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

Optimal. Leaf size=167 \[ \frac{\sqrt{a+b x} \left (c \left (3 a^2 d^2-2 a b c d+3 b^2 c^2\right )+d x (b c-3 a d) (b c-a d)\right )}{b^2 d^2 \sqrt{c+d x} (b c-a d)^2}-\frac{3 (a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{5/2} d^{5/2}}+\frac{2 a x^2}{b \sqrt{a+b x} \sqrt{c+d x} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.129546, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {98, 143, 63, 217, 206} \[ \frac{\sqrt{a+b x} \left (c \left (3 a^2 d^2-2 a b c d+3 b^2 c^2\right )+d x (b c-3 a d) (b c-a d)\right )}{b^2 d^2 \sqrt{c+d x} (b c-a d)^2}-\frac{3 (a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{5/2} d^{5/2}}+\frac{2 a x^2}{b \sqrt{a+b x} \sqrt{c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 98

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

Rule 143

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x
)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)), x] + Dist[(a*d*f*h*m + b*(d*(f*g + e*h) - c*f*h*(m +
2)))/(b^2*d), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m
+ n + 2, 0] && NeQ[m, -1] &&  !(SumSimplerQ[n, 1] &&  !SumSimplerQ[m, 1])

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.705951, size = 195, normalized size = 1.17 \[ \frac{\frac{\sqrt{d} \left (a^2 b d \left (-2 c^2-c d x+d^2 x^2\right )+3 a^3 d^2 (c+d x)+a b^2 c \left (3 c^2-c d x-2 d^2 x^2\right )+b^3 c^2 x (3 c+d x)\right )}{\sqrt{a+b x} (b c-a d)^2}-\frac{3 \sqrt{b c-a d} (a d+b c) \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{b}}{b^2 d^{5/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.023, size = 906, normalized size = 5.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.57263, size = 1832, normalized size = 10.97 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [B]  time = 3.29654, size = 462, normalized size = 2.77 \begin{align*} \frac{4 \, \sqrt{b d} a^{3}}{{\left (b^{2} c{\left | b \right |} - a b d{\left | b \right |}\right )}{\left (b^{2} c - a b d -{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}} + \frac{\sqrt{b x + a}{\left (\frac{{\left (b^{6} c^{2} d^{2} - 2 \, a b^{5} c d^{3} + a^{2} b^{4} d^{4}\right )}{\left (b x + a\right )}}{b^{7} c^{2} d^{3}{\left | b \right |} - 2 \, a b^{6} c d^{4}{\left | b \right |} + a^{2} b^{5} d^{5}{\left | b \right |}} + \frac{3 \, b^{7} c^{3} d - 3 \, a b^{6} c^{2} d^{2} + 3 \, a^{2} b^{5} c d^{3} - a^{3} b^{4} d^{4}}{b^{7} c^{2} d^{3}{\left | b \right |} - 2 \, a b^{6} c d^{4}{\left | b \right |} + a^{2} b^{5} d^{5}{\left | b \right |}}\right )}}{\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}} + \frac{3 \,{\left (\sqrt{b d} b c + \sqrt{b d} a d\right )} \log \left ({\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{2 \, b^{2} d^{3}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*sqrt(b*d)*a^3/((b^2*c*abs(b) - a*b*d*abs(b))*(b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x +
a)*b*d - a*b*d))^2)) + sqrt(b*x + a)*((b^6*c^2*d^2 - 2*a*b^5*c*d^3 + a^2*b^4*d^4)*(b*x + a)/(b^7*c^2*d^3*abs(
b) - 2*a*b^6*c*d^4*abs(b) + a^2*b^5*d^5*abs(b)) + (3*b^7*c^3*d - 3*a*b^6*c^2*d^2 + 3*a^2*b^5*c*d^3 - a^3*b^4*d
^4)/(b^7*c^2*d^3*abs(b) - 2*a*b^6*c*d^4*abs(b) + a^2*b^5*d^5*abs(b)))/sqrt(b^2*c + (b*x + a)*b*d - a*b*d) + 3/
2*(sqrt(b*d)*b*c + sqrt(b*d)*a*d)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(b^2*
d^3*abs(b))