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

Optimal. Leaf size=99 \[ -\frac {3 d}{\sqrt {c+d x} (b c-a d)^2}-\frac {1}{(a+b x) \sqrt {c+d x} (b c-a d)}+\frac {3 \sqrt {b} d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {3 d}{\sqrt {c+d x} (b c-a d)^2}-\frac {1}{(a+b x) \sqrt {c+d x} (b c-a d)}+\frac {3 \sqrt {b} d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d)/((b*c - a*d)^2*Sqrt[c + d*x]) - 1/((b*c - a*d)*(a + b*x)*Sqrt[c + d*x]) + (3*Sqrt[b]*d*ArcTanh[(Sqrt[b]
*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b*c - a*d)^(5/2)

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)^2 (c+d x)^{3/2}} \, dx &=-\frac {1}{(b c-a d) (a+b x) \sqrt {c+d x}}-\frac {(3 d) \int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx}{2 (b c-a d)}\\ &=-\frac {3 d}{(b c-a d)^2 \sqrt {c+d x}}-\frac {1}{(b c-a d) (a+b x) \sqrt {c+d x}}-\frac {(3 b d) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{2 (b c-a d)^2}\\ &=-\frac {3 d}{(b c-a d)^2 \sqrt {c+d x}}-\frac {1}{(b c-a d) (a+b x) \sqrt {c+d x}}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{(b c-a d)^2}\\ &=-\frac {3 d}{(b c-a d)^2 \sqrt {c+d x}}-\frac {1}{(b c-a d) (a+b x) \sqrt {c+d x}}+\frac {3 \sqrt {b} d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 48, normalized size = 0.48 \begin {gather*} -\frac {2 d \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};-\frac {b (c+d x)}{a d-b c}\right )}{\sqrt {c+d x} (a d-b c)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.29, size = 115, normalized size = 1.16 \begin {gather*} \frac {d (2 a d+3 b (c+d x)-2 b c)}{\sqrt {c+d x} (b c-a d)^2 (-a d-b (c+d x)+b c)}+\frac {3 \sqrt {b} d \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{(a d-b c)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(-2*b*c + 2*a*d + 3*b*(c + d*x)))/((b*c - a*d)^2*Sqrt[c + d*x]*(b*c - a*d - b*(c + d*x))) + (3*Sqrt[b]*d*Ar
cTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c + d*x])/(b*c - a*d)])/(-(b*c) + a*d)^(5/2)

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fricas [B]  time = 1.32, size = 423, normalized size = 4.27 \begin {gather*} \left [\frac {3 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\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 \, {\left (3 \, b d x + b c + 2 \, a d\right )} \sqrt {d x + c}}{2 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{2} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x\right )}}, \frac {3 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\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 ) - {\left (3 \, b d x + b c + 2 \, a d\right )} \sqrt {d x + c}}{a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{2} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)*sqrt(b/(b*c - a*d))*log((b*d*x + 2*b*c - a*d + 2*(b*c - a*d)*s
qrt(d*x + c)*sqrt(b/(b*c - a*d)))/(b*x + a)) - 2*(3*b*d*x + b*c + 2*a*d)*sqrt(d*x + c))/(a*b^2*c^3 - 2*a^2*b*c
^2*d + a^3*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^2 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^
3)*x), (3*(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)*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)) - (3*b*d*x + b*c + 2*a*d)*sqrt(d*x + c))/(a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2
+ (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^2 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x)]

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giac [A]  time = 1.03, size = 143, normalized size = 1.44 \begin {gather*} -\frac {3 \, b d \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d}} - \frac {3 \, {\left (d x + c\right )} b d - 2 \, b c d + 2 \, a d^{2}}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left ({\left (d x + c\right )}^{\frac {3}{2}} b - \sqrt {d x + c} b c + \sqrt {d x + c} a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*b*d*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-b^2*c + a*b*d)) - (
3*(d*x + c)*b*d - 2*b*c*d + 2*a*d^2)/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*((d*x + c)^(3/2)*b - sqrt(d*x + c)*b*c +
sqrt(d*x + c)*a*d))

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maple [A]  time = 0.01, size = 101, normalized size = 1.02 \begin {gather*} -\frac {3 b d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{2} \sqrt {\left (a d -b c \right ) b}}-\frac {\sqrt {d x +c}\, b d}{\left (a d -b c \right )^{2} \left (b d x +a d \right )}-\frac {2 d}{\left (a d -b c \right )^{2} \sqrt {d x +c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*d/(a*d-b*c)^2/(d*x+c)^(1/2)-d*b/(a*d-b*c)^2*(d*x+c)^(1/2)/(b*d*x+a*d)-3*d*b/(a*d-b*c)^2/((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)^2/(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 = 0.19, size = 123, normalized size = 1.24 \begin {gather*} -\frac {\frac {2\,d}{a\,d-b\,c}+\frac {3\,b\,d\,\left (c+d\,x\right )}{{\left (a\,d-b\,c\right )}^2}}{b\,{\left (c+d\,x\right )}^{3/2}+\left (a\,d-b\,c\right )\,\sqrt {c+d\,x}}-\frac {3\,\sqrt {b}\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{{\left (a\,d-b\,c\right )}^{5/2}}\right )}{{\left (a\,d-b\,c\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*d)/(a*d - b*c) + (3*b*d*(c + d*x))/(a*d - b*c)^2)/(b*(c + d*x)^(3/2) + (a*d - b*c)*(c + d*x)^(1/2)) - (3
*b^(1/2)*d*atan((b^(1/2)*(c + d*x)^(1/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(a*d - b*c)^(5/2)))/(a*d - b*c)^(5/2
)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x\right )^{2} \left (c + d x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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