3.645 \(\int \frac{(a+b x)^{3/2}}{x^2 (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=149 \[ \frac{(a+b x)^{3/2} (3 b c-5 a d)}{3 a c^2 (c+d x)^{3/2}}+\frac{\sqrt{a+b x} (3 b c-5 a d)}{c^3 \sqrt{c+d x}}-\frac{\sqrt{a} (3 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{7/2}}-\frac{(a+b x)^{5/2}}{a c x (c+d x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0638717, antiderivative size = 149, 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} \[ \frac{(a+b x)^{3/2} (3 b c-5 a d)}{3 a c^2 (c+d x)^{3/2}}+\frac{\sqrt{a+b x} (3 b c-5 a d)}{c^3 \sqrt{c+d x}}-\frac{\sqrt{a} (3 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{7/2}}-\frac{(a+b x)^{5/2}}{a c x (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)/(x^2*(c + d*x)^(5/2)),x]

[Out]

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

Mathematica [A]  time = 0.165192, size = 128, normalized size = 0.86 \[ \frac{x (3 b c-5 a d) \left (\sqrt{c} \sqrt{a+b x} (4 a c+3 a d x+b c x)-3 a^{3/2} (c+d x)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )\right )-3 c^{5/2} (a+b x)^{5/2}}{3 a c^{7/2} x (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)/(x^2*(c + d*x)^(5/2)),x]

[Out]

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

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Maple [B]  time = 0.024, size = 459, normalized size = 3.1 \begin{align*}{\frac{1}{6\,{c}^{3}x}\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}-9\,\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}+30\,\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+15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) x{a}^{2}{c}^{2}d-9\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) xab{c}^{3}-30\,{x}^{2}a{d}^{2}\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+8\,{x}^{2}bcd\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }-40\,xacd\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+12\,xb{c}^{2}\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }-6\,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 ) }}} \left ( dx+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)/x^2/(d*x+c)^(5/2),x)

[Out]

1/6*(b*x+a)^(1/2)/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^2*d^3-9*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*c*d^2+30*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+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*a^2*c^2*d-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*a*b*c^3-30*x^2*a*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+8*
x^2*b*c*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-40*x*a*c*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+12*x*b*c^2*(a*c)^
(1/2)*((b*x+a)*(d*x+c))^(1/2)-6*a*c^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x/(a*c)^(1/2)/((b*x+a)*(d*x+c))^(1/
2)/(d*x+c)^(3/2)

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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^2/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 8.7684, size = 1017, normalized size = 6.83 \begin{align*} \left [-\frac{3 \,{\left ({\left (3 \, b c d^{2} - 5 \, a d^{3}\right )} x^{3} + 2 \,{\left (3 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2} +{\left (3 \, b c^{3} - 5 \, a c^{2} d\right )} x\right )} \sqrt{\frac{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^{2} +{\left (b c^{2} + a c d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c} \sqrt{\frac{a}{c}} + 8 \,{\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \,{\left (3 \, a c^{2} -{\left (4 \, b c d - 15 \, a d^{2}\right )} x^{2} - 2 \,{\left (3 \, b c^{2} - 10 \, a c d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{12 \,{\left (c^{3} d^{2} x^{3} + 2 \, c^{4} d x^{2} + c^{5} x\right )}}, \frac{3 \,{\left ({\left (3 \, b c d^{2} - 5 \, a d^{3}\right )} x^{3} + 2 \,{\left (3 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2} +{\left (3 \, b c^{3} - 5 \, a c^{2} d\right )} x\right )} \sqrt{-\frac{a}{c}} \arctan \left (\frac{{\left (2 \, a c +{\left (b c + a d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c} \sqrt{-\frac{a}{c}}}{2 \,{\left (a b d x^{2} + a^{2} c +{\left (a b c + a^{2} d\right )} x\right )}}\right ) - 2 \,{\left (3 \, a c^{2} -{\left (4 \, b c d - 15 \, a d^{2}\right )} x^{2} - 2 \,{\left (3 \, b c^{2} - 10 \, a c d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{6 \,{\left (c^{3} d^{2} x^{3} + 2 \, c^{4} d x^{2} + c^{5} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^2/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*((3*b*c*d^2 - 5*a*d^3)*x^3 + 2*(3*b*c^2*d - 5*a*c*d^2)*x^2 + (3*b*c^3 - 5*a*c^2*d)*x)*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^2 + (b*c^2 + a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)
*sqrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(3*a*c^2 - (4*b*c*d - 15*a*d^2)*x^2 - 2*(3*b*c^2 - 10*a*c*d)*x)
*sqrt(b*x + a)*sqrt(d*x + c))/(c^3*d^2*x^3 + 2*c^4*d*x^2 + c^5*x), 1/6*(3*((3*b*c*d^2 - 5*a*d^3)*x^3 + 2*(3*b*
c^2*d - 5*a*c*d^2)*x^2 + (3*b*c^3 - 5*a*c^2*d)*x)*sqrt(-a/c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*
sqrt(d*x + c)*sqrt(-a/c)/(a*b*d*x^2 + a^2*c + (a*b*c + a^2*d)*x)) - 2*(3*a*c^2 - (4*b*c*d - 15*a*d^2)*x^2 - 2*
(3*b*c^2 - 10*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^3*d^2*x^3 + 2*c^4*d*x^2 + c^5*x)]

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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**2/(d*x+c)**(5/2),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^2/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError