3.18.47 \(\int \frac {A+B x}{(a+b x) (d+e x)^{3/2}} \, dx\) [1747]

Optimal. Leaf size=88 \[ -\frac {2 (B d-A e)}{e (b d-a e) \sqrt {d+e x}}-\frac {2 (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {79, 65, 214} \begin {gather*} -\frac {2 (B d-A e)}{e \sqrt {d+e x} (b d-a e)}-\frac {2 (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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]))))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rubi steps

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

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Mathematica [A]
time = 0.25, size = 88, normalized size = 1.00 \begin {gather*} \frac {2 B d-2 A e}{e (-b d+a e) \sqrt {d+e x}}-\frac {2 (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{\sqrt {b} (-b d+a e)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 89, normalized size = 1.01

method	result	size
derivativedivides	\(\frac {-\frac {2 e \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}}-\frac {2 \left (A e -B d \right )}{\left (a e -b d \right ) \sqrt {e x +d}}}{e}\)	\(89\)
default	\(\frac {-\frac {2 e \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}}-\frac {2 \left (A e -B d \right )}{\left (a e -b d \right ) \sqrt {e x +d}}}{e}\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/e*(-e*(A*b-B*a)/(a*e-b*d)/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))-(A*e-B*d)/(a*e-b*d
)/(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)/(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 [B] Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (81) = 162\).
time = 1.14, size = 370, normalized size = 4.20 \begin {gather*} \left [\frac {\sqrt {b^{2} d - a b e} {\left ({\left (B a - A b\right )} x e^{2} + {\left (B a - A b\right )} d e\right )} \log \left (\frac {2 \, b d + {\left (b x - a\right )} e + 2 \, \sqrt {b^{2} d - a b e} \sqrt {x e + d}}{b x + a}\right ) - 2 \, {\left (B b^{2} d^{2} + A a b e^{2} - {\left (B a b + A b^{2}\right )} d e\right )} \sqrt {x e + d}}{b^{3} d^{3} e + a^{2} b x e^{4} - {\left (2 \, a b^{2} d x - a^{2} b d\right )} e^{3} + {\left (b^{3} d^{2} x - 2 \, a b^{2} d^{2}\right )} e^{2}}, -\frac {2 \, {\left (\sqrt {-b^{2} d + a b e} {\left ({\left (B a - A b\right )} x e^{2} + {\left (B a - A b\right )} d e\right )} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {x e + d}}{b x e + b d}\right ) + {\left (B b^{2} d^{2} + A a b e^{2} - {\left (B a b + A b^{2}\right )} d e\right )} \sqrt {x e + d}\right )}}{b^{3} d^{3} e + a^{2} b x e^{4} - {\left (2 \, a b^{2} d x - a^{2} b d\right )} e^{3} + {\left (b^{3} d^{2} x - 2 \, a b^{2} d^{2}\right )} e^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(sqrt(b^2*d - a*b*e)*((B*a - A*b)*x*e^2 + (B*a - A*b)*d*e)*log((2*b*d + (b*x - a)*e + 2*sqrt(b^2*d - a*b*e)*s
qrt(x*e + d))/(b*x + a)) - 2*(B*b^2*d^2 + A*a*b*e^2 - (B*a*b + A*b^2)*d*e)*sqrt(x*e + d))/(b^3*d^3*e + a^2*b*x
*e^4 - (2*a*b^2*d*x - a^2*b*d)*e^3 + (b^3*d^2*x - 2*a*b^2*d^2)*e^2), -2*(sqrt(-b^2*d + a*b*e)*((B*a - A*b)*x*e
^2 + (B*a - A*b)*d*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(x*e + d)/(b*x*e + b*d)) + (B*b^2*d^2 + A*a*b*e^2 - (B*a
*b + A*b^2)*d*e)*sqrt(x*e + d))/(b^3*d^3*e + a^2*b*x*e^4 - (2*a*b^2*d*x - a^2*b*d)*e^3 + (b^3*d^2*x - 2*a*b^2*
d^2)*e^2)]

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Sympy [A]
time = 12.35, size = 76, normalized size = 0.86 \begin {gather*} \frac {2 \left (- A e + B d\right )}{e \sqrt {d + e x} \left (a e - b d\right )} + \frac {2 \left (- A b + B a\right ) \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e - b d}{b}}} \right )}}{b \sqrt {\frac {a e - b d}{b}} \left (a e - b d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.87, size = 93, normalized size = 1.06 \begin {gather*} -\frac {2 \, {\left (B a - A b\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} {\left (b d - a e\right )}} - \frac {2 \, {\left (B d - A e\right )}}{{\left (b d e - a e^{2}\right )} \sqrt {x e + d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.11, size = 96, normalized size = 1.09 \begin {gather*} -\frac {2\,\mathrm {atan}\left (\frac {2\,\sqrt {b}\,\left (A\,b-B\,a\right )\,\sqrt {d+e\,x}}{\left (2\,A\,b-2\,B\,a\right )\,\sqrt {a\,e-b\,d}}\right )\,\left (A\,b-B\,a\right )}{\sqrt {b}\,{\left (a\,e-b\,d\right )}^{3/2}}-\frac {2\,\left (A\,e-B\,d\right )}{e\,\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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