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

Optimal. Leaf size=142 \[ \frac{5 c^{3/2} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{7/2}}-\frac{5 (c+d x)^{3/2} (b c-a d)}{3 a^2 (a+b x)^{3/2}}-\frac{5 c \sqrt{c+d x} (b c-a d)}{a^3 \sqrt{a+b x}}-\frac{(c+d x)^{5/2}}{a x (a+b x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0562996, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {94, 93, 208} \[ \frac{5 c^{3/2} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{7/2}}-\frac{5 (c+d x)^{3/2} (b c-a d)}{3 a^2 (a+b x)^{3/2}}-\frac{5 c \sqrt{c+d x} (b c-a d)}{a^3 \sqrt{a+b x}}-\frac{(c+d x)^{5/2}}{a x (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.157774, size = 125, normalized size = 0.88 \[ -\frac{3 a^{5/2} (c+d x)^{5/2}+5 x (b c-a d) \left (\sqrt{a} \sqrt{c+d x} (4 a c+a d x+3 b c x)-3 c^{3/2} (a+b x)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )\right )}{3 a^{7/2} x (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.023, size = 502, normalized size = 3.5 \begin{align*} -{\frac{1}{6\,{a}^{3}x}\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{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}^{3}{b}^{3}{c}^{3}+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}b{c}^{2}d-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{b}^{2}{c}^{3}+15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) x{a}^{3}{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}b{c}^{3}-4\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}{x}^{2}{a}^{2}{d}^{2}-20\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}{x}^{2}abcd+30\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}{x}^{2}{b}^{2}{c}^{2}-28\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}x{a}^{2}cd+40\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}xab{c}^{2}+6\,\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 ) }}} \left ( bx+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(d*x+c)^(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*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^3*b^3*c^3+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*b*c^2*d-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*b^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*a^3*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*b*c^3-4*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x^2*a^2*
d^2-20*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x^2*a*b*c*d+30*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x^2*b^2*c^2-28*(
(b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x*a^2*c*d+40*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x*a*b*c^2+6*((b*x+a)*(d*x+
c))^(1/2)*(a*c)^(1/2)*a^2*c^2)/a^3/((b*x+a)*(d*x+c))^(1/2)/(a*c)^(1/2)/x/(b*x+a)^(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((d*x+c)^(5/2)/x^2/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 7.38676, size = 1083, normalized size = 7.63 \begin{align*} \left [-\frac{15 \,{\left ({\left (b^{3} c^{2} - a b^{2} c d\right )} x^{3} + 2 \,{\left (a b^{2} c^{2} - a^{2} b c d\right )} x^{2} +{\left (a^{2} b c^{2} - a^{3} c d\right )} x\right )} \sqrt{\frac{c}{a}} \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^{2} c +{\left (a b c + a^{2} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c} \sqrt{\frac{c}{a}} + 8 \,{\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \,{\left (3 \, a^{2} c^{2} +{\left (15 \, b^{2} c^{2} - 10 \, a b c d - 2 \, a^{2} d^{2}\right )} x^{2} + 2 \,{\left (10 \, a b c^{2} - 7 \, a^{2} c d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{12 \,{\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}, -\frac{15 \,{\left ({\left (b^{3} c^{2} - a b^{2} c d\right )} x^{3} + 2 \,{\left (a b^{2} c^{2} - a^{2} b c d\right )} x^{2} +{\left (a^{2} b c^{2} - a^{3} c d\right )} x\right )} \sqrt{-\frac{c}{a}} \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{c}{a}}}{2 \,{\left (b c d x^{2} + a c^{2} +{\left (b c^{2} + a c d\right )} x\right )}}\right ) + 2 \,{\left (3 \, a^{2} c^{2} +{\left (15 \, b^{2} c^{2} - 10 \, a b c d - 2 \, a^{2} d^{2}\right )} x^{2} + 2 \,{\left (10 \, a b c^{2} - 7 \, a^{2} c d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{6 \,{\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(15*((b^3*c^2 - a*b^2*c*d)*x^3 + 2*(a*b^2*c^2 - a^2*b*c*d)*x^2 + (a^2*b*c^2 - a^3*c*d)*x)*sqrt(c/a)*log
((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c
)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(3*a^2*c^2 + (15*b^2*c^2 - 10*a*b*c*d - 2*a^2*d^2)*x^2 + 2*(10
*a*b*c^2 - 7*a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x), -1/6*(15*((b^3*c^2
- a*b^2*c*d)*x^3 + 2*(a*b^2*c^2 - a^2*b*c*d)*x^2 + (a^2*b*c^2 - a^3*c*d)*x)*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*
c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) + 2*(3*a^2*c^2 + (
15*b^2*c^2 - 10*a*b*c*d - 2*a^2*d^2)*x^2 + 2*(10*a*b*c^2 - 7*a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*b^2
*x^3 + 2*a^4*b*x^2 + a^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((d*x+c)**(5/2)/x**2/(b*x+a)**(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((d*x+c)^(5/2)/x^2/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError