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

Optimal. Leaf size=189 \[ \frac{\left (-a^2 d^2-6 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{3/2} d^{7/2}}+\frac{\sqrt{a+b x} \sqrt{c+d x} \left (\frac{a^2 d}{b}+6 a c-\frac{15 b c^2}{d}\right )}{4 d^2 (b c-a d)}+\frac{2 c^2 (a+b x)^{3/2}}{d^2 \sqrt{c+d x} (b c-a d)}+\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 b d^2} \]

[Out]

(2*c^2*(a + b*x)^(3/2))/(d^2*(b*c - a*d)*Sqrt[c + d*x]) + ((6*a*c - (15*b*c^2)/d + (a^2*d)/b)*Sqrt[a + b*x]*Sq
rt[c + d*x])/(4*d^2*(b*c - a*d)) + ((a + b*x)^(3/2)*Sqrt[c + d*x])/(2*b*d^2) + ((15*b^2*c^2 - 6*a*b*c*d - a^2*
d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(3/2)*d^(7/2))

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Rubi [A]  time = 0.185242, antiderivative size = 189, 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 = {89, 80, 50, 63, 217, 206} \[ \frac{\left (-a^2 d^2-6 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{3/2} d^{7/2}}+\frac{\sqrt{a+b x} \sqrt{c+d x} \left (\frac{a^2 d}{b}+6 a c-\frac{15 b c^2}{d}\right )}{4 d^2 (b c-a d)}+\frac{2 c^2 (a+b x)^{3/2}}{d^2 \sqrt{c+d x} (b c-a d)}+\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^2*(a + b*x)^(3/2))/(d^2*(b*c - a*d)*Sqrt[c + d*x]) + ((6*a*c - (15*b*c^2)/d + (a^2*d)/b)*Sqrt[a + b*x]*Sq
rt[c + d*x])/(4*d^2*(b*c - a*d)) + ((a + b*x)^(3/2)*Sqrt[c + d*x])/(2*b*d^2) + ((15*b^2*c^2 - 6*a*b*c*d - a^2*
d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(3/2)*d^(7/2))

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 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}}{(c+d x)^{3/2}} \, dx &=\frac{2 c^2 (a+b x)^{3/2}}{d^2 (b c-a d) \sqrt{c+d x}}-\frac{2 \int \frac{\sqrt{a+b x} \left (\frac{1}{2} c (3 b c-a d)-\frac{1}{2} d (b c-a d) x\right )}{\sqrt{c+d x}} \, dx}{d^2 (b c-a d)}\\ &=\frac{2 c^2 (a+b x)^{3/2}}{d^2 (b c-a d) \sqrt{c+d x}}+\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 b d^2}-\frac{\left (15 b^2 c^2-6 a b c d-a^2 d^2\right ) \int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx}{4 b d^2 (b c-a d)}\\ &=\frac{2 c^2 (a+b x)^{3/2}}{d^2 (b c-a d) \sqrt{c+d x}}-\frac{\left (15 b^2 c^2-6 a b c d-a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{4 b d^3 (b c-a d)}+\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 b d^2}+\frac{\left (15 b^2 c^2-6 a b c d-a^2 d^2\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 b d^3}\\ &=\frac{2 c^2 (a+b x)^{3/2}}{d^2 (b c-a d) \sqrt{c+d x}}-\frac{\left (15 b^2 c^2-6 a b c d-a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{4 b d^3 (b c-a d)}+\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 b d^2}+\frac{\left (15 b^2 c^2-6 a b c d-a^2 d^2\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 )}{4 b^2 d^3}\\ &=\frac{2 c^2 (a+b x)^{3/2}}{d^2 (b c-a d) \sqrt{c+d x}}-\frac{\left (15 b^2 c^2-6 a b c d-a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{4 b d^3 (b c-a d)}+\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 b d^2}+\frac{\left (15 b^2 c^2-6 a b c d-a^2 d^2\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 )}{4 b^2 d^3}\\ &=\frac{2 c^2 (a+b x)^{3/2}}{d^2 (b c-a d) \sqrt{c+d x}}-\frac{\left (15 b^2 c^2-6 a b c d-a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{4 b d^3 (b c-a d)}+\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 b d^2}+\frac{\left (15 b^2 c^2-6 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 )}{4 b^{3/2} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.350918, size = 193, normalized size = 1.02 \[ \frac{\frac{b \sqrt{d} \left (a^2 d (c+d x)+a b \left (-15 c^2-4 c d x+3 d^2 x^2\right )+b^2 x \left (-15 c^2-5 c d x+2 d^2 x^2\right )\right )}{\sqrt{a+b x}}+\frac{\left (5 a^2 b c d^2+a^3 d^3-21 a b^2 c^2 d+15 b^3 c^3\right ) \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{\sqrt{b c-a d}}}{4 b^2 d^{7/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Sqrt[d]*(a^2*d*(c + d*x) + b^2*x*(-15*c^2 - 5*c*d*x + 2*d^2*x^2) + a*b*(-15*c^2 - 4*c*d*x + 3*d^2*x^2)))/S
qrt[a + b*x] + ((15*b^3*c^3 - 21*a*b^2*c^2*d + 5*a^2*b*c*d^2 + a^3*d^3)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSin
h[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/Sqrt[b*c - a*d])/(4*b^2*d^(7/2)*Sqrt[c + d*x])

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Maple [B]  time = 0.023, size = 456, normalized size = 2.4 \begin{align*} -{\frac{1}{8\,b{d}^{3}}\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}+6\,\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}-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 ) x{b}^{2}{c}^{2}d-4\,{x}^{2}b{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+\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}+6\,\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-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}^{2}{c}^{3}-2\,xa{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+10\,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}+30\,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*(b*x+a)^(1/2)/(d*x+c)^(3/2),x)

[Out]

-1/8*(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+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*b*c*d^2-15*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))*x*b^2*c^2*d-4*x^2*b*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(
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))*a^2*c*d^2+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))*a*b*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^2*c^3-2*x*a*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+10*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)+30*b*c^2*((b*x+a)*(d*x+c))^(1/2)*(b*d
)^(1/2))/(b*d)^(1/2)/b/((b*x+a)*(d*x+c))^(1/2)/d^3/(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*(b*x+a)^(1/2)/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.91925, size = 975, normalized size = 5.16 \begin{align*} \left [-\frac{{\left (15 \, b^{2} c^{3} - 6 \, a b c^{2} d - a^{2} c d^{2} +{\left (15 \, b^{2} c^{2} d - 6 \, 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 (2 \, b^{2} d^{3} x^{2} - 15 \, b^{2} c^{2} d + a b c d^{2} -{\left (5 \, b^{2} c d^{2} - a b d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{16 \,{\left (b^{2} d^{5} x + b^{2} c d^{4}\right )}}, -\frac{{\left (15 \, b^{2} c^{3} - 6 \, a b c^{2} d - a^{2} c d^{2} +{\left (15 \, b^{2} c^{2} d - 6 \, 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 (2 \, b^{2} d^{3} x^{2} - 15 \, b^{2} c^{2} d + a b c d^{2} -{\left (5 \, b^{2} c d^{2} - a b d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{8 \,{\left (b^{2} d^{5} x + b^{2} c d^{4}\right )}}\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)^(3/2),x, algorithm="fricas")

[Out]

[-1/16*((15*b^2*c^3 - 6*a*b*c^2*d - a^2*c*d^2 + (15*b^2*c^2*d - 6*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*(2*b^2*d^3*x^2 - 15*b^2*c^2*d + a*b*c*d^2 - (5*b^2*c*d^2 - a*b*d^3)*x)*sqrt(b*x + a)*
sqrt(d*x + c))/(b^2*d^5*x + b^2*c*d^4), -1/8*((15*b^2*c^3 - 6*a*b*c^2*d - a^2*c*d^2 + (15*b^2*c^2*d - 6*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*(2*b^2*d^3*x^2 - 15*b^2*c^2*d + a*b*c*d^2 - (5*b^2*c*d^2 - a*b*d^3
)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^5*x + b^2*c*d^4)]

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

[Out]

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

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Giac [A]  time = 2.31747, size = 346, normalized size = 1.83 \begin{align*} -\frac{{\left (15 \, b^{2} c^{2} - 6 \, a b c d - a^{2} d^{2}\right )} \sqrt{b d} \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 )}{384 \,{\left (b^{7} c d^{5} - a b^{6} d^{6}\right )}} + \frac{{\left ({\left (\frac{2 \,{\left (b x + a\right )} b^{2} d^{4}}{b^{8} c d^{6} - a b^{7} d^{7}} - \frac{5 \, b^{3} c d^{3} + 3 \, a b^{2} d^{4}}{b^{8} c d^{6} - a b^{7} d^{7}}\right )}{\left (b x + a\right )} - \frac{15 \, b^{4} c^{2} d^{2} - 6 \, a b^{3} c d^{3} - a^{2} b^{2} d^{4}}{b^{8} c d^{6} - a b^{7} d^{7}}\right )} \sqrt{b x + a}}{384 \, \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*(15*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*sqrt(b*d)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*
b*d - a*b*d)))/(b^7*c*d^5 - a*b^6*d^6) + 1/384*((2*(b*x + a)*b^2*d^4/(b^8*c*d^6 - a*b^7*d^7) - (5*b^3*c*d^3 +
3*a*b^2*d^4)/(b^8*c*d^6 - a*b^7*d^7))*(b*x + a) - (15*b^4*c^2*d^2 - 6*a*b^3*c*d^3 - a^2*b^2*d^4)/(b^8*c*d^6 -
a*b^7*d^7))*sqrt(b*x + a)/sqrt(b^2*c + (b*x + a)*b*d - a*b*d)