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

Optimal. Leaf size=154 \[ \frac{5 (c+d x)^{3/2} (b c-a d)}{4 a^2 x \sqrt{a+b x}}+\frac{15 \sqrt{c+d x} (b c-a d)^2}{4 a^3 \sqrt{a+b x}}-\frac{15 \sqrt{c} (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{7/2}}-\frac{(c+d x)^{5/2}}{2 a x^2 \sqrt{a+b x}} \]

[Out]

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

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Rubi [A]  time = 0.0741547, antiderivative size = 154, 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} \[ \frac{5 (c+d x)^{3/2} (b c-a d)}{4 a^2 x \sqrt{a+b x}}+\frac{15 \sqrt{c+d x} (b c-a d)^2}{4 a^3 \sqrt{a+b x}}-\frac{15 \sqrt{c} (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{7/2}}-\frac{(c+d x)^{5/2}}{2 a x^2 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(b*c - a*d)^2*Sqrt[c + d*x])/(4*a^3*Sqrt[a + b*x]) + (5*(b*c - a*d)*(c + d*x)^(3/2))/(4*a^2*x*Sqrt[a + b*x
]) - (c + d*x)^(5/2)/(2*a*x^2*Sqrt[a + b*x]) - (15*Sqrt[c]*(b*c - a*d)^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt
[a]*Sqrt[c + d*x])])/(4*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^3 (a+b x)^{3/2}} \, dx &=-\frac{(c+d x)^{5/2}}{2 a x^2 \sqrt{a+b x}}-\frac{(5 (b c-a d)) \int \frac{(c+d x)^{3/2}}{x^2 (a+b x)^{3/2}} \, dx}{4 a}\\ &=\frac{5 (b c-a d) (c+d x)^{3/2}}{4 a^2 x \sqrt{a+b x}}-\frac{(c+d x)^{5/2}}{2 a x^2 \sqrt{a+b x}}+\frac{\left (15 (b c-a d)^2\right ) \int \frac{\sqrt{c+d x}}{x (a+b x)^{3/2}} \, dx}{8 a^2}\\ &=\frac{15 (b c-a d)^2 \sqrt{c+d x}}{4 a^3 \sqrt{a+b x}}+\frac{5 (b c-a d) (c+d x)^{3/2}}{4 a^2 x \sqrt{a+b x}}-\frac{(c+d x)^{5/2}}{2 a x^2 \sqrt{a+b x}}+\frac{\left (15 c (b c-a d)^2\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 a^3}\\ &=\frac{15 (b c-a d)^2 \sqrt{c+d x}}{4 a^3 \sqrt{a+b x}}+\frac{5 (b c-a d) (c+d x)^{3/2}}{4 a^2 x \sqrt{a+b x}}-\frac{(c+d x)^{5/2}}{2 a x^2 \sqrt{a+b x}}+\frac{\left (15 c (b c-a d)^2\right ) \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{15 (b c-a d)^2 \sqrt{c+d x}}{4 a^3 \sqrt{a+b x}}+\frac{5 (b c-a d) (c+d x)^{3/2}}{4 a^2 x \sqrt{a+b x}}-\frac{(c+d x)^{5/2}}{2 a x^2 \sqrt{a+b x}}-\frac{15 \sqrt{c} (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0976674, size = 130, normalized size = 0.84 \[ \frac{\sqrt{c+d x} \left (a^2 \left (-2 c^2-9 c d x+8 d^2 x^2\right )+5 a b c x (c-5 d x)+15 b^2 c^2 x^2\right )}{4 a^3 x^2 \sqrt{a+b x}}-\frac{15 \sqrt{c} (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*(15*b^2*c^2*x^2 + 5*a*b*c*x*(c - 5*d*x) + a^2*(-2*c^2 - 9*c*d*x + 8*d^2*x^2)))/(4*a^3*x^2*Sqrt[
a + b*x]) - (15*Sqrt[c]*(b*c - a*d)^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(7/2))

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Maple [B]  time = 0.023, size = 507, normalized size = 3.3 \begin{align*} -{\frac{1}{8\,{a}^{3}{x}^{2}}\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}^{2}bc{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}^{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}+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}^{3}c{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}b{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{b}^{2}{c}^{3}-16\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}{x}^{2}{a}^{2}{d}^{2}+50\,\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}+18\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}x{a}^{2}cd-10\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}xab{c}^{2}+4\,\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 ) }}}{\frac{1}{\sqrt{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(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^2*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^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+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^3*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*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^2*a*b^2*c^3-16*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x
^2*a^2*d^2+50*((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+18*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x*a^2*c*d-10*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x*a*b*c^2+4*((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^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)^(5/2)/x^3/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 6.18424, size = 1029, normalized size = 6.68 \begin{align*} \left [\frac{15 \,{\left ({\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{3} +{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2}\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 (2 \, a^{2} c^{2} -{\left (15 \, b^{2} c^{2} - 25 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{2} -{\left (5 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{16 \,{\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}, \frac{15 \,{\left ({\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{3} +{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2}\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 (2 \, a^{2} c^{2} -{\left (15 \, b^{2} c^{2} - 25 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{2} -{\left (5 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{8 \,{\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(15*((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*x^3 + (a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x^2)*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)*s
qrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*a^2*c^2 - (15*b^2*c^2 - 25*a*b*c*d + 8*a^2*d^2)*x^2 - (5*a*b*c
^2 - 9*a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*b*x^3 + a^4*x^2), 1/8*(15*((b^3*c^2 - 2*a*b^2*c*d + a^2*b
*d^2)*x^3 + (a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x^2)*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*(2*a^2*c^2 - (15*b^2*c^2 - 25*a*b*c*d
+ 8*a^2*d^2)*x^2 - (5*a*b*c^2 - 9*a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*b*x^3 + a^4*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)**(5/2)/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)^(5/2)/x^3/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError