3.1432 \(\int \frac{1}{(a+b x)^3 (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=140 \[ \frac{15 d^2}{4 \sqrt{c+d x} (b c-a d)^3}-\frac{15 \sqrt{b} d^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{4 (b c-a d)^{7/2}}+\frac{5 d}{4 (a+b x) \sqrt{c+d x} (b c-a d)^2}-\frac{1}{2 (a+b x)^2 \sqrt{c+d x} (b c-a d)} \]

[Out]

(15*d^2)/(4*(b*c - a*d)^3*Sqrt[c + d*x]) - 1/(2*(b*c - a*d)*(a + b*x)^2*Sqrt[c + d*x]) + (5*d)/(4*(b*c - a*d)^
2*(a + b*x)*Sqrt[c + d*x]) - (15*Sqrt[b]*d^2*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(4*(b*c - a*d)^
(7/2))

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Rubi [A]  time = 0.0515628, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {51, 63, 208} \[ \frac{15 d^2}{4 \sqrt{c+d x} (b c-a d)^3}-\frac{15 \sqrt{b} d^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{4 (b c-a d)^{7/2}}+\frac{5 d}{4 (a+b x) \sqrt{c+d x} (b c-a d)^2}-\frac{1}{2 (a+b x)^2 \sqrt{c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*d^2)/(4*(b*c - a*d)^3*Sqrt[c + d*x]) - 1/(2*(b*c - a*d)*(a + b*x)^2*Sqrt[c + d*x]) + (5*d)/(4*(b*c - a*d)^
2*(a + b*x)*Sqrt[c + d*x]) - (15*Sqrt[b]*d^2*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(4*(b*c - a*d)^
(7/2))

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 (c+d x)^{3/2}} \, dx &=-\frac{1}{2 (b c-a d) (a+b x)^2 \sqrt{c+d x}}-\frac{(5 d) \int \frac{1}{(a+b x)^2 (c+d x)^{3/2}} \, dx}{4 (b c-a d)}\\ &=-\frac{1}{2 (b c-a d) (a+b x)^2 \sqrt{c+d x}}+\frac{5 d}{4 (b c-a d)^2 (a+b x) \sqrt{c+d x}}+\frac{\left (15 d^2\right ) \int \frac{1}{(a+b x) (c+d x)^{3/2}} \, dx}{8 (b c-a d)^2}\\ &=\frac{15 d^2}{4 (b c-a d)^3 \sqrt{c+d x}}-\frac{1}{2 (b c-a d) (a+b x)^2 \sqrt{c+d x}}+\frac{5 d}{4 (b c-a d)^2 (a+b x) \sqrt{c+d x}}+\frac{\left (15 b d^2\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{8 (b c-a d)^3}\\ &=\frac{15 d^2}{4 (b c-a d)^3 \sqrt{c+d x}}-\frac{1}{2 (b c-a d) (a+b x)^2 \sqrt{c+d x}}+\frac{5 d}{4 (b c-a d)^2 (a+b x) \sqrt{c+d x}}+\frac{(15 b 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)^3}\\ &=\frac{15 d^2}{4 (b c-a d)^3 \sqrt{c+d x}}-\frac{1}{2 (b c-a d) (a+b x)^2 \sqrt{c+d x}}+\frac{5 d}{4 (b c-a d)^2 (a+b x) \sqrt{c+d x}}-\frac{15 \sqrt{b} d^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{4 (b c-a d)^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0127035, size = 50, normalized size = 0.36 \[ -\frac{2 d^2 \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};-\frac{b (c+d x)}{a d-b c}\right )}{\sqrt{c+d x} (a d-b c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 179, normalized size = 1.3 \begin{align*} -2\,{\frac{{d}^{2}}{ \left ( ad-bc \right ) ^{3}\sqrt{dx+c}}}-{\frac{7\,{d}^{2}{b}^{2}}{4\, \left ( ad-bc \right ) ^{3} \left ( bdx+ad \right ) ^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-{\frac{9\,{d}^{3}ba}{4\, \left ( ad-bc \right ) ^{3} \left ( bdx+ad \right ) ^{2}}\sqrt{dx+c}}+{\frac{9\,{d}^{2}{b}^{2}c}{4\, \left ( ad-bc \right ) ^{3} \left ( bdx+ad \right ) ^{2}}\sqrt{dx+c}}-{\frac{15\,{d}^{2}b}{4\, \left ( ad-bc \right ) ^{3}}\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)^(3/2),x)

[Out]

-2*d^2/(a*d-b*c)^3/(d*x+c)^(1/2)-7/4*d^2/(a*d-b*c)^3*b^2/(b*d*x+a*d)^2*(d*x+c)^(3/2)-9/4*d^3/(a*d-b*c)^3*b/(b*
d*x+a*d)^2*(d*x+c)^(1/2)*a+9/4*d^2/(a*d-b*c)^3*b^2/(b*d*x+a*d)^2*(d*x+c)^(1/2)*c-15/4*d^2/(a*d-b*c)^3*b/((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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.25975, size = 1582, normalized size = 11.3 \begin{align*} \left [-\frac{15 \,{\left (b^{2} d^{3} x^{3} + a^{2} c d^{2} +{\left (b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{2} +{\left (2 \, a b c d^{2} + a^{2} d^{3}\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 (15 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} + 9 \, a b c d + 8 \, a^{2} d^{2} + 5 \,{\left (b^{2} c d + 5 \, a b d^{2}\right )} x\right )} \sqrt{d x + c}}{8 \,{\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} +{\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} +{\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} +{\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}}, -\frac{15 \,{\left (b^{2} d^{3} x^{3} + a^{2} c d^{2} +{\left (b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{2} +{\left (2 \, a b c d^{2} + a^{2} d^{3}\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 (15 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} + 9 \, a b c d + 8 \, a^{2} d^{2} + 5 \,{\left (b^{2} c d + 5 \, a b d^{2}\right )} x\right )} \sqrt{d x + c}}{4 \,{\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} +{\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} +{\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} +{\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\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)^(3/2),x, algorithm="fricas")

[Out]

[-1/8*(15*(b^2*d^3*x^3 + a^2*c*d^2 + (b^2*c*d^2 + 2*a*b*d^3)*x^2 + (2*a*b*c*d^2 + a^2*d^3)*x)*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*(15*b^2*d^2*x^2
- 2*b^2*c^2 + 9*a*b*c*d + 8*a^2*d^2 + 5*(b^2*c*d + 5*a*b*d^2)*x)*sqrt(d*x + c))/(a^2*b^3*c^4 - 3*a^3*b^2*c^3*
d + 3*a^4*b*c^2*d^2 - a^5*c*d^3 + (b^5*c^3*d - 3*a*b^4*c^2*d^2 + 3*a^2*b^3*c*d^3 - a^3*b^2*d^4)*x^3 + (b^5*c^4
- a*b^4*c^3*d - 3*a^2*b^3*c^2*d^2 + 5*a^3*b^2*c*d^3 - 2*a^4*b*d^4)*x^2 + (2*a*b^4*c^4 - 5*a^2*b^3*c^3*d + 3*a
^3*b^2*c^2*d^2 + a^4*b*c*d^3 - a^5*d^4)*x), -1/4*(15*(b^2*d^3*x^3 + a^2*c*d^2 + (b^2*c*d^2 + 2*a*b*d^3)*x^2 +
(2*a*b*c*d^2 + a^2*d^3)*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)) - (15*b^2*d^2*x^2 - 2*b^2*c^2 + 9*a*b*c*d + 8*a^2*d^2 + 5*(b^2*c*d + 5*a*b*d^2)*x)*sqrt(d*x + c))/(a^2
*b^3*c^4 - 3*a^3*b^2*c^3*d + 3*a^4*b*c^2*d^2 - a^5*c*d^3 + (b^5*c^3*d - 3*a*b^4*c^2*d^2 + 3*a^2*b^3*c*d^3 - a^
3*b^2*d^4)*x^3 + (b^5*c^4 - a*b^4*c^3*d - 3*a^2*b^3*c^2*d^2 + 5*a^3*b^2*c*d^3 - 2*a^4*b*d^4)*x^2 + (2*a*b^4*c^
4 - 5*a^2*b^3*c^3*d + 3*a^3*b^2*c^2*d^2 + a^4*b*c*d^3 - a^5*d^4)*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)**(3/2),x)

[Out]

Exception raised: ValueError

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Giac [B]  time = 1.08341, size = 316, normalized size = 2.26 \begin{align*} \frac{15 \, b d^{2} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{4 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-b^{2} c + a b d}} + \frac{2 \, d^{2}}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{d x + c}} + \frac{7 \,{\left (d x + c\right )}^{\frac{3}{2}} b^{2} d^{2} - 9 \, \sqrt{d x + c} b^{2} c d^{2} + 9 \, \sqrt{d x + c} a b d^{3}}{4 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\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)^(3/2),x, algorithm="giac")

[Out]

15/4*b*d^2*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*s
qrt(-b^2*c + a*b*d)) + 2*d^2/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(d*x + c)) + 1/4*(7*(d*x
+ c)^(3/2)*b^2*d^2 - 9*sqrt(d*x + c)*b^2*c*d^2 + 9*sqrt(d*x + c)*a*b*d^3)/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b
*c*d^2 - a^3*d^3)*((d*x + c)*b - b*c + a*d)^2)