3.577 \(\int \frac{x^2 \sqrt{a+b x}}{\sqrt{c+d x}} \, dx\)

Optimal. Leaf size=191 \[ \frac{\sqrt{a+b x} \sqrt{c+d x} \left (a^2 d^2+2 a b c d+5 b^2 c^2\right )}{8 b^2 d^3}-\frac{(b c-a d) \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{5/2} d^{7/2}}-\frac{(a+b x)^{3/2} \sqrt{c+d x} (3 a d+5 b c)}{12 b^2 d^2}+\frac{x (a+b x)^{3/2} \sqrt{c+d x}}{3 b d} \]

[Out]

((5*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b^2*d^3) - ((5*b*c + 3*a*d)*(a + b*x)^(3/2)
*Sqrt[c + d*x])/(12*b^2*d^2) + (x*(a + b*x)^(3/2)*Sqrt[c + d*x])/(3*b*d) - ((b*c - a*d)*(5*b^2*c^2 + 2*a*b*c*d
+ a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(5/2)*d^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.163678, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {90, 80, 50, 63, 217, 206} \[ \frac{\sqrt{a+b x} \sqrt{c+d x} \left (a^2 d^2+2 a b c d+5 b^2 c^2\right )}{8 b^2 d^3}-\frac{(b c-a d) \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{5/2} d^{7/2}}-\frac{(a+b x)^{3/2} \sqrt{c+d x} (3 a d+5 b c)}{12 b^2 d^2}+\frac{x (a+b x)^{3/2} \sqrt{c+d x}}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b^2*d^3) - ((5*b*c + 3*a*d)*(a + b*x)^(3/2)
*Sqrt[c + d*x])/(12*b^2*d^2) + (x*(a + b*x)^(3/2)*Sqrt[c + d*x])/(3*b*d) - ((b*c - a*d)*(5*b^2*c^2 + 2*a*b*c*d
+ a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(5/2)*d^(7/2))

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 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^2 \sqrt{a+b x}}{\sqrt{c+d x}} \, dx &=\frac{x (a+b x)^{3/2} \sqrt{c+d x}}{3 b d}+\frac{\int \frac{\sqrt{a+b x} \left (-a c-\frac{1}{2} (5 b c+3 a d) x\right )}{\sqrt{c+d x}} \, dx}{3 b d}\\ &=-\frac{(5 b c+3 a d) (a+b x)^{3/2} \sqrt{c+d x}}{12 b^2 d^2}+\frac{x (a+b x)^{3/2} \sqrt{c+d x}}{3 b d}+\frac{\left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx}{8 b^2 d^2}\\ &=\frac{\left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 b^2 d^3}-\frac{(5 b c+3 a d) (a+b x)^{3/2} \sqrt{c+d x}}{12 b^2 d^2}+\frac{x (a+b x)^{3/2} \sqrt{c+d x}}{3 b d}-\frac{\left ((b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right )\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{16 b^2 d^3}\\ &=\frac{\left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 b^2 d^3}-\frac{(5 b c+3 a d) (a+b x)^{3/2} \sqrt{c+d x}}{12 b^2 d^2}+\frac{x (a+b x)^{3/2} \sqrt{c+d x}}{3 b d}-\frac{\left ((b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{8 b^3 d^3}\\ &=\frac{\left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 b^2 d^3}-\frac{(5 b c+3 a d) (a+b x)^{3/2} \sqrt{c+d x}}{12 b^2 d^2}+\frac{x (a+b x)^{3/2} \sqrt{c+d x}}{3 b d}-\frac{\left ((b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{8 b^3 d^3}\\ &=\frac{\left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 b^2 d^3}-\frac{(5 b c+3 a d) (a+b x)^{3/2} \sqrt{c+d x}}{12 b^2 d^2}+\frac{x (a+b x)^{3/2} \sqrt{c+d x}}{3 b d}-\frac{(b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{5/2} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.381404, size = 173, normalized size = 0.91 \[ \frac{-b \sqrt{d} \sqrt{a+b x} (c+d x) \left (3 a^2 d^2-2 a b d (d x-2 c)+b^2 \left (-15 c^2+10 c d x-8 d^2 x^2\right )\right )-3 \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) (b c-a d)^{3/2} \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{24 b^3 d^{7/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*Sqrt[d]*Sqrt[a + b*x]*(c + d*x)*(3*a^2*d^2 - 2*a*b*d*(-2*c + d*x) + b^2*(-15*c^2 + 10*c*d*x - 8*d^2*x^2))
) - 3*(b*c - a*d)^(3/2)*(5*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqr
t[a + b*x])/Sqrt[b*c - a*d]])/(24*b^3*d^(7/2)*Sqrt[c + d*x])

________________________________________________________________________________________

Maple [B]  time = 0.018, size = 395, normalized size = 2.1 \begin{align*}{\frac{1}{48\,{d}^{3}{b}^{2}}\sqrt{bx+a}\sqrt{dx+c} \left ( 16\,{x}^{2}{b}^{2}{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{3}{d}^{3}+3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}bc{d}^{2}+9\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) a{b}^{2}{c}^{2}d-15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{3}{c}^{3}+4\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}xab{d}^{2}-20\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}x{b}^{2}cd-6\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}{a}^{2}{d}^{2}-8\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}abcd+30\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }{b}^{2}{c}^{2} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(16*x^2*b^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(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))*a^3*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^2*b*c*d^2+9*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^2*c^2*d-15*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))*b^3*c^3+4*
((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x*a*b*d^2-20*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x*b^2*c*d-6*((b*x+a)*(d*x
+c))^(1/2)*(b*d)^(1/2)*a^2*d^2-8*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*b*c*d+30*(b*d)^(1/2)*((b*x+a)*(d*x+c))^
(1/2)*b^2*c^2)/d^3/((b*x+a)*(d*x+c))^(1/2)/b^2/(b*d)^(1/2)

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.06095, size = 921, normalized size = 4.82 \begin{align*} \left [-\frac{3 \,{\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \,{\left (2 \, b d x + b c + a d\right )} \sqrt{b d} \sqrt{b x + a} \sqrt{d x + c} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \,{\left (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d - 4 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3} - 2 \,{\left (5 \, b^{3} c d^{2} - a b^{2} d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{96 \, b^{3} d^{4}}, \frac{3 \,{\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-b d} \arctan \left (\frac{{\left (2 \, b d x + b c + a d\right )} \sqrt{-b d} \sqrt{b x + a} \sqrt{d x + c}}{2 \,{\left (b^{2} d^{2} x^{2} + a b c d +{\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \,{\left (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d - 4 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3} - 2 \,{\left (5 \, b^{3} c d^{2} - a b^{2} d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{48 \, b^{3} d^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*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*(8*
b^3*d^3*x^2 + 15*b^3*c^2*d - 4*a*b^2*c*d^2 - 3*a^2*b*d^3 - 2*(5*b^3*c*d^2 - a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d
*x + c))/(b^3*d^4), 1/48*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*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*(8*b
^3*d^3*x^2 + 15*b^3*c^2*d - 4*a*b^2*c*d^2 - 3*a^2*b*d^3 - 2*(5*b^3*c*d^2 - a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*
x + c))/(b^3*d^4)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.48108, size = 289, normalized size = 1.51 \begin{align*} \frac{{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a}{\left (2 \,{\left (b x + a\right )}{\left (\frac{4 \,{\left (b x + a\right )}}{b^{3} d} - \frac{5 \, b^{7} c d^{3} + 7 \, a b^{6} d^{4}}{b^{9} d^{5}}\right )} + \frac{3 \,{\left (5 \, b^{8} c^{2} d^{2} + 2 \, a b^{7} c d^{3} + a^{2} b^{6} d^{4}\right )}}{b^{9} d^{5}}\right )} + \frac{3 \,{\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\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 |}\right )}{\sqrt{b d} b^{2} d^{3}}\right )} b}{24 \,{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/(b^3*d) - (5*b^7*c*d^3 + 7*a
*b^6*d^4)/(b^9*d^5)) + 3*(5*b^8*c^2*d^2 + 2*a*b^7*c*d^3 + a^2*b^6*d^4)/(b^9*d^5)) + 3*(5*b^3*c^3 - 3*a*b^2*c^2
*d - a^2*b*c*d^2 - a^3*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d
)*b^2*d^3))*b/abs(b)