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

Optimal. Leaf size=185 \[ -\frac{d \sqrt{a+b x} \left (3 a^2 d^2-2 a b c d+3 b^2 c^2\right )}{a^2 c^2 \sqrt{c+d x} (b c-a d)^2}+\frac{3 (a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{5/2} c^{5/2}}-\frac{b (3 b c-a d)}{a^2 c \sqrt{a+b x} \sqrt{c+d x} (b c-a d)}-\frac{1}{a c x \sqrt{a+b x} \sqrt{c+d x}} \]

[Out]

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

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Rubi [A]  time = 0.154501, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {103, 152, 12, 93, 208} \[ -\frac{d \sqrt{a+b x} \left (3 a^2 d^2-2 a b c d+3 b^2 c^2\right )}{a^2 c^2 \sqrt{c+d x} (b c-a d)^2}+\frac{3 (a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{5/2} c^{5/2}}-\frac{b (3 b c-a d)}{a^2 c \sqrt{a+b x} \sqrt{c+d x} (b c-a d)}-\frac{1}{a c x \sqrt{a+b x} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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}{x^2 (a+b x)^{3/2} (c+d x)^{3/2}} \, dx &=-\frac{1}{a c x \sqrt{a+b x} \sqrt{c+d x}}-\frac{\int \frac{\frac{3}{2} (b c+a d)+2 b d x}{x (a+b x)^{3/2} (c+d x)^{3/2}} \, dx}{a c}\\ &=-\frac{b (3 b c-a d)}{a^2 c (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}-\frac{1}{a c x \sqrt{a+b x} \sqrt{c+d x}}-\frac{2 \int \frac{\frac{3}{4} (b c-a d) (b c+a d)+\frac{1}{2} b d (3 b c-a d) x}{x \sqrt{a+b x} (c+d x)^{3/2}} \, dx}{a^2 c (b c-a d)}\\ &=-\frac{b (3 b c-a d)}{a^2 c (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}-\frac{1}{a c x \sqrt{a+b x} \sqrt{c+d x}}-\frac{d \left (3 b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt{a+b x}}{a^2 c^2 (b c-a d)^2 \sqrt{c+d x}}+\frac{4 \int -\frac{3 (b c-a d)^2 (b c+a d)}{8 x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{a^2 c^2 (b c-a d)^2}\\ &=-\frac{b (3 b c-a d)}{a^2 c (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}-\frac{1}{a c x \sqrt{a+b x} \sqrt{c+d x}}-\frac{d \left (3 b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt{a+b x}}{a^2 c^2 (b c-a d)^2 \sqrt{c+d x}}-\frac{(3 (b c+a d)) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 a^2 c^2}\\ &=-\frac{b (3 b c-a d)}{a^2 c (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}-\frac{1}{a c x \sqrt{a+b x} \sqrt{c+d x}}-\frac{d \left (3 b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt{a+b x}}{a^2 c^2 (b c-a d)^2 \sqrt{c+d x}}-\frac{(3 (b c+a d)) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{a^2 c^2}\\ &=-\frac{b (3 b c-a d)}{a^2 c (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}-\frac{1}{a c x \sqrt{a+b x} \sqrt{c+d x}}-\frac{d \left (3 b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt{a+b x}}{a^2 c^2 (b c-a d)^2 \sqrt{c+d x}}+\frac{3 (b c+a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{5/2} c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.195159, size = 165, normalized size = 0.89 \[ \frac{a^2 b d \left (2 c^2+c d x-3 d^2 x^2\right )+a^3 \left (-d^2\right ) (c+3 d x)+a b^2 c \left (-c^2+c d x+2 d^2 x^2\right )-3 b^3 c^2 x (c+d x)}{a^2 c^2 x \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^2}+\frac{3 (a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{5/2} c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.034, size = 897, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a^2/c^2*(3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^3*a^3*b*d^4-3*ln((a*d*x+b*c*x
+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^3*a^2*b^2*c*d^3-3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d
*x+c))^(1/2)+2*a*c)/x)*x^3*a*b^3*c^2*d^2+3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^3
*b^4*c^3*d+3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a^4*d^4-6*ln((a*d*x+b*c*x+2*(
a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a^2*b^2*c^2*d^2+3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x
+c))^(1/2)+2*a*c)/x)*x^2*b^4*c^4+3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x*a^4*c*d^3
-3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x*a^3*b*c^2*d^2-3*ln((a*d*x+b*c*x+2*(a*c)^(
1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x*a^2*b^2*c^3*d+3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)
+2*a*c)/x)*x*a*b^3*c^4-6*x^2*a^2*b*d^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+4*x^2*a*b^2*c*d^2*((b*x+a)*(d*x+c))
^(1/2)*(a*c)^(1/2)-6*x^2*b^3*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-6*x*a^3*d^3*((b*x+a)*(d*x+c))^(1/2)*(a*
c)^(1/2)+2*x*a^2*b*c*d^2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+2*x*a*b^2*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/
2)-6*x*b^3*c^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-2*a^3*c*d^2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+4*a^2*b*c^2
*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-2*a*b^2*c^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2))/((b*x+a)*(d*x+c))^(1/2
)/(a*d-b*c)^2/(a*c)^(1/2)/x/(b*x+a)^(1/2)/(d*x+c)^(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/x^2/(b*x+a)^(3/2)/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 8.64111, size = 1882, normalized size = 10.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError