3.23.49 \(\int \frac {A+B x}{(a+b x)^{3/2} (d+e x)^{3/2}} \, dx\) [2249]

Optimal. Leaf size=89 \[ -\frac {2 (B d-A e)}{e (b d-a e) \sqrt {a+b x} \sqrt {d+e x}}+\frac {2 (b B d-2 A b e+a B e) \sqrt {d+e x}}{e (b d-a e)^2 \sqrt {a+b x}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 89, 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 = {79, 37} \begin {gather*} \frac {2 \sqrt {d+e x} (a B e-2 A b e+b B d)}{e \sqrt {a+b x} (b d-a e)^2}-\frac {2 (B d-A e)}{e \sqrt {a+b x} \sqrt {d+e x} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(B*d - A*e))/(e*(b*d - a*e)*Sqrt[a + b*x]*Sqrt[d + e*x]) + (2*(b*B*d - 2*A*b*e + a*B*e)*Sqrt[d + e*x])/(e*
(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 79

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] || L
tQ[p, n]))))

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 61, normalized size = 0.69 \begin {gather*} \frac {2 B (2 a d+b d x+a e x)-2 A (a e+b (d+2 e x))}{(b d-a e)^2 \sqrt {a+b x} \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 59, normalized size = 0.66

method	result	size
default	\(-\frac {2 \left (2 A b e x -B a e x -B b d x +A a e +A b d -2 B a d \right )}{\left (a e -b d \right )^{2} \sqrt {b x +a}\, \sqrt {e x +d}}\)	\(59\)
gosper	\(-\frac {2 \left (2 A b e x -B a e x -B b d x +A a e +A b d -2 B a d \right )}{\sqrt {b x +a}\, \sqrt {e x +d}\, \left (a^{2} e^{2}-2 b e a d +b^{2} d^{2}\right )}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^(3/2)/(e*x+d)^(3/2),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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 1.12, size = 150, normalized size = 1.69 \begin {gather*} \frac {2 \, {\left (B b d x + {\left (2 \, B a - A b\right )} d - {\left (A a - {\left (B a - 2 \, A b\right )} x\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d}}{b^{3} d^{3} x + a b^{2} d^{3} + {\left (a^{2} b x^{2} + a^{3} x\right )} e^{3} - {\left (2 \, a b^{2} d x^{2} + a^{2} b d x - a^{3} d\right )} e^{2} + {\left (b^{3} d^{2} x^{2} - a b^{2} d^{2} x - 2 \, a^{2} b d^{2}\right )} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{\left (a + b x\right )^{\frac {3}{2}} \left (d + e x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)**(3/2)/(e*x+d)**(3/2),x)

[Out]

Integral((A + B*x)/((a + b*x)**(3/2)*(d + e*x)**(3/2)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (86) = 172\).
time = 0.55, size = 264, normalized size = 2.97 \begin {gather*} \frac {2 \, {\left (B b^{2} d - A b^{2} e\right )} \sqrt {b x + a}}{{\left (b^{2} d^{2} {\left | b \right |} - 2 \, a b d {\left | b \right |} e + a^{2} {\left | b \right |} e^{2}\right )} \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}} + \frac {4 \, {\left (B^{2} a^{2} b^{3} e - 2 \, A B a b^{4} e + A^{2} b^{5} e\right )}}{{\left (B a b^{\frac {7}{2}} d e^{\frac {1}{2}} - A b^{\frac {9}{2}} d e^{\frac {1}{2}} - B a^{2} b^{\frac {5}{2}} e^{\frac {3}{2}} + A a b^{\frac {7}{2}} e^{\frac {3}{2}} - {\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 b^{\frac {3}{2}} e^{\frac {1}{2}} + {\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 {1}{2}}\right )} {\left (b d {\left | b \right |} - a {\left | b \right |} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.00, size = 96, normalized size = 1.08 \begin {gather*} -\frac {\left (\frac {2\,A\,a\,e+2\,A\,b\,d-4\,B\,a\,d}{e\,{\left (a\,e-b\,d\right )}^2}-\frac {x\,\left (2\,B\,a\,e-4\,A\,b\,e+2\,B\,b\,d\right )}{e\,{\left (a\,e-b\,d\right )}^2}\right )\,\sqrt {d+e\,x}}{x\,\sqrt {a+b\,x}+\frac {d\,\sqrt {a+b\,x}}{e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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