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

Optimal. Leaf size=112 \[ -\frac{(a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{3/2} d^{5/2}}+\frac{2 c^2 \sqrt{a+b x}}{d^2 \sqrt{c+d x} (b c-a d)}+\frac{\sqrt{a+b x} \sqrt{c+d x}}{b d^2} \]

[Out]

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

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Rubi [A]  time = 0.0924784, antiderivative size = 112, 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 = {89, 80, 63, 217, 206} \[ -\frac{(a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{3/2} d^{5/2}}+\frac{2 c^2 \sqrt{a+b x}}{d^2 \sqrt{c+d x} (b c-a d)}+\frac{\sqrt{a+b x} \sqrt{c+d x}}{b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 89

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

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

Mathematica [A]  time = 0.212374, size = 151, normalized size = 1.35 \[ \frac{b \sqrt{d} \sqrt{a+b x} (b c (3 c+d x)-a d (c+d x))-\sqrt{b c-a d} \left (-a^2 d^2-2 a b c d+3 b^2 c^2\right ) \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{b^2 d^{5/2} \sqrt{c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.021, size = 439, normalized size = 3.9 \begin{align*} -{\frac{1}{2\, \left ( ad-bc \right ) b{d}^{2}}\sqrt{bx+a} \left ( \ln \left ({\frac{1}{2} \left ( 2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ) x{a}^{2}{d}^{3}+2\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) xabc{d}^{2}-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 ) x{b}^{2}{c}^{2}d+\ln \left ({\frac{1}{2} \left ( 2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ){a}^{2}c{d}^{2}+2\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) ab{c}^{2}d-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 ){b}^{2}{c}^{3}-2\,xa{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+2\,xbcd\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}-2\,acd\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+6\,b{c}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{dx+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.7458, size = 1000, normalized size = 8.93 \begin{align*} \left [\frac{{\left (3 \, b^{2} c^{3} - 2 \, a b c^{2} d - a^{2} c d^{2} +{\left (3 \, b^{2} c^{2} d - 2 \, a b c d^{2} - a^{2} d^{3}\right )} x\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 (3 \, b^{2} c^{2} d - a b c d^{2} +{\left (b^{2} c d^{2} - a b d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{4 \,{\left (b^{3} c^{2} d^{3} - a b^{2} c d^{4} +{\left (b^{3} c d^{4} - a b^{2} d^{5}\right )} x\right )}}, \frac{{\left (3 \, b^{2} c^{3} - 2 \, a b c^{2} d - a^{2} c d^{2} +{\left (3 \, b^{2} c^{2} d - 2 \, a b c d^{2} - a^{2} d^{3}\right )} x\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 (3 \, b^{2} c^{2} d - a b c d^{2} +{\left (b^{2} c d^{2} - a b d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{2 \,{\left (b^{3} c^{2} d^{3} - a b^{2} c d^{4} +{\left (b^{3} c d^{4} - a b^{2} d^{5}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.29053, size = 251, normalized size = 2.24 \begin{align*} \frac{\sqrt{b x + a}{\left (\frac{{\left (b^{3} c d^{2} - a b^{2} d^{3}\right )}{\left (b x + a\right )}}{b^{6} c d^{4} - a b^{5} d^{5}} + \frac{3 \, b^{4} c^{2} d - 2 \, a b^{3} c d^{2} + a^{2} b^{2} d^{3}}{b^{6} c d^{4} - a b^{5} d^{5}}\right )}}{8 \, \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}} + \frac{{\left (3 \, b c + 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 |}\right )}{8 \, \sqrt{b d} b^{3} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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