3.23.40 \(\int \frac {A+B x}{\sqrt {a+b x} (d+e x)^{3/2}} \, dx\) [2240]

Optimal. Leaf size=85 \[ -\frac {2 (B d-A e) \sqrt {a+b x}}{e (b d-a e) \sqrt {d+e x}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} e^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {79, 65, 223, 212} \begin {gather*} \frac {2 B \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} e^{3/2}}-\frac {2 \sqrt {a+b x} (B d-A e)}{e \sqrt {d+e x} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(B*d - A*e)*Sqrt[a + b*x])/(e*(b*d - a*e)*Sqrt[d + e*x]) + (2*B*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*S
qrt[d + e*x])])/(Sqrt[b]*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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

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

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Mathematica [A]
time = 0.18, size = 85, normalized size = 1.00 \begin {gather*} -\frac {2 (-B d+A e) \sqrt {a+b x}}{e (-b d+a e) \sqrt {d+e x}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(277\) vs. \(2(69)=138\).
time = 0.11, size = 278, normalized size = 3.27

method	result	size
default	\(\frac {\left (B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a \,e^{2} x -B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b d e x +B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a d e -B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b \,d^{2}-2 A e \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+2 B d \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\right ) \sqrt {b x +a}}{\sqrt {b e}\, \left (a e -b d \right ) \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, e \sqrt {e x +d}}\)	\(278\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(e*x+d)^(3/2)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*e^2*x-B*ln(1/2*(2*b*e*x+2*((b
*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b*d*e*x+B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*
e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*d*e-B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1
/2))*b*d^2-2*A*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+2*B*d*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))*(b*x+a)^(1/2)/
(b*e)^(1/2)/(a*e-b*d)/((b*x+a)*(e*x+d))^(1/2)/e/(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)/(e*x+d)^(3/2)/(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(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. 173 vs. \(2 (70) = 140\).
time = 0.74, size = 356, normalized size = 4.19 \begin {gather*} \left [\frac {{\left (B b d^{2} - B a x e^{2} + {\left (B b d x - B a d\right )} e\right )} \sqrt {b} e^{\frac {1}{2}} \log \left (b^{2} d^{2} + 4 \, {\left (b d + {\left (2 \, b x + a\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d} \sqrt {b} e^{\frac {1}{2}} + {\left (8 \, b^{2} x^{2} + 8 \, a b x + a^{2}\right )} e^{2} + 2 \, {\left (4 \, b^{2} d x + 3 \, a b d\right )} e\right ) - 4 \, {\left (B b d e - A b e^{2}\right )} \sqrt {b x + a} \sqrt {x e + d}}{2 \, {\left (b^{2} d^{2} e^{2} - a b x e^{4} + {\left (b^{2} d x - a b d\right )} e^{3}\right )}}, -\frac {{\left (B b d^{2} - B a x e^{2} + {\left (B b d x - B a d\right )} e\right )} \sqrt {-b e} \arctan \left (\frac {{\left (b d + {\left (2 \, b x + a\right )} e\right )} \sqrt {b x + a} \sqrt {-b e} \sqrt {x e + d}}{2 \, {\left ({\left (b^{2} x^{2} + a b x\right )} e^{2} + {\left (b^{2} d x + a b d\right )} e\right )}}\right ) + 2 \, {\left (B b d e - A b e^{2}\right )} \sqrt {b x + a} \sqrt {x e + d}}{b^{2} d^{2} e^{2} - a b x e^{4} + {\left (b^{2} d x - a b d\right )} e^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^(3/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/2*((B*b*d^2 - B*a*x*e^2 + (B*b*d*x - B*a*d)*e)*sqrt(b)*e^(1/2)*log(b^2*d^2 + 4*(b*d + (2*b*x + a)*e)*sqrt(b
*x + a)*sqrt(x*e + d)*sqrt(b)*e^(1/2) + (8*b^2*x^2 + 8*a*b*x + a^2)*e^2 + 2*(4*b^2*d*x + 3*a*b*d)*e) - 4*(B*b*
d*e - A*b*e^2)*sqrt(b*x + a)*sqrt(x*e + d))/(b^2*d^2*e^2 - a*b*x*e^4 + (b^2*d*x - a*b*d)*e^3), -((B*b*d^2 - B*
a*x*e^2 + (B*b*d*x - B*a*d)*e)*sqrt(-b*e)*arctan(1/2*(b*d + (2*b*x + a)*e)*sqrt(b*x + a)*sqrt(-b*e)*sqrt(x*e +
d)/((b^2*x^2 + a*b*x)*e^2 + (b^2*d*x + a*b*d)*e)) + 2*(B*b*d*e - A*b*e^2)*sqrt(b*x + a)*sqrt(x*e + d))/(b^2*d
^2*e^2 - a*b*x*e^4 + (b^2*d*x - a*b*d)*e^3)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{\sqrt {a + b x} \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)/(e*x+d)**(3/2)/(b*x+a)**(1/2),x)

[Out]

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

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Giac [A]
time = 1.06, size = 120, normalized size = 1.41 \begin {gather*} -\frac {2 \, B {\left | b \right |} e^{\left (-\frac {3}{2}\right )} \log \left ({\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 |}\right )}{b^{\frac {3}{2}}} - \frac {2 \, {\left (B b^{2} d {\left | b \right |} - A b^{2} {\left | b \right |} e\right )} \sqrt {b x + a}}{{\left (b^{3} d e - a b^{2} e^{2}\right )} \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^(3/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,x}{\sqrt {a+b\,x}\,{\left (d+e\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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