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

Optimal. Leaf size=125 \[ -\frac{(b c-a d) (3 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{5/2} d^{3/2}}-\frac{\sqrt{a+b x} \sqrt{c+d x} (3 a d+b c)}{4 b^2 d}+\frac{\sqrt{a+b x} (c+d x)^{3/2}}{2 b d} \]

[Out]

-((b*c + 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*b^2*d) + (Sqrt[a + b*x]*(c + d*x)^(3/2))/(2*b*d) - ((b*c - a*d
)*(b*c + 3*a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(5/2)*d^(3/2))

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Rubi [A]  time = 0.0638365, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {80, 50, 63, 217, 206} \[ -\frac{(b c-a d) (3 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{5/2} d^{3/2}}-\frac{\sqrt{a+b x} \sqrt{c+d x} (3 a d+b c)}{4 b^2 d}+\frac{\sqrt{a+b x} (c+d x)^{3/2}}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c + 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*b^2*d) + (Sqrt[a + b*x]*(c + d*x)^(3/2))/(2*b*d) - ((b*c - a*d
)*(b*c + 3*a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(5/2)*d^(3/2))

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 \sqrt{c+d x}}{\sqrt{a+b x}} \, dx &=\frac{\sqrt{a+b x} (c+d x)^{3/2}}{2 b d}-\frac{(b c+3 a d) \int \frac{\sqrt{c+d x}}{\sqrt{a+b x}} \, dx}{4 b d}\\ &=-\frac{(b c+3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 b^2 d}+\frac{\sqrt{a+b x} (c+d x)^{3/2}}{2 b d}-\frac{((b c-a d) (b c+3 a d)) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 b^2 d}\\ &=-\frac{(b c+3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 b^2 d}+\frac{\sqrt{a+b x} (c+d x)^{3/2}}{2 b d}-\frac{((b c-a d) (b c+3 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 )}{4 b^3 d}\\ &=-\frac{(b c+3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 b^2 d}+\frac{\sqrt{a+b x} (c+d x)^{3/2}}{2 b d}-\frac{((b c-a d) (b c+3 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 )}{4 b^3 d}\\ &=-\frac{(b c+3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 b^2 d}+\frac{\sqrt{a+b x} (c+d x)^{3/2}}{2 b d}-\frac{(b c-a d) (b c+3 a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{5/2} d^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.277263, size = 121, normalized size = 0.97 \[ \frac{\sqrt{c+d x} \left (\sqrt{d} \sqrt{a+b x} (b (c+2 d x)-3 a d)-\frac{\sqrt{b c-a d} (3 a d+b c) \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{\sqrt{\frac{b (c+d x)}{b c-a d}}}\right )}{4 b^2 d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*(Sqrt[d]*Sqrt[a + b*x]*(-3*a*d + b*(c + 2*d*x)) - (Sqrt[b*c - a*d]*(b*c + 3*a*d)*ArcSinh[(Sqrt[
d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/Sqrt[(b*(c + d*x))/(b*c - a*d)]))/(4*b^2*d^(3/2))

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Maple [B]  time = 0.013, size = 251, normalized size = 2. \begin{align*}{\frac{1}{8\,{b}^{2}d}\sqrt{bx+a}\sqrt{dx+c} \left ( 4\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}xbd+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}{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 ) abcd-\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 ){b}^{2}{c}^{2}-6\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}ad+2\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }bc \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*(d*x+c)^(1/2)/(b*x+a)^(1/2),x)

[Out]

1/8*(d*x+c)^(1/2)*(b*x+a)^(1/2)*(4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x*b*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^2*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*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))*b^2*c
^2-6*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*d+2*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b*c)/((b*x+a)*(d*x+c))^(1/2
)/b^2/d/(b*d)^(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*(d*x+c)^(1/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.16683, size = 711, normalized size = 5.69 \begin{align*} \left [-\frac{{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\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^{2} x + b^{2} c d - 3 \, a b d^{2}\right )} \sqrt{b x + a} \sqrt{d x + c}}{16 \, b^{3} d^{2}}, \frac{{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\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^{2} x + b^{2} c d - 3 \, a b d^{2}\right )} \sqrt{b x + a} \sqrt{d x + c}}{8 \, b^{3} d^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 2.6957, size = 190, normalized size = 1.52 \begin{align*} \frac{{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a}{\left (\frac{2 \,{\left (b x + a\right )}}{b^{4} d^{2}} + \frac{b c d - 5 \, a d^{2}}{b^{4} d^{4}}\right )} + \frac{{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\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^{3} d^{3}}\right )}{\left | b \right |}}{48 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)/(b^4*d^2) + (b*c*d - 5*a*d^2)/(b^4*d^4))
+ (b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*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))*abs(b)/b^4