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

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

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Rubi [A]  time = 0.04, 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 = {78, 63, 217, 206} \begin {gather*} \frac {2 B \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{3/2} \sqrt {e}}-\frac {2 \sqrt {d+e x} (A b-a B)}{b \sqrt {a+b x} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

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 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]))))

Rule 206

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

Rule 217

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

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Mathematica [A]  time = 0.59, size = 117, normalized size = 1.38 \begin {gather*} \frac {2 \left (\frac {b (d+e x) (a B-A b)}{\sqrt {a+b x} (b d-a e)}+\frac {B \sqrt {b d-a e} \sqrt {\frac {b (d+e x)}{b d-a e}} \sinh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b d-a e}}\right )}{\sqrt {e}}\right )}{b^2 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 2.24, size = 360, normalized size = 4.24 \begin {gather*} \left [\frac {4 \, {\left (B a b - A b^{2}\right )} \sqrt {b x + a} \sqrt {e x + d} e + {\left (B a b d - B a^{2} e + {\left (B b^{2} d - B a b e\right )} x\right )} \sqrt {b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e x + b d + a e\right )} \sqrt {b e} \sqrt {b x + a} \sqrt {e x + d} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right )}{2 \, {\left (a b^{3} d e - a^{2} b^{2} e^{2} + {\left (b^{4} d e - a b^{3} e^{2}\right )} x\right )}}, \frac {2 \, {\left (B a b - A b^{2}\right )} \sqrt {b x + a} \sqrt {e x + d} e - {\left (B a b d - B a^{2} e + {\left (B b^{2} d - B a b e\right )} x\right )} \sqrt {-b e} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {-b e} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right )}{a b^{3} d e - a^{2} b^{2} e^{2} + {\left (b^{4} d e - a b^{3} e^{2}\right )} x}\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)^(1/2),x, algorithm="fricas")

[Out]

[1/2*(4*(B*a*b - A*b^2)*sqrt(b*x + a)*sqrt(e*x + d)*e + (B*a*b*d - B*a^2*e + (B*b^2*d - B*a*b*e)*x)*sqrt(b*e)*
log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*b*e*x + b*d + a*e)*sqrt(b*e)*sqrt(b*x + a)*sqrt(e*x +
d) + 8*(b^2*d*e + a*b*e^2)*x))/(a*b^3*d*e - a^2*b^2*e^2 + (b^4*d*e - a*b^3*e^2)*x), (2*(B*a*b - A*b^2)*sqrt(b
*x + a)*sqrt(e*x + d)*e - (B*a*b*d - B*a^2*e + (B*b^2*d - B*a*b*e)*x)*sqrt(-b*e)*arctan(1/2*(2*b*e*x + b*d + a
*e)*sqrt(-b*e)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*e^2*x^2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x)))/(a*b^3*d*e - a^2*
b^2*e^2 + (b^4*d*e - a*b^3*e^2)*x)]

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giac [A]  time = 1.20, size = 135, normalized size = 1.59 \begin {gather*} -\frac {B e^{\left (-\frac {1}{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 )}^{2}\right )}{\sqrt {b} {\left | b \right |}} + \frac {4 \, {\left (B a \sqrt {b} e^{\frac {1}{2}} - A b^{\frac {3}{2}} e^{\frac {1}{2}}\right )}}{{\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 )} {\left | b \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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.02, size = 278, normalized size = 3.27 \begin {gather*} \frac {\sqrt {e x +d}\, \left (B a b e x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-B \,b^{2} d x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+B \,a^{2} e \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-B a b d \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A b -2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a \right )}{\sqrt {b e}\, \left (a e -b d \right ) \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b x +a}\, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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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}} \sqrt {d + e x}}\, 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)**(1/2),x)

[Out]

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

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