3.1421 \(\int \frac{1}{(a+b x)^3 \sqrt{c+d x}} \, dx\)

Optimal. Leaf size=114 \[ -\frac{3 d^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{4 \sqrt{b} (b c-a d)^{5/2}}+\frac{3 d \sqrt{c+d x}}{4 (a+b x) (b c-a d)^2}-\frac{\sqrt{c+d x}}{2 (a+b x)^2 (b c-a d)} \]

[Out]

-Sqrt[c + d*x]/(2*(b*c - a*d)*(a + b*x)^2) + (3*d*Sqrt[c + d*x])/(4*(b*c - a*d)^2*(a + b*x)) - (3*d^2*ArcTanh[
(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(4*Sqrt[b]*(b*c - a*d)^(5/2))

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Rubi [A]  time = 0.0372377, antiderivative size = 114, normalized size of antiderivative = 1., 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} \[ -\frac{3 d^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{4 \sqrt{b} (b c-a d)^{5/2}}+\frac{3 d \sqrt{c+d x}}{4 (a+b x) (b c-a d)^2}-\frac{\sqrt{c+d x}}{2 (a+b x)^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^3*Sqrt[c + d*x]),x]

[Out]

-Sqrt[c + d*x]/(2*(b*c - a*d)*(a + b*x)^2) + (3*d*Sqrt[c + d*x])/(4*(b*c - a*d)^2*(a + b*x)) - (3*d^2*ArcTanh[
(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(4*Sqrt[b]*(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)^3 \sqrt{c+d x}} \, dx &=-\frac{\sqrt{c+d x}}{2 (b c-a d) (a+b x)^2}-\frac{(3 d) \int \frac{1}{(a+b x)^2 \sqrt{c+d x}} \, dx}{4 (b c-a d)}\\ &=-\frac{\sqrt{c+d x}}{2 (b c-a d) (a+b x)^2}+\frac{3 d \sqrt{c+d x}}{4 (b c-a d)^2 (a+b x)}+\frac{\left (3 d^2\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{8 (b c-a d)^2}\\ &=-\frac{\sqrt{c+d x}}{2 (b c-a d) (a+b x)^2}+\frac{3 d \sqrt{c+d x}}{4 (b c-a d)^2 (a+b x)}+\frac{(3 d) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{4 (b c-a d)^2}\\ &=-\frac{\sqrt{c+d x}}{2 (b c-a d) (a+b x)^2}+\frac{3 d \sqrt{c+d x}}{4 (b c-a d)^2 (a+b x)}-\frac{3 d^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{4 \sqrt{b} (b c-a d)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0110318, size = 50, normalized size = 0.44 \[ \frac{2 d^2 \sqrt{c+d x} \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};-\frac{b (c+d x)}{a d-b c}\right )}{(a d-b c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^2*Sqrt[c + d*x]*Hypergeometric2F1[1/2, 3, 3/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(-(b*c) + a*d)^3

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Maple [A]  time = 0.008, size = 115, normalized size = 1. \begin{align*}{\frac{{d}^{2}}{ \left ( 2\,ad-2\,bc \right ) \left ( bdx+ad \right ) ^{2}}\sqrt{dx+c}}+{\frac{3\,{d}^{2}}{4\, \left ( ad-bc \right ) ^{2} \left ( bdx+ad \right ) }\sqrt{dx+c}}+{\frac{3\,{d}^{2}}{4\, \left ( ad-bc \right ) ^{2}}\arctan \left ({b\sqrt{dx+c}{\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*d^2*(d*x+c)^(1/2)/(a*d-b*c)/(b*d*x+a*d)^2+3/4*d^2/(a*d-b*c)^2*(d*x+c)^(1/2)/(b*d*x+a*d)+3/4*d^2/(a*d-b*c)^
2/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.35511, size = 1119, normalized size = 9.82 \begin{align*} \left [\frac{3 \,{\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \sqrt{b^{2} c - a b d} \log \left (\frac{b d x + 2 \, b c - a d - 2 \, \sqrt{b^{2} c - a b d} \sqrt{d x + c}}{b x + a}\right ) - 2 \,{\left (2 \, b^{3} c^{2} - 7 \, a b^{2} c d + 5 \, a^{2} b d^{2} - 3 \,{\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt{d x + c}}{8 \,{\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3} +{\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{2} + 2 \,{\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x\right )}}, \frac{3 \,{\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \sqrt{-b^{2} c + a b d} \arctan \left (\frac{\sqrt{-b^{2} c + a b d} \sqrt{d x + c}}{b d x + b c}\right ) -{\left (2 \, b^{3} c^{2} - 7 \, a b^{2} c d + 5 \, a^{2} b d^{2} - 3 \,{\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt{d x + c}}{4 \,{\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3} +{\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{2} + 2 \,{\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*(b^2*d^2*x^2 + 2*a*b*d^2*x + a^2*d^2)*sqrt(b^2*c - a*b*d)*log((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*
b*d)*sqrt(d*x + c))/(b*x + a)) - 2*(2*b^3*c^2 - 7*a*b^2*c*d + 5*a^2*b*d^2 - 3*(b^3*c*d - a*b^2*d^2)*x)*sqrt(d*
x + c))/(a^2*b^4*c^3 - 3*a^3*b^3*c^2*d + 3*a^4*b^2*c*d^2 - a^5*b*d^3 + (b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*
d^2 - a^3*b^3*d^3)*x^2 + 2*(a*b^5*c^3 - 3*a^2*b^4*c^2*d + 3*a^3*b^3*c*d^2 - a^4*b^2*d^3)*x), 1/4*(3*(b^2*d^2*x
^2 + 2*a*b*d^2*x + a^2*d^2)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (2
*b^3*c^2 - 7*a*b^2*c*d + 5*a^2*b*d^2 - 3*(b^3*c*d - a*b^2*d^2)*x)*sqrt(d*x + c))/(a^2*b^4*c^3 - 3*a^3*b^3*c^2*
d + 3*a^4*b^2*c*d^2 - a^5*b*d^3 + (b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*d^2 - a^3*b^3*d^3)*x^2 + 2*(a*b^5*c^3
- 3*a^2*b^4*c^2*d + 3*a^3*b^3*c*d^2 - a^4*b^2*d^3)*x)]

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.0649, size = 200, normalized size = 1.75 \begin{align*} \frac{3 \, d^{2} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{4 \,{\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 )}^{\frac{3}{2}} b d^{2} - 5 \, \sqrt{d x + c} b c d^{2} + 5 \, \sqrt{d x + c} a d^{3}}{4 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}{\left ({\left (d x + c\right )} b - b c + a d\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*d^2*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)) +
1/4*(3*(d*x + c)^(3/2)*b*d^2 - 5*sqrt(d*x + c)*b*c*d^2 + 5*sqrt(d*x + c)*a*d^3)/((b^2*c^2 - 2*a*b*c*d + a^2*d^
2)*((d*x + c)*b - b*c + a*d)^2)