3.790 \(\int \frac{1}{x (a+b x)^{3/2} (c+d x)^{5/2}} \, dx\)
Optimal. Leaf size=180 \[ -\frac{2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{3/2} c^{5/2}}+\frac{2 d \sqrt{a+b x} (3 b c-a d) (3 a d+b c)}{3 a c^2 \sqrt{c+d x} (b c-a d)^3}+\frac{2 b}{a \sqrt{a+b x} (c+d x)^{3/2} (b c-a d)}+\frac{2 d \sqrt{a+b x} (a d+3 b c)}{3 a c (c+d x)^{3/2} (b c-a d)^2} \]
[Out]
(2*b)/(a*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) + (2*d*(3*b*c + a*d)*Sqrt[a + b*x])/(3*a*c*(b*c - a*d)^2*(
c + d*x)^(3/2)) + (2*d*(3*b*c - a*d)*(b*c + 3*a*d)*Sqrt[a + b*x])/(3*a*c^2*(b*c - a*d)^3*Sqrt[c + d*x]) - (2*A
rcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(a^(3/2)*c^(5/2))
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Rubi [A] time = 0.14542, antiderivative size = 180, 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 =
{104, 152, 12, 93, 208} \[ -\frac{2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{3/2} c^{5/2}}+\frac{2 d \sqrt{a+b x} (3 b c-a d) (3 a d+b c)}{3 a c^2 \sqrt{c+d x} (b c-a d)^3}+\frac{2 b}{a \sqrt{a+b x} (c+d x)^{3/2} (b c-a d)}+\frac{2 d \sqrt{a+b x} (a d+3 b c)}{3 a c (c+d x)^{3/2} (b c-a d)^2} \]
Antiderivative was successfully verified.
[In]
Int[1/(x*(a + b*x)^(3/2)*(c + d*x)^(5/2)),x]
[Out]
(2*b)/(a*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) + (2*d*(3*b*c + a*d)*Sqrt[a + b*x])/(3*a*c*(b*c - a*d)^2*(
c + d*x)^(3/2)) + (2*d*(3*b*c - a*d)*(b*c + 3*a*d)*Sqrt[a + b*x])/(3*a*c^2*(b*c - a*d)^3*Sqrt[c + d*x]) - (2*A
rcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(a^(3/2)*c^(5/2))
Rule 104
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] &&
IntegersQ[2*m, 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 (a+b x)^{3/2} (c+d x)^{5/2}} \, dx &=\frac{2 b}{a (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}+\frac{2 \int \frac{\frac{1}{2} (b c-a d)+2 b d x}{x \sqrt{a+b x} (c+d x)^{5/2}} \, dx}{a (b c-a d)}\\ &=\frac{2 b}{a (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}+\frac{2 d (3 b c+a d) \sqrt{a+b x}}{3 a c (b c-a d)^2 (c+d x)^{3/2}}-\frac{4 \int \frac{-\frac{3}{4} (b c-a d)^2-\frac{1}{2} b d (3 b c+a d) x}{x \sqrt{a+b x} (c+d x)^{3/2}} \, dx}{3 a c (b c-a d)^2}\\ &=\frac{2 b}{a (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}+\frac{2 d (3 b c+a d) \sqrt{a+b x}}{3 a c (b c-a d)^2 (c+d x)^{3/2}}+\frac{2 d (3 b c-a d) (b c+3 a d) \sqrt{a+b x}}{3 a c^2 (b c-a d)^3 \sqrt{c+d x}}+\frac{8 \int \frac{3 (b c-a d)^3}{8 x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{3 a c^2 (b c-a d)^3}\\ &=\frac{2 b}{a (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}+\frac{2 d (3 b c+a d) \sqrt{a+b x}}{3 a c (b c-a d)^2 (c+d x)^{3/2}}+\frac{2 d (3 b c-a d) (b c+3 a d) \sqrt{a+b x}}{3 a c^2 (b c-a d)^3 \sqrt{c+d x}}+\frac{\int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{a c^2}\\ &=\frac{2 b}{a (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}+\frac{2 d (3 b c+a d) \sqrt{a+b x}}{3 a c (b c-a d)^2 (c+d x)^{3/2}}+\frac{2 d (3 b c-a d) (b c+3 a d) \sqrt{a+b x}}{3 a c^2 (b c-a d)^3 \sqrt{c+d x}}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{a c^2}\\ &=\frac{2 b}{a (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}+\frac{2 d (3 b c+a d) \sqrt{a+b x}}{3 a c (b c-a d)^2 (c+d x)^{3/2}}+\frac{2 d (3 b c-a d) (b c+3 a d) \sqrt{a+b x}}{3 a c^2 (b c-a d)^3 \sqrt{c+d x}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{3/2} c^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.387574, size = 158, normalized size = 0.88 \[ \frac{2 \left (a^2 b d^2 \left (9 c^2+4 c d x-3 d^2 x^2\right )-a^3 d^3 (4 c+3 d x)+a b^2 c d^2 x (9 c+8 d x)+3 b^3 c^2 (c+d x)^2\right )}{3 a c^2 \sqrt{a+b x} (c+d x)^{3/2} (b c-a d)^3}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{3/2} c^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(x*(a + b*x)^(3/2)*(c + d*x)^(5/2)),x]
[Out]
(2*(3*b^3*c^2*(c + d*x)^2 - a^3*d^3*(4*c + 3*d*x) + a*b^2*c*d^2*x*(9*c + 8*d*x) + a^2*b*d^2*(9*c^2 + 4*c*d*x -
3*d^2*x^2)))/(3*a*c^2*(b*c - a*d)^3*Sqrt[a + b*x]*(c + d*x)^(3/2)) - (2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt
[a]*Sqrt[c + d*x])])/(a^(3/2)*c^(5/2))
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Maple [B] time = 0.038, size = 1253, normalized size = 7. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/x/(b*x+a)^(3/2)/(d*x+c)^(5/2),x)
[Out]
-1/3*(9*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^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^2-6*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*b^4*c^4*d+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*a^4*c*d^4-
9*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*a^3*b*c^3*d^2+9*ln((a*d*x+b*c*x+2*(a*c)^(1/2
)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*a^2*b^2*c^4*d-6*x*a^3*d^4*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-8*a^3*c*d^3*
(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-9*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^4+9*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^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^2*a^3*b*c*d^4-9*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^3+15*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*b^3*c^3*d^2-15*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^3+6*b^3*c^4*
(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+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*a^4*
d^5-3*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*b^4*c^5+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)*a^4*c^2*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)*a*b^3*c^5+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^5+16*x^2*a*b^2*c*d
^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+8*x*a^2*b*c*d^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+18*x*a*b^2*c^2*d^2*
(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+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*b^3*
c^4*d-6*x^2*a^2*b*d^4*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+6*x^2*b^3*c^2*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2
)+12*x*b^3*c^3*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+18*a^2*b*c^2*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/a/c
^2/(a*c)^(1/2)/(a*d-b*c)^3/((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(3/2)/(b*x+a)^(1/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x + a\right )}^{\frac{3}{2}}{\left (d x + c\right )}^{\frac{5}{2}} x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="maxima")
[Out]
integrate(1/((b*x + a)^(3/2)*(d*x + c)^(5/2)*x), x)
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Fricas [B] time = 11.073, size = 2669, normalized size = 14.83 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="fricas")
[Out]
[1/6*(3*(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*
b^2*c*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^2 +
(b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*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*(3*a*b^3*c^5 + 9*a^3*b*c^3*d^2 - 4*a^4*c^2*d^3 + (3*a*b^3*c^3*d^2 + 8*a^2*b^2*c^2*d^3
- 3*a^3*b*c*d^4)*x^2 + (6*a*b^3*c^4*d + 9*a^2*b^2*c^3*d^2 + 4*a^3*b*c^2*d^3 - 3*a^4*c*d^4)*x)*sqrt(b*x + a)*s
qrt(d*x + c))/(a^3*b^3*c^8 - 3*a^4*b^2*c^7*d + 3*a^5*b*c^6*d^2 - a^6*c^5*d^3 + (a^2*b^4*c^6*d^2 - 3*a^3*b^3*c^
5*d^3 + 3*a^4*b^2*c^4*d^4 - a^5*b*c^3*d^5)*x^3 + (2*a^2*b^4*c^7*d - 5*a^3*b^3*c^6*d^2 + 3*a^4*b^2*c^5*d^3 + a^
5*b*c^4*d^4 - a^6*c^3*d^5)*x^2 + (a^2*b^4*c^8 - a^3*b^3*c^7*d - 3*a^4*b^2*c^6*d^2 + 5*a^5*b*c^5*d^3 - 2*a^6*c^
4*d^4)*x), 1/3*(3*(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^
3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*
d^5)*x^2 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*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*(3*a*b^3*c^5 + 9*a^3*b*c^3*d^2 - 4*a^4*c^2*d^3 + (3*a*b^3*c^3*d^2 + 8*a^2*b^2*c^2*d^3 - 3*a^3*b*c*d^4)
*x^2 + (6*a*b^3*c^4*d + 9*a^2*b^2*c^3*d^2 + 4*a^3*b*c^2*d^3 - 3*a^4*c*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^
3*b^3*c^8 - 3*a^4*b^2*c^7*d + 3*a^5*b*c^6*d^2 - a^6*c^5*d^3 + (a^2*b^4*c^6*d^2 - 3*a^3*b^3*c^5*d^3 + 3*a^4*b^2
*c^4*d^4 - a^5*b*c^3*d^5)*x^3 + (2*a^2*b^4*c^7*d - 5*a^3*b^3*c^6*d^2 + 3*a^4*b^2*c^5*d^3 + a^5*b*c^4*d^4 - a^6
*c^3*d^5)*x^2 + (a^2*b^4*c^8 - a^3*b^3*c^7*d - 3*a^4*b^2*c^6*d^2 + 5*a^5*b*c^5*d^3 - 2*a^6*c^4*d^4)*x)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (a + b x\right )^{\frac{3}{2}} \left (c + d x\right )^{\frac{5}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(b*x+a)**(3/2)/(d*x+c)**(5/2),x)
[Out]
Integral(1/(x*(a + b*x)**(3/2)*(c + d*x)**(5/2)), 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/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="giac")
[Out]
Exception raised: TypeError