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

Optimal. Leaf size=175 \[ -\frac{(a+b x)^{3/2} (b c-5 a d)}{4 a c^2 x \sqrt{c+d x}}+\frac{3 \sqrt{a+b x} (b c-5 a d) (b c-a d)}{4 a c^3 \sqrt{c+d x}}-\frac{3 (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 \sqrt{a} c^{7/2}}-\frac{(a+b x)^{5/2}}{2 a c x^2 \sqrt{c+d x}} \]

[Out]

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

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Rubi [A]  time = 0.0770323, antiderivative size = 175, 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} \[ -\frac{(a+b x)^{3/2} (b c-5 a d)}{4 a c^2 x \sqrt{c+d x}}+\frac{3 \sqrt{a+b x} (b c-5 a d) (b c-a d)}{4 a c^3 \sqrt{c+d x}}-\frac{3 (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 \sqrt{a} c^{7/2}}-\frac{(a+b x)^{5/2}}{2 a c x^2 \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0937708, size = 130, normalized size = 0.74 \[ \frac{\sqrt{a+b x} \left (a \left (-2 c^2+5 c d x+15 d^2 x^2\right )-b c x (5 c+13 d x)\right )}{4 c^3 x^2 \sqrt{c+d x}}-\frac{3 \left (5 a^2 d^2-6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 \sqrt{a} c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.023, size = 464, normalized size = 2.7 \begin{align*} -{\frac{1}{8\,{c}^{3}{x}^{2}}\sqrt{bx+a} \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}^{2}{d}^{3}-18\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}abc{d}^{2}+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}^{2}{c}^{2}d+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}^{2}c{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}ab{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}^{2}{b}^{2}{c}^{3}-30\,{x}^{2}a{d}^{2}\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+26\,{x}^{2}bcd\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }-10\,xacd\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+10\,xb{c}^{2}\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+4\,a{c}^{2}\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{dx+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(b*x+a)^(1/2)*(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^2*d^3-18*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*c*d^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*b^2*c^2*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^2*c*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*b*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^2*b^2*c^3-30*x^2*a*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2
6*x^2*b*c*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-10*x*a*c*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+10*x*b*c^2*(a*c
)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+4*a*c^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/c^3/((b*x+a)*(d*x+c))^(1/2)/(a*c)
^(1/2)/x^2/(d*x+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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 7.99146, size = 1029, normalized size = 5.88 \begin{align*} \left [\frac{3 \,{\left ({\left (b^{2} c^{2} d - 6 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} x^{3} +{\left (b^{2} c^{3} - 6 \, a b c^{2} d + 5 \, a^{2} c 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^{2} c^{3} +{\left (13 \, a b c^{2} d - 15 \, a^{2} c d^{2}\right )} x^{2} + 5 \,{\left (a b c^{3} - a^{2} c^{2} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{16 \,{\left (a c^{4} d x^{3} + a c^{5} x^{2}\right )}}, \frac{3 \,{\left ({\left (b^{2} c^{2} d - 6 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} x^{3} +{\left (b^{2} c^{3} - 6 \, a b c^{2} d + 5 \, a^{2} c 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^{2} c^{3} +{\left (13 \, a b c^{2} d - 15 \, a^{2} c d^{2}\right )} x^{2} + 5 \,{\left (a b c^{3} - a^{2} c^{2} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{8 \,{\left (a c^{4} d x^{3} + a c^{5} x^{2}\right )}}\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)^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: TypeError