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

Optimal. Leaf size=178 \[ \frac{(c+d x)^{3/2} (5 b c-a d)}{4 a^2 c x \sqrt{a+b x}}+\frac{3 \sqrt{c+d x} (b c-a d) (5 b c-a d)}{4 a^3 c \sqrt{a+b x}}-\frac{3 (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{7/2} \sqrt{c}}-\frac{(c+d x)^{5/2}}{2 a c x^2 \sqrt{a+b x}} \]

[Out]

(3*(b*c - a*d)*(5*b*c - a*d)*Sqrt[c + d*x])/(4*a^3*c*Sqrt[a + b*x]) + ((5*b*c - a*d)*(c + d*x)^(3/2))/(4*a^2*c
*x*Sqrt[a + b*x]) - (c + d*x)^(5/2)/(2*a*c*x^2*Sqrt[a + b*x]) - (3*(b*c - a*d)*(5*b*c - a*d)*ArcTanh[(Sqrt[c]*
Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(7/2)*Sqrt[c])

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Rubi [A]  time = 0.0763655, antiderivative size = 178, 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} \[ \frac{(c+d x)^{3/2} (5 b c-a d)}{4 a^2 c x \sqrt{a+b x}}+\frac{3 \sqrt{c+d x} (b c-a d) (5 b c-a d)}{4 a^3 c \sqrt{a+b x}}-\frac{3 (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{7/2} \sqrt{c}}-\frac{(c+d x)^{5/2}}{2 a c x^2 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*c - a*d)*(5*b*c - a*d)*Sqrt[c + d*x])/(4*a^3*c*Sqrt[a + b*x]) + ((5*b*c - a*d)*(c + d*x)^(3/2))/(4*a^2*c
*x*Sqrt[a + b*x]) - (c + d*x)^(5/2)/(2*a*c*x^2*Sqrt[a + b*x]) - (3*(b*c - a*d)*(5*b*c - a*d)*ArcTanh[(Sqrt[c]*
Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(7/2)*Sqrt[c])

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

Mathematica [A]  time = 0.0968543, size = 130, normalized size = 0.73 \[ \frac{\sqrt{c+d x} \left (a^2 (-(2 c+5 d x))+a b x (5 c-13 d x)+15 b^2 c x^2\right )}{4 a^3 x^2 \sqrt{a+b x}}-\frac{3 \left (a^2 d^2-6 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{7/2} \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*(15*b^2*c*x^2 + a*b*x*(5*c - 13*d*x) - a^2*(2*c + 5*d*x)))/(4*a^3*x^2*Sqrt[a + b*x]) - (3*(5*b^
2*c^2 - 6*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(7/2)*Sqrt[c])

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Maple [B]  time = 0.022, size = 464, normalized size = 2.6 \begin{align*} -{\frac{1}{8\,{a}^{3}{x}^{2}}\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}^{3}{a}^{2}b{d}^{2}-18\,\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}cd+15\,\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}^{2}+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}^{3}{d}^{2}-18\,\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}bcd+15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}a{b}^{2}{c}^{2}+26\,{x}^{2}abd\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }-30\,{x}^{2}{b}^{2}c\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+10\,x{a}^{2}d\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }-10\,xabc\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+4\,{a}^{2}c\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) } \right ){\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(d*x+c)^(1/2)*(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^3*a^2*b*d^2-18*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*d+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*b^3*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^3*d^2-18*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*b*c*d+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^2*a*b^2*c^2+26*x^2*a*b*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-3
0*x^2*b^2*c*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+10*x*a^2*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-10*x*a*b*c*(a*c
)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+4*a^2*c*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/a^3/((b*x+a)*(d*x+c))^(1/2)/(a*c)
^(1/2)/x^2/(b*x+a)^(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)/x^3/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 6.58829, size = 1029, normalized size = 5.78 \begin{align*} \left [\frac{3 \,{\left ({\left (5 \, b^{3} c^{2} - 6 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{3} +{\left (5 \, a b^{2} c^{2} - 6 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2}\right )} \sqrt{a c} \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^{3} c^{2} -{\left (15 \, a b^{2} c^{2} - 13 \, a^{2} b c d\right )} x^{2} - 5 \,{\left (a^{2} b c^{2} - a^{3} c d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{16 \,{\left (a^{4} b c x^{3} + a^{5} c x^{2}\right )}}, \frac{3 \,{\left ({\left (5 \, b^{3} c^{2} - 6 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{3} +{\left (5 \, a b^{2} c^{2} - 6 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2}\right )} \sqrt{-a c} \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^{3} c^{2} -{\left (15 \, a b^{2} c^{2} - 13 \, a^{2} b c d\right )} x^{2} - 5 \,{\left (a^{2} b c^{2} - a^{3} c d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{8 \,{\left (a^{4} b c x^{3} + a^{5} c x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*((5*b^3*c^2 - 6*a*b^2*c*d + a^2*b*d^2)*x^3 + (5*a*b^2*c^2 - 6*a^2*b*c*d + a^3*d^2)*x^2)*sqrt(a*c)*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^3*c^2 - (15*a*b^2*c^2 - 13*a^2*b*c*d)*x^2 - 5*(a^2*b*c^2 - a^3*
c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^4*b*c*x^3 + a^5*c*x^2), 1/8*(3*((5*b^3*c^2 - 6*a*b^2*c*d + a^2*b*d^2)*
x^3 + (5*a*b^2*c^2 - 6*a^2*b*c*d + a^3*d^2)*x^2)*sqrt(-a*c)*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^3*c^2 - (15*a*b^2*c^2 - 13*a
^2*b*c*d)*x^2 - 5*(a^2*b*c^2 - a^3*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^4*b*c*x^3 + a^5*c*x^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((d*x+c)**(3/2)/x**3/(b*x+a)**(3/2),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((d*x+c)^(3/2)/x^3/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError