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

Optimal. Leaf size=178 \[ \frac{(c+d x)^{5/2} (2 b c-7 a d)}{5 b^2 (b c-a d)}+\frac{(c+d x)^{3/2} (2 b c-7 a d)}{3 b^3}+\frac{\sqrt{c+d x} (2 b c-7 a d) (b c-a d)}{b^4}-\frac{(2 b c-7 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{9/2}}+\frac{a (c+d x)^{7/2}}{b (a+b x) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.101726, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {78, 50, 63, 208} \[ \frac{(c+d x)^{5/2} (2 b c-7 a d)}{5 b^2 (b c-a d)}+\frac{(c+d x)^{3/2} (2 b c-7 a d)}{3 b^3}+\frac{\sqrt{c+d x} (2 b c-7 a d) (b c-a d)}{b^4}-\frac{(2 b c-7 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{9/2}}+\frac{a (c+d x)^{7/2}}{b (a+b x) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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{x (c+d x)^{5/2}}{(a+b x)^2} \, dx &=\frac{a (c+d x)^{7/2}}{b (b c-a d) (a+b x)}+\frac{(2 b c-7 a d) \int \frac{(c+d x)^{5/2}}{a+b x} \, dx}{2 b (b c-a d)}\\ &=\frac{(2 b c-7 a d) (c+d x)^{5/2}}{5 b^2 (b c-a d)}+\frac{a (c+d x)^{7/2}}{b (b c-a d) (a+b x)}+\frac{(2 b c-7 a d) \int \frac{(c+d x)^{3/2}}{a+b x} \, dx}{2 b^2}\\ &=\frac{(2 b c-7 a d) (c+d x)^{3/2}}{3 b^3}+\frac{(2 b c-7 a d) (c+d x)^{5/2}}{5 b^2 (b c-a d)}+\frac{a (c+d x)^{7/2}}{b (b c-a d) (a+b x)}+\frac{((2 b c-7 a d) (b c-a d)) \int \frac{\sqrt{c+d x}}{a+b x} \, dx}{2 b^3}\\ &=\frac{(2 b c-7 a d) (b c-a d) \sqrt{c+d x}}{b^4}+\frac{(2 b c-7 a d) (c+d x)^{3/2}}{3 b^3}+\frac{(2 b c-7 a d) (c+d x)^{5/2}}{5 b^2 (b c-a d)}+\frac{a (c+d x)^{7/2}}{b (b c-a d) (a+b x)}+\frac{\left ((2 b c-7 a d) (b c-a d)^2\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{2 b^4}\\ &=\frac{(2 b c-7 a d) (b c-a d) \sqrt{c+d x}}{b^4}+\frac{(2 b c-7 a d) (c+d x)^{3/2}}{3 b^3}+\frac{(2 b c-7 a d) (c+d x)^{5/2}}{5 b^2 (b c-a d)}+\frac{a (c+d x)^{7/2}}{b (b c-a d) (a+b x)}+\frac{\left ((2 b c-7 a d) (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{b^4 d}\\ &=\frac{(2 b c-7 a d) (b c-a d) \sqrt{c+d x}}{b^4}+\frac{(2 b c-7 a d) (c+d x)^{3/2}}{3 b^3}+\frac{(2 b c-7 a d) (c+d x)^{5/2}}{5 b^2 (b c-a d)}+\frac{a (c+d x)^{7/2}}{b (b c-a d) (a+b x)}-\frac{(2 b c-7 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.281862, size = 150, normalized size = 0.84 \[ \frac{\frac{2 \left (b c-\frac{7 a d}{2}\right ) \left (5 (b c-a d) \left (\sqrt{b} \sqrt{c+d x} (-3 a d+4 b c+b d x)-3 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )\right )+3 b^{5/2} (c+d x)^{5/2}\right )}{15 b^{7/2}}+\frac{a (c+d x)^{7/2}}{a+b x}}{b (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(c + d*x)^(7/2))/(a + b*x) + (2*(b*c - (7*a*d)/2)*(3*b^(5/2)*(c + d*x)^(5/2) + 5*(b*c - a*d)*(Sqrt[b]*Sqrt
[c + d*x]*(4*b*c - 3*a*d + b*d*x) - 3*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])))/(1
5*b^(7/2)))/(b*(b*c - a*d))

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Maple [B]  time = 0.016, size = 348, normalized size = 2. \begin{align*}{\frac{2}{5\,{b}^{2}} \left ( dx+c \right ) ^{{\frac{5}{2}}}}-{\frac{4\,ad}{3\,{b}^{3}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+{\frac{2\,c}{3\,{b}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+6\,{\frac{{a}^{2}{d}^{2}\sqrt{dx+c}}{{b}^{4}}}-8\,{\frac{acd\sqrt{dx+c}}{{b}^{3}}}+2\,{\frac{{c}^{2}\sqrt{dx+c}}{{b}^{2}}}+{\frac{{a}^{3}{d}^{3}}{{b}^{4} \left ( bdx+ad \right ) }\sqrt{dx+c}}-2\,{\frac{\sqrt{dx+c}{a}^{2}c{d}^{2}}{{b}^{3} \left ( bdx+ad \right ) }}+{\frac{a{c}^{2}d}{{b}^{2} \left ( bdx+ad \right ) }\sqrt{dx+c}}-7\,{\frac{{a}^{3}{d}^{3}}{{b}^{4}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+16\,{\frac{c{a}^{2}{d}^{2}}{{b}^{3}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-11\,{\frac{a{c}^{2}d}{{b}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+2\,{\frac{{c}^{3}}{b\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5/b^2*(d*x+c)^(5/2)-4/3/b^3*(d*x+c)^(3/2)*a*d+2/3/b^2*(d*x+c)^(3/2)*c+6/b^4*a^2*d^2*(d*x+c)^(1/2)-8/b^3*a*c*
d*(d*x+c)^(1/2)+2/b^2*c^2*(d*x+c)^(1/2)+1/b^4*(d*x+c)^(1/2)/(b*d*x+a*d)*a^3*d^3-2/b^3*(d*x+c)^(1/2)/(b*d*x+a*d
)*a^2*c*d^2+1/b^2*(d*x+c)^(1/2)/(b*d*x+a*d)*a*c^2*d-7/b^4/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c
)*b)^(1/2))*a^3*d^3+16/b^3/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*a^2*c*d^2-11/b^2/((
a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*a*c^2*d+2/b/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)
^(1/2)/((a*d-b*c)*b)^(1/2))*c^3

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.83125, size = 981, normalized size = 5.51 \begin{align*} \left [\frac{15 \,{\left (2 \, a b^{2} c^{2} - 9 \, a^{2} b c d + 7 \, a^{3} d^{2} +{\left (2 \, b^{3} c^{2} - 9 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x\right )} \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b d x + 2 \, b c - a d - 2 \, \sqrt{d x + c} b \sqrt{\frac{b c - a d}{b}}}{b x + a}\right ) + 2 \,{\left (6 \, b^{3} d^{2} x^{3} + 61 \, a b^{2} c^{2} - 170 \, a^{2} b c d + 105 \, a^{3} d^{2} + 2 \,{\left (11 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{2} + 2 \,{\left (23 \, b^{3} c^{2} - 59 \, a b^{2} c d + 35 \, a^{2} b d^{2}\right )} x\right )} \sqrt{d x + c}}{30 \,{\left (b^{5} x + a b^{4}\right )}}, -\frac{15 \,{\left (2 \, a b^{2} c^{2} - 9 \, a^{2} b c d + 7 \, a^{3} d^{2} +{\left (2 \, b^{3} c^{2} - 9 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x\right )} \sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{\sqrt{d x + c} b \sqrt{-\frac{b c - a d}{b}}}{b c - a d}\right ) -{\left (6 \, b^{3} d^{2} x^{3} + 61 \, a b^{2} c^{2} - 170 \, a^{2} b c d + 105 \, a^{3} d^{2} + 2 \,{\left (11 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{2} + 2 \,{\left (23 \, b^{3} c^{2} - 59 \, a b^{2} c d + 35 \, a^{2} b d^{2}\right )} x\right )} \sqrt{d x + c}}{15 \,{\left (b^{5} x + a b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(15*(2*a*b^2*c^2 - 9*a^2*b*c*d + 7*a^3*d^2 + (2*b^3*c^2 - 9*a*b^2*c*d + 7*a^2*b*d^2)*x)*sqrt((b*c - a*d)
/b)*log((b*d*x + 2*b*c - a*d - 2*sqrt(d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) + 2*(6*b^3*d^2*x^3 + 61*a*b^2
*c^2 - 170*a^2*b*c*d + 105*a^3*d^2 + 2*(11*b^3*c*d - 7*a*b^2*d^2)*x^2 + 2*(23*b^3*c^2 - 59*a*b^2*c*d + 35*a^2*
b*d^2)*x)*sqrt(d*x + c))/(b^5*x + a*b^4), -1/15*(15*(2*a*b^2*c^2 - 9*a^2*b*c*d + 7*a^3*d^2 + (2*b^3*c^2 - 9*a*
b^2*c*d + 7*a^2*b*d^2)*x)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (6*
b^3*d^2*x^3 + 61*a*b^2*c^2 - 170*a^2*b*c*d + 105*a^3*d^2 + 2*(11*b^3*c*d - 7*a*b^2*d^2)*x^2 + 2*(23*b^3*c^2 -
59*a*b^2*c*d + 35*a^2*b*d^2)*x)*sqrt(d*x + c))/(b^5*x + a*b^4)]

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

[Out]

Timed out

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Giac [A]  time = 1.20341, size = 324, normalized size = 1.82 \begin{align*} \frac{{\left (2 \, b^{3} c^{3} - 11 \, a b^{2} c^{2} d + 16 \, a^{2} b c d^{2} - 7 \, a^{3} d^{3}\right )} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} b^{4}} + \frac{\sqrt{d x + c} a b^{2} c^{2} d - 2 \, \sqrt{d x + c} a^{2} b c d^{2} + \sqrt{d x + c} a^{3} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b^{4}} + \frac{2 \,{\left (3 \,{\left (d x + c\right )}^{\frac{5}{2}} b^{8} + 5 \,{\left (d x + c\right )}^{\frac{3}{2}} b^{8} c + 15 \, \sqrt{d x + c} b^{8} c^{2} - 10 \,{\left (d x + c\right )}^{\frac{3}{2}} a b^{7} d - 60 \, \sqrt{d x + c} a b^{7} c d + 45 \, \sqrt{d x + c} a^{2} b^{6} d^{2}\right )}}{15 \, b^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b^3*c^3 - 11*a*b^2*c^2*d + 16*a^2*b*c*d^2 - 7*a^3*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-
b^2*c + a*b*d)*b^4) + (sqrt(d*x + c)*a*b^2*c^2*d - 2*sqrt(d*x + c)*a^2*b*c*d^2 + sqrt(d*x + c)*a^3*d^3)/(((d*x
+ c)*b - b*c + a*d)*b^4) + 2/15*(3*(d*x + c)^(5/2)*b^8 + 5*(d*x + c)^(3/2)*b^8*c + 15*sqrt(d*x + c)*b^8*c^2 -
10*(d*x + c)^(3/2)*a*b^7*d - 60*sqrt(d*x + c)*a*b^7*c*d + 45*sqrt(d*x + c)*a^2*b^6*d^2)/b^10