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

Optimal. Leaf size=180 \[ \frac{(b c-a d)^2 (5 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{8 a^{3/2} c^{7/2}}+\frac{(a+b x)^{3/2} \sqrt{c+d x} (5 a d+b c)}{12 a c^2 x^2}+\frac{\sqrt{a+b x} \sqrt{c+d x} (b c-a d) (5 a d+b c)}{8 a c^3 x}-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 a c x^3} \]

[Out]

((b*c - a*d)*(b*c + 5*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*a*c^3*x) + ((b*c + 5*a*d)*(a + b*x)^(3/2)*Sqrt[c +
d*x])/(12*a*c^2*x^2) - ((a + b*x)^(5/2)*Sqrt[c + d*x])/(3*a*c*x^3) + ((b*c - a*d)^2*(b*c + 5*a*d)*ArcTanh[(Sqr
t[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(8*a^(3/2)*c^(7/2))

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Rubi [A]  time = 0.0822984, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {96, 94, 93, 208} \[ \frac{(b c-a d)^2 (5 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{8 a^{3/2} c^{7/2}}+\frac{(a+b x)^{3/2} \sqrt{c+d x} (5 a d+b c)}{12 a c^2 x^2}+\frac{\sqrt{a+b x} \sqrt{c+d x} (b c-a d) (5 a d+b c)}{8 a c^3 x}-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 a c x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)*(b*c + 5*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*a*c^3*x) + ((b*c + 5*a*d)*(a + b*x)^(3/2)*Sqrt[c +
d*x])/(12*a*c^2*x^2) - ((a + b*x)^(5/2)*Sqrt[c + d*x])/(3*a*c*x^3) + ((b*c - a*d)^2*(b*c + 5*a*d)*ArcTanh[(Sqr
t[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(8*a^(3/2)*c^(7/2))

Rule 96

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

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^4 \sqrt{c+d x}} \, dx &=-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 a c x^3}-\frac{\left (\frac{b c}{2}+\frac{5 a d}{2}\right ) \int \frac{(a+b x)^{3/2}}{x^3 \sqrt{c+d x}} \, dx}{3 a c}\\ &=\frac{(b c+5 a d) (a+b x)^{3/2} \sqrt{c+d x}}{12 a c^2 x^2}-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 a c x^3}-\frac{((b c-a d) (b c+5 a d)) \int \frac{\sqrt{a+b x}}{x^2 \sqrt{c+d x}} \, dx}{8 a c^2}\\ &=\frac{(b c-a d) (b c+5 a d) \sqrt{a+b x} \sqrt{c+d x}}{8 a c^3 x}+\frac{(b c+5 a d) (a+b x)^{3/2} \sqrt{c+d x}}{12 a c^2 x^2}-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 a c x^3}-\frac{\left ((b c-a d)^2 (b c+5 a d)\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{16 a c^3}\\ &=\frac{(b c-a d) (b c+5 a d) \sqrt{a+b x} \sqrt{c+d x}}{8 a c^3 x}+\frac{(b c+5 a d) (a+b x)^{3/2} \sqrt{c+d x}}{12 a c^2 x^2}-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 a c x^3}-\frac{\left ((b c-a d)^2 (b c+5 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{8 a c^3}\\ &=\frac{(b c-a d) (b c+5 a d) \sqrt{a+b x} \sqrt{c+d x}}{8 a c^3 x}+\frac{(b c+5 a d) (a+b x)^{3/2} \sqrt{c+d x}}{12 a c^2 x^2}-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 a c x^3}+\frac{(b c-a d)^2 (b c+5 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{8 a^{3/2} c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.173926, size = 144, normalized size = 0.8 \[ \frac{\frac{x (5 a d+b c) \left (3 x^2 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\sqrt{a} \sqrt{c} \sqrt{a+b x} \sqrt{c+d x} (2 a c-3 a d x+5 b c x)\right )}{\sqrt{a} c^{5/2}}-8 (a+b x)^{5/2} \sqrt{c+d x}}{24 a c x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.021, size = 408, normalized size = 2.3 \begin{align*}{\frac{1}{48\,a{c}^{3}{x}^{3}}\sqrt{bx+a}\sqrt{dx+c} \left ( 15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}{a}^{3}{d}^{3}-27\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}{a}^{2}bc{d}^{2}+9\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}a{b}^{2}{c}^{2}d+3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}{b}^{3}{c}^{3}-30\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}{x}^{2}{a}^{2}{d}^{2}+44\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}{x}^{2}abcd-6\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}{x}^{2}{b}^{2}{c}^{2}+20\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}x{a}^{2}cd-28\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}xab{c}^{2}-16\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}{a}^{2}{c}^{2} \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^4/(d*x+c)^(1/2),x)

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a/c^3*(15*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^3
*a^3*d^3-27*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^3*a^2*b*c*d^2+9*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^3*a*b^2*c^2*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^3*b^3*c^3-30*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x^2*a^2*d^2+44*((b*x+a)*(d*x+c))^(1/2)*
(a*c)^(1/2)*x^2*a*b*c*d-6*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x^2*b^2*c^2+20*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/
2)*x*a^2*c*d-28*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x*a*b*c^2-16*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*c^2)/
((b*x+a)*(d*x+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((b*x+a)^(3/2)/x^4/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 9.88145, size = 977, normalized size = 5.43 \begin{align*} \left [\frac{3 \,{\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + 5 \, 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} - 22 \, a^{2} b c^{2} d + 15 \, a^{3} c d^{2}\right )} x^{2} + 2 \,{\left (7 \, a^{2} b c^{3} - 5 \, a^{3} c^{2} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{96 \, a^{2} c^{4} x^{3}}, -\frac{3 \,{\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + 5 \, 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} - 22 \, a^{2} b c^{2} d + 15 \, a^{3} c d^{2}\right )} x^{2} + 2 \,{\left (7 \, a^{2} b c^{3} - 5 \, a^{3} c^{2} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{48 \, a^{2} c^{4} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*(b^3*c^3 + 3*a*b^2*c^2*d - 9*a^2*b*c*d^2 + 5*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 - 22*a^2*b*c^2*d + 15*a^3*c*d^2)*x^2 + 2*(7*a^2*b*c^3 - 5*a^3*c^2*d)*x)*
sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^4*x^3), -1/48*(3*(b^3*c^3 + 3*a*b^2*c^2*d - 9*a^2*b*c*d^2 + 5*a^3*d^3)*sqr
t(-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 - 22*a^2*b*c^2*d + 15*a^3*c*d^2)*x^2 + 2*(7*a^2*b*c^3 -
5*a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^4*x^3)]

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

[Out]

Timed out

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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^4/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError