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

Optimal. Leaf size=135 \[ -\frac{2 a (c+d x)^{5/2}}{5 b^2}-\frac{2 a (c+d x)^{3/2} (b c-a d)}{3 b^3}-\frac{2 a \sqrt{c+d x} (b c-a d)^2}{b^4}+\frac{2 a (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{9/2}}+\frac{2 (c+d x)^{7/2}}{7 b d} \]

[Out]

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

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Rubi [A]  time = 0.0746933, antiderivative size = 135, 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 = {80, 50, 63, 208} \[ -\frac{2 a (c+d x)^{5/2}}{5 b^2}-\frac{2 a (c+d x)^{3/2} (b c-a d)}{3 b^3}-\frac{2 a \sqrt{c+d x} (b c-a d)^2}{b^4}+\frac{2 a (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{9/2}}+\frac{2 (c+d x)^{7/2}}{7 b d} \]

Antiderivative was successfully verified.

[In]

Int[(x*(c + d*x)^(5/2))/(a + b*x),x]

[Out]

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

Rule 80

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

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} \, dx &=\frac{2 (c+d x)^{7/2}}{7 b d}-\frac{a \int \frac{(c+d x)^{5/2}}{a+b x} \, dx}{b}\\ &=-\frac{2 a (c+d x)^{5/2}}{5 b^2}+\frac{2 (c+d x)^{7/2}}{7 b d}-\frac{(a (b c-a d)) \int \frac{(c+d x)^{3/2}}{a+b x} \, dx}{b^2}\\ &=-\frac{2 a (b c-a d) (c+d x)^{3/2}}{3 b^3}-\frac{2 a (c+d x)^{5/2}}{5 b^2}+\frac{2 (c+d x)^{7/2}}{7 b d}-\frac{\left (a (b c-a d)^2\right ) \int \frac{\sqrt{c+d x}}{a+b x} \, dx}{b^3}\\ &=-\frac{2 a (b c-a d)^2 \sqrt{c+d x}}{b^4}-\frac{2 a (b c-a d) (c+d x)^{3/2}}{3 b^3}-\frac{2 a (c+d x)^{5/2}}{5 b^2}+\frac{2 (c+d x)^{7/2}}{7 b d}-\frac{\left (a (b c-a d)^3\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{b^4}\\ &=-\frac{2 a (b c-a d)^2 \sqrt{c+d x}}{b^4}-\frac{2 a (b c-a d) (c+d x)^{3/2}}{3 b^3}-\frac{2 a (c+d x)^{5/2}}{5 b^2}+\frac{2 (c+d x)^{7/2}}{7 b d}-\frac{\left (2 a (b c-a d)^3\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 a (b c-a d)^2 \sqrt{c+d x}}{b^4}-\frac{2 a (b c-a d) (c+d x)^{3/2}}{3 b^3}-\frac{2 a (c+d x)^{5/2}}{5 b^2}+\frac{2 (c+d x)^{7/2}}{7 b d}+\frac{2 a (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.23325, size = 126, normalized size = 0.93 \[ -\frac{2 a (c+d x)^{5/2}}{5 b^2}-\frac{2 a (b c-a d) \left (\sqrt{b} \sqrt{c+d x} (-3 a d+4 b c+b d x)-3 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )\right )}{3 b^{9/2}}+\frac{2 (c+d x)^{7/2}}{7 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.009, size = 291, normalized size = 2.2 \begin{align*}{\frac{2}{7\,bd} \left ( dx+c \right ) ^{{\frac{7}{2}}}}-{\frac{2\,a}{5\,{b}^{2}} \left ( dx+c \right ) ^{{\frac{5}{2}}}}+{\frac{2\,d{a}^{2}}{3\,{b}^{3}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-{\frac{2\,ac}{3\,{b}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-2\,{\frac{{d}^{2}{a}^{3}\sqrt{dx+c}}{{b}^{4}}}+4\,{\frac{d{a}^{2}c\sqrt{dx+c}}{{b}^{3}}}-2\,{\frac{a{c}^{2}\sqrt{dx+c}}{{b}^{2}}}+2\,{\frac{{d}^{3}{a}^{4}}{{b}^{4}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-6\,{\frac{{d}^{2}{a}^{3}c}{{b}^{3}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+6\,{\frac{d{a}^{2}{c}^{2}}{{b}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-2\,{\frac{a{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),x)

[Out]

2/7*(d*x+c)^(7/2)/b/d-2/5*a*(d*x+c)^(5/2)/b^2+2/3*d/b^3*(d*x+c)^(3/2)*a^2-2/3/b^2*(d*x+c)^(3/2)*a*c-2*d^2/b^4*
a^3*(d*x+c)^(1/2)+4*d/b^3*a^2*c*(d*x+c)^(1/2)-2/b^2*a*c^2*(d*x+c)^(1/2)+2*d^3*a^4/b^4/((a*d-b*c)*b)^(1/2)*arct
an(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))-6*d^2*a^3/b^3/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)
^(1/2))*c+6*d*a^2/b^2/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*c^2-2*a/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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.59758, size = 907, normalized size = 6.72 \begin{align*} \left [\frac{105 \,{\left (a b^{2} c^{2} d - 2 \, a^{2} b c d^{2} + a^{3} d^{3}\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 (15 \, b^{3} d^{3} x^{3} + 15 \, b^{3} c^{3} - 161 \, a b^{2} c^{2} d + 245 \, a^{2} b c d^{2} - 105 \, a^{3} d^{3} + 3 \,{\left (15 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{2} +{\left (45 \, b^{3} c^{2} d - 77 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x\right )} \sqrt{d x + c}}{105 \, b^{4} d}, \frac{2 \,{\left (105 \,{\left (a b^{2} c^{2} d - 2 \, a^{2} b c d^{2} + a^{3} d^{3}\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 (15 \, b^{3} d^{3} x^{3} + 15 \, b^{3} c^{3} - 161 \, a b^{2} c^{2} d + 245 \, a^{2} b c d^{2} - 105 \, a^{3} d^{3} + 3 \,{\left (15 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{2} +{\left (45 \, b^{3} c^{2} d - 77 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x\right )} \sqrt{d x + c}\right )}}{105 \, b^{4} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/105*(105*(a*b^2*c^2*d - 2*a^2*b*c*d^2 + a^3*d^3)*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*(15*b^3*d^3*x^3 + 15*b^3*c^3 - 161*a*b^2*c^2*d + 245*a^2*b*c*d^2 -
105*a^3*d^3 + 3*(15*b^3*c*d^2 - 7*a*b^2*d^3)*x^2 + (45*b^3*c^2*d - 77*a*b^2*c*d^2 + 35*a^2*b*d^3)*x)*sqrt(d*x
+ c))/(b^4*d), 2/105*(105*(a*b^2*c^2*d - 2*a^2*b*c*d^2 + a^3*d^3)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b
*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + (15*b^3*d^3*x^3 + 15*b^3*c^3 - 161*a*b^2*c^2*d + 245*a^2*b*c*d^2 - 105*a^
3*d^3 + 3*(15*b^3*c*d^2 - 7*a*b^2*d^3)*x^2 + (45*b^3*c^2*d - 77*a*b^2*c*d^2 + 35*a^2*b*d^3)*x)*sqrt(d*x + c))/
(b^4*d)]

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Sympy [A]  time = 41.9367, size = 148, normalized size = 1.1 \begin{align*} - \frac{2 a \left (c + d x\right )^{\frac{5}{2}}}{5 b^{2}} + \frac{2 a \left (a d - b c\right )^{3} \operatorname{atan}{\left (\frac{\sqrt{c + d x}}{\sqrt{\frac{a d - b c}{b}}} \right )}}{b^{5} \sqrt{\frac{a d - b c}{b}}} + \frac{2 \left (c + d x\right )^{\frac{7}{2}}}{7 b d} + \frac{\left (c + d x\right )^{\frac{3}{2}} \left (2 a^{2} d - 2 a b c\right )}{3 b^{3}} + \frac{\sqrt{c + d x} \left (- 2 a^{3} d^{2} + 4 a^{2} b c d - 2 a b^{2} c^{2}\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*(c + d*x)**(5/2)/(5*b**2) + 2*a*(a*d - b*c)**3*atan(sqrt(c + d*x)/sqrt((a*d - b*c)/b))/(b**5*sqrt((a*d -
b*c)/b)) + 2*(c + d*x)**(7/2)/(7*b*d) + (c + d*x)**(3/2)*(2*a**2*d - 2*a*b*c)/(3*b**3) + sqrt(c + d*x)*(-2*a**
3*d**2 + 4*a**2*b*c*d - 2*a*b**2*c**2)/b**4

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Giac [A]  time = 1.25262, size = 286, normalized size = 2.12 \begin{align*} -\frac{2 \,{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} 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{2 \,{\left (15 \,{\left (d x + c\right )}^{\frac{7}{2}} b^{6} d^{6} - 21 \,{\left (d x + c\right )}^{\frac{5}{2}} a b^{5} d^{7} - 35 \,{\left (d x + c\right )}^{\frac{3}{2}} a b^{5} c d^{7} - 105 \, \sqrt{d x + c} a b^{5} c^{2} d^{7} + 35 \,{\left (d x + c\right )}^{\frac{3}{2}} a^{2} b^{4} d^{8} + 210 \, \sqrt{d x + c} a^{2} b^{4} c d^{8} - 105 \, \sqrt{d x + c} a^{3} b^{3} d^{9}\right )}}{105 \, b^{7} d^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(
-b^2*c + a*b*d)*b^4) + 2/105*(15*(d*x + c)^(7/2)*b^6*d^6 - 21*(d*x + c)^(5/2)*a*b^5*d^7 - 35*(d*x + c)^(3/2)*a
*b^5*c*d^7 - 105*sqrt(d*x + c)*a*b^5*c^2*d^7 + 35*(d*x + c)^(3/2)*a^2*b^4*d^8 + 210*sqrt(d*x + c)*a^2*b^4*c*d^
8 - 105*sqrt(d*x + c)*a^3*b^3*d^9)/(b^7*d^7)