3.14.24 \(\int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=69 \[ \frac {2}{\sqrt {c+d x} (b c-a d)}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {51, 63, 208} \begin {gather*} \frac {2}{\sqrt {c+d x} (b c-a d)}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)*(c + d*x)^(3/2)),x]

[Out]

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

Rule 51

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] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 208

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

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 46, normalized size = 0.67 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b (c+d x)}{b c-a d}\right )}{\sqrt {c+d x} (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)*(c + d*x)^(3/2)),x]

[Out]

(-2*Hypergeometric2F1[-1/2, 1, 1/2, (b*(c + d*x))/(b*c - a*d)])/((-(b*c) + a*d)*Sqrt[c + d*x])

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IntegrateAlgebraic [A]  time = 0.09, size = 79, normalized size = 1.14 \begin {gather*} \frac {2}{\sqrt {c+d x} (b c-a d)}+\frac {2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{(a d-b c)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)*(c + d*x)^(3/2)),x]

[Out]

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

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fricas [A]  time = 1.43, size = 214, normalized size = 3.10 \begin {gather*} \left [-\frac {{\left (d x + c\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, {\left (b c - a d\right )} \sqrt {d x + c} \sqrt {\frac {b}{b c - a d}}}{b x + a}\right ) - 2 \, \sqrt {d x + c}}{b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x}, -\frac {2 \, {\left ({\left (d x + c\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {{\left (b c - a d\right )} \sqrt {d x + c} \sqrt {-\frac {b}{b c - a d}}}{b d x + b c}\right ) - \sqrt {d x + c}\right )}}{b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-((d*x + c)*sqrt(b/(b*c - a*d))*log((b*d*x + 2*b*c - a*d + 2*(b*c - a*d)*sqrt(d*x + c)*sqrt(b/(b*c - a*d)))/(
b*x + a)) - 2*sqrt(d*x + c))/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x), -2*((d*x + c)*sqrt(-b/(b*c - a*d))*arctan(-(
b*c - a*d)*sqrt(d*x + c)*sqrt(-b/(b*c - a*d))/(b*d*x + b*c)) - sqrt(d*x + c))/(b*c^2 - a*c*d + (b*c*d - a*d^2)
*x)]

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giac [A]  time = 0.94, size = 69, normalized size = 1.00 \begin {gather*} \frac {2 \, b \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} {\left (b c - a d\right )}} + \frac {2}{{\left (b c - a d\right )} \sqrt {d x + c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*(b*c - a*d)) + 2/((b*c - a*d)*sqrt(d*x
+ c))

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maple [A]  time = 0.01, size = 68, normalized size = 0.99 \begin {gather*} -\frac {2 b \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}}-\frac {2}{\left (a d -b c \right ) \sqrt {d x +c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)/(d*x+c)^(3/2),x)

[Out]

-2/(a*d-b*c)/(d*x+c)^(1/2)-2*b/(a*d-b*c)/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)

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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(1/(b*x+a)/(d*x+c)^(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(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c positive or negative?

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mupad [B]  time = 0.27, size = 57, normalized size = 0.83 \begin {gather*} -\frac {2}{\left (a\,d-b\,c\right )\,\sqrt {c+d\,x}}-\frac {2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{{\left (a\,d-b\,c\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)*(c + d*x)^(3/2)),x)

[Out]

- 2/((a*d - b*c)*(c + d*x)^(1/2)) - (2*b^(1/2)*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(a*d - b*c)^
(3/2)

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sympy [A]  time = 11.49, size = 60, normalized size = 0.87 \begin {gather*} - \frac {2}{\sqrt {c + d x} \left (a d - b c\right )} - \frac {2 \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{\sqrt {\frac {a d - b c}{b}} \left (a d - b c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)**(3/2),x)

[Out]

-2/(sqrt(c + d*x)*(a*d - b*c)) - 2*atan(sqrt(c + d*x)/sqrt((a*d - b*c)/b))/(sqrt((a*d - b*c)/b)*(a*d - b*c))

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