3.563 \(\int \frac{\sqrt{a+b x} (c+d x)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=163 \[ -\frac{(b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{8 a^{5/2} c^{3/2}}+\frac{\sqrt{a+b x} \sqrt{c+d x} (b c-a d)^2}{8 a^2 c x}-\frac{\sqrt{a+b x} (c+d x)^{3/2} (b c-a d)}{12 a c x^2}-\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 c x^3} \]

[Out]

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

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Rubi [A]  time = 0.0733824, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {94, 93, 208} \[ -\frac{(b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{8 a^{5/2} c^{3/2}}+\frac{\sqrt{a+b x} \sqrt{c+d x} (b c-a d)^2}{8 a^2 c x}-\frac{\sqrt{a+b x} (c+d x)^{3/2} (b c-a d)}{12 a c x^2}-\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 c x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 94

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\sqrt{a+b x} (c+d x)^{3/2}}{x^4} \, dx &=-\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 c x^3}+\frac{(b c-a d) \int \frac{(c+d x)^{3/2}}{x^3 \sqrt{a+b x}} \, dx}{6 c}\\ &=-\frac{(b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}{12 a c x^2}-\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 c x^3}-\frac{(b c-a d)^2 \int \frac{\sqrt{c+d x}}{x^2 \sqrt{a+b x}} \, dx}{8 a c}\\ &=\frac{(b c-a d)^2 \sqrt{a+b x} \sqrt{c+d x}}{8 a^2 c x}-\frac{(b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}{12 a c x^2}-\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 c x^3}+\frac{(b c-a d)^3 \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{16 a^2 c}\\ &=\frac{(b c-a d)^2 \sqrt{a+b x} \sqrt{c+d x}}{8 a^2 c x}-\frac{(b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}{12 a c x^2}-\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 c x^3}+\frac{(b c-a d)^3 \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{8 a^2 c}\\ &=\frac{(b c-a d)^2 \sqrt{a+b x} \sqrt{c+d x}}{8 a^2 c x}-\frac{(b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}{12 a c x^2}-\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 c x^3}-\frac{(b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{8 a^{5/2} c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.216033, size = 141, normalized size = 0.87 \[ -\frac{\frac{x (b c-a d) \left (3 x^2 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\sqrt{a} \sqrt{c} \sqrt{a+b x} \sqrt{c+d x} (2 a c+5 a d x-3 b c x)\right )}{a^{5/2} \sqrt{c}}+8 \sqrt{a+b x} (c+d x)^{5/2}}{24 c x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(8*Sqrt[a + b*x]*(c + d*x)^(5/2) + ((b*c - a*d)*x*(Sqrt[a]*Sqrt[c]*Sqrt[a + b*x]*Sqrt[c + d*x]*(2*a*c - 3*b*c
*x + 5*a*d*x) + 3*(b*c - a*d)^2*x^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(a^(5/2)*Sqrt[c
]))/(24*c*x^3)

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Maple [B]  time = 0.015, size = 485, normalized size = 3. \begin{align*}{\frac{1}{48\,{a}^{2}c{x}^{3}}\sqrt{bx+a}\sqrt{dx+c} \left ( 3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{3}{a}^{3}{d}^{3}-9\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{3}{a}^{2}bc{d}^{2}+9\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{3}a{b}^{2}{c}^{2}d-3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{3}{b}^{3}{c}^{3}-6\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{ac}{x}^{2}{a}^{2}{d}^{2}-16\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{2}abcd+6\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{ac}{x}^{2}{b}^{2}{c}^{2}-28\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}x{a}^{2}cd-4\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}xab{c}^{2}-16\,\sqrt{d{x}^{2}b+adx+bcx+ac}{a}^{2}{c}^{2}\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{d{x}^{2}b+adx+bcx+ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(3/2)*(b*x+a)^(1/2)/x^4,x)

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^2/c*(3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)
/x)*x^3*a^3*d^3-9*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x^3*a^2*b*c*d^2+9*ln
((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x^3*a*b^2*c^2*d-3*ln((a*d*x+b*c*x+2*(a*c
)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x^3*b^3*c^3-6*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(a*c)^(1/2)*x^
2*a^2*d^2-16*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x^2*a*b*c*d+6*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(a*c)^(
1/2)*x^2*b^2*c^2-28*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a^2*c*d-4*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a
*c)^(1/2)*x*a*b*c^2-16*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*c^2*(a*c)^(1/2))/(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)/x^
3/(a*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((d*x+c)^(3/2)*(b*x+a)^(1/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 6.93558, size = 961, normalized size = 5.9 \begin{align*} \left [-\frac{3 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{a c} x^{3} \log \left (\frac{8 \, a^{2} c^{2} +{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} + 4 \,{\left (2 \, a c +{\left (b c + a d\right )} x\right )} \sqrt{a c} \sqrt{b x + a} \sqrt{d x + c} + 8 \,{\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \,{\left (8 \, a^{3} c^{3} -{\left (3 \, a b^{2} c^{3} - 8 \, a^{2} b c^{2} d - 3 \, a^{3} c d^{2}\right )} x^{2} + 2 \,{\left (a^{2} b c^{3} + 7 \, a^{3} c^{2} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{96 \, a^{3} c^{2} x^{3}}, \frac{3 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-a c} x^{3} \arctan \left (\frac{{\left (2 \, a c +{\left (b c + a d\right )} x\right )} \sqrt{-a c} \sqrt{b x + a} \sqrt{d x + c}}{2 \,{\left (a b c d x^{2} + a^{2} c^{2} +{\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 2 \,{\left (8 \, a^{3} c^{3} -{\left (3 \, a b^{2} c^{3} - 8 \, a^{2} b c^{2} d - 3 \, a^{3} c d^{2}\right )} x^{2} + 2 \,{\left (a^{2} b c^{3} + 7 \, a^{3} c^{2} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{48 \, a^{3} c^{2} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)*(b*x+a)^(1/2)/x^4,x, algorithm="fricas")

[Out]

[-1/96*(3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(a*c)*x^3*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*
c*d + a^2*d^2)*x^2 + 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x
)/x^2) + 4*(8*a^3*c^3 - (3*a*b^2*c^3 - 8*a^2*b*c^2*d - 3*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 + 7*a^3*c^2*d)*x)*sqrt(
b*x + a)*sqrt(d*x + c))/(a^3*c^2*x^3), 1/48*(3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-a*c)*
x^3*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^
2 + a^2*c*d)*x)) - 2*(8*a^3*c^3 - (3*a*b^2*c^3 - 8*a^2*b*c^2*d - 3*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 + 7*a^3*c^2*d
)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*c^2*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + b x} \left (c + d x\right )^{\frac{3}{2}}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(3/2)*(b*x+a)**(1/2)/x**4,x)

[Out]

Integral(sqrt(a + b*x)*(c + d*x)**(3/2)/x**4, x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)*(b*x+a)^(1/2)/x^4,x, algorithm="giac")

[Out]

Exception raised: TypeError