\,e\right )}{3\,e^2\,{\left (a\,e-b\,d\right )}^2}\right )}{x^2\,\sqrt {a+b\,x}+\frac {d^2\,\sqrt {a+b\,x}}{e^2}+\frac {2\,d\,x\,\sqrt {a+b\,x}}{e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/((a + b*x)^(1/2)*(d + e*x)^(5/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x}{\sqrt {a + b x} \left (d + e x\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)**(5/2)/(b*x+a)**(1/2),x)

[Out]

Integral((A + B*x)/(sqrt(a + b*x)*(d + e*x)**(5/2)), x)

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