3.2257 \(\int \frac {A+B x}{(a+b x)^{5/2} \sqrt {d+e x}} \, dx\)

Optimal. Leaf size=95 \[ -\frac {2 \sqrt {d+e x} (A b-a B)}{3 b (a+b x)^{3/2} (b d-a e)}-\frac {2 \sqrt {d+e x} (-a B e-2 A b e+3 b B d)}{3 b \sqrt {a+b x} (b d-a e)^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 95, 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} \[ -\frac {2 \sqrt {d+e x} (A b-a B)}{3 b (a+b x)^{3/2} (b d-a e)}-\frac {2 \sqrt {d+e x} (-a B e-2 A b e+3 b B d)}{3 b \sqrt {a+b x} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 64, normalized size = 0.67 \[ -\frac {2 \sqrt {d+e x} (-3 a A e+B (2 a d-a e x+3 b d x)+A b (d-2 e x))}{3 (a+b x)^{3/2} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 2.05, size = 141, normalized size = 1.48 \[ \frac {2 \, {\left (3 \, A a e - {\left (2 \, B a + A b\right )} d - {\left (3 \, B b d - {\left (B a + 2 \, A b\right )} e\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{3 \, {\left (a^{2} b^{2} d^{2} - 2 \, a^{3} b d e + a^{4} e^{2} + {\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} x^{2} + 2 \, {\left (a b^{3} d^{2} - 2 \, a^{2} b^{2} d e + a^{3} b e^{2}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.41, size = 261, normalized size = 2.75 \[ -\frac {4 \, {\left (3 \, B b^{\frac {9}{2}} d^{2} e^{\frac {1}{2}} - 4 \, B a b^{\frac {7}{2}} d e^{\frac {3}{2}} - 2 \, A b^{\frac {9}{2}} d e^{\frac {3}{2}} - 6 \, {\left (\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2} B b^{\frac {5}{2}} d e^{\frac {1}{2}} + B a^{2} b^{\frac {5}{2}} e^{\frac {5}{2}} + 2 \, A a b^{\frac {7}{2}} e^{\frac {5}{2}} + 6 \, {\left (\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2} A b^{\frac {5}{2}} e^{\frac {3}{2}} + 3 \, {\left (\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{4} B \sqrt {b} e^{\frac {1}{2}}\right )}}{3 \, {\left (b^{2} d - a b e - {\left (\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2}\right )}^{3} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 73, normalized size = 0.77 \[ \frac {2 \sqrt {e x +d}\, \left (2 A b e x +B a e x -3 B b d x +3 A a e -A b d -2 B a d \right )}{3 \left (b x +a \right )^{\frac {3}{2}} \left (a^{2} e^{2}-2 b e a d +b^{2} d^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(e*x+d)^(1/2)*(2*A*b*e*x+B*a*e*x-3*B*b*d*x+3*A*a*e-A*b*d-2*B*a*d)/(b*x+a)^(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)/(b*x+a)^(5/2)/(e*x+d)^(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 = 2.02, size = 97, normalized size = 1.02 \[ -\frac {\left (\frac {2\,A\,b\,d-6\,A\,a\,e+4\,B\,a\,d}{3\,b\,{\left (a\,e-b\,d\right )}^2}-\frac {x\,\left (4\,A\,b\,e+2\,B\,a\,e-6\,B\,b\,d\right )}{3\,b\,{\left (a\,e-b\,d\right )}^2}\right )\,\sqrt {d+e\,x}}{x\,\sqrt {a+b\,x}+\frac {a\,\sqrt {a+b\,x}}{b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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