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

Optimal. Leaf size=119 \[ -\frac{3 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)}{4 c^2 x}-\frac{3 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 \sqrt{a} c^{5/2}}-\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 c x^2} \]

[Out]

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

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Rubi [A]  time = 0.0432698, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {94, 93, 208} \[ -\frac{3 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)}{4 c^2 x}-\frac{3 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 \sqrt{a} c^{5/2}}-\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0644395, size = 98, normalized size = 0.82 \[ \frac{\sqrt{a+b x} \sqrt{c+d x} (-2 a c+3 a d x-5 b c x)}{4 c^2 x^2}-\frac{3 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 \sqrt{a} c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.018, size = 255, normalized size = 2.1 \begin{align*} -{\frac{1}{8\,{c}^{2}{x}^{2}}\sqrt{bx+a}\sqrt{dx+c} \left ( 3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{a}^{2}{d}^{2}-6\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}abcd+3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{b}^{2}{c}^{2}-6\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}xad+10\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}xbc+4\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }ac\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c^2*(3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a^
2*d^2-6*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a*b*c*d+3*ln((a*d*x+b*c*x+2*(a*c)^
(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*b^2*c^2-6*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x*a*d+10*((b*x+a)*(d
*x+c))^(1/2)*(a*c)^(1/2)*x*b*c+4*((b*x+a)*(d*x+c))^(1/2)*a*c*(a*c)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/x^2/(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((b*x+a)^(3/2)/x^3/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.23565, size = 753, normalized size = 6.33 \begin{align*} \left [\frac{3 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{a c} x^{2} \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 (2 \, a^{2} c^{2} +{\left (5 \, a b c^{2} - 3 \, a^{2} c d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{16 \, a c^{3} x^{2}}, \frac{3 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{-a c} x^{2} \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 (2 \, a^{2} c^{2} +{\left (5 \, a b c^{2} - 3 \, a^{2} c d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{8 \, a c^{3} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: TypeError