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

Optimal. Leaf size=220 \[ -\frac{(c+d x)^{5/2} \left (-693 a^2 d^2-5 b d x (10 b c-99 a d)+180 a b c d+20 b^2 c^2\right )}{315 b^4 d^2}+\frac{a^2 (c+d x)^{3/2} (6 b c-11 a d)}{3 b^5}+\frac{a^2 \sqrt{c+d x} (6 b c-11 a d) (b c-a d)}{b^6}-\frac{a^2 (6 b c-11 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{13/2}}-\frac{x^3 (c+d x)^{5/2}}{b (a+b x)}+\frac{11 x^2 (c+d x)^{5/2}}{9 b^2} \]

[Out]

(a^2*(6*b*c - 11*a*d)*(b*c - a*d)*Sqrt[c + d*x])/b^6 + (a^2*(6*b*c - 11*a*d)*(c + d*x)^(3/2))/(3*b^5) + (11*x^
2*(c + d*x)^(5/2))/(9*b^2) - (x^3*(c + d*x)^(5/2))/(b*(a + b*x)) - ((c + d*x)^(5/2)*(20*b^2*c^2 + 180*a*b*c*d
- 693*a^2*d^2 - 5*b*d*(10*b*c - 99*a*d)*x))/(315*b^4*d^2) - (a^2*(6*b*c - 11*a*d)*(b*c - a*d)^(3/2)*ArcTanh[(S
qrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/b^(13/2)

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Rubi [A]  time = 0.221488, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {97, 153, 147, 50, 63, 208} \[ -\frac{(c+d x)^{5/2} \left (-693 a^2 d^2-5 b d x (10 b c-99 a d)+180 a b c d+20 b^2 c^2\right )}{315 b^4 d^2}+\frac{a^2 (c+d x)^{3/2} (6 b c-11 a d)}{3 b^5}+\frac{a^2 \sqrt{c+d x} (6 b c-11 a d) (b c-a d)}{b^6}-\frac{a^2 (6 b c-11 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{13/2}}-\frac{x^3 (c+d x)^{5/2}}{b (a+b x)}+\frac{11 x^2 (c+d x)^{5/2}}{9 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(6*b*c - 11*a*d)*(b*c - a*d)*Sqrt[c + d*x])/b^6 + (a^2*(6*b*c - 11*a*d)*(c + d*x)^(3/2))/(3*b^5) + (11*x^
2*(c + d*x)^(5/2))/(9*b^2) - (x^3*(c + d*x)^(5/2))/(b*(a + b*x)) - ((c + d*x)^(5/2)*(20*b^2*c^2 + 180*a*b*c*d
- 693*a^2*d^2 - 5*b*d*(10*b*c - 99*a*d)*x))/(315*b^4*d^2) - (a^2*(6*b*c - 11*a*d)*(b*c - a*d)^(3/2)*ArcTanh[(S
qrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/b^(13/2)

Rule 97

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)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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^3 (c+d x)^{5/2}}{(a+b x)^2} \, dx &=-\frac{x^3 (c+d x)^{5/2}}{b (a+b x)}+\frac{\int \frac{x^2 (c+d x)^{3/2} \left (3 c+\frac{11 d x}{2}\right )}{a+b x} \, dx}{b}\\ &=\frac{11 x^2 (c+d x)^{5/2}}{9 b^2}-\frac{x^3 (c+d x)^{5/2}}{b (a+b x)}+\frac{2 \int \frac{x (c+d x)^{3/2} \left (-11 a c d+\frac{1}{4} d (10 b c-99 a d) x\right )}{a+b x} \, dx}{9 b^2 d}\\ &=\frac{11 x^2 (c+d x)^{5/2}}{9 b^2}-\frac{x^3 (c+d x)^{5/2}}{b (a+b x)}-\frac{(c+d x)^{5/2} \left (20 b^2 c^2+180 a b c d-693 a^2 d^2-5 b d (10 b c-99 a d) x\right )}{315 b^4 d^2}+\frac{\left (a^2 (6 b c-11 a d)\right ) \int \frac{(c+d x)^{3/2}}{a+b x} \, dx}{2 b^4}\\ &=\frac{a^2 (6 b c-11 a d) (c+d x)^{3/2}}{3 b^5}+\frac{11 x^2 (c+d x)^{5/2}}{9 b^2}-\frac{x^3 (c+d x)^{5/2}}{b (a+b x)}-\frac{(c+d x)^{5/2} \left (20 b^2 c^2+180 a b c d-693 a^2 d^2-5 b d (10 b c-99 a d) x\right )}{315 b^4 d^2}+\frac{\left (a^2 (6 b c-11 a d) (b c-a d)\right ) \int \frac{\sqrt{c+d x}}{a+b x} \, dx}{2 b^5}\\ &=\frac{a^2 (6 b c-11 a d) (b c-a d) \sqrt{c+d x}}{b^6}+\frac{a^2 (6 b c-11 a d) (c+d x)^{3/2}}{3 b^5}+\frac{11 x^2 (c+d x)^{5/2}}{9 b^2}-\frac{x^3 (c+d x)^{5/2}}{b (a+b x)}-\frac{(c+d x)^{5/2} \left (20 b^2 c^2+180 a b c d-693 a^2 d^2-5 b d (10 b c-99 a d) x\right )}{315 b^4 d^2}+\frac{\left (a^2 (6 b c-11 a d) (b c-a d)^2\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{2 b^6}\\ &=\frac{a^2 (6 b c-11 a d) (b c-a d) \sqrt{c+d x}}{b^6}+\frac{a^2 (6 b c-11 a d) (c+d x)^{3/2}}{3 b^5}+\frac{11 x^2 (c+d x)^{5/2}}{9 b^2}-\frac{x^3 (c+d x)^{5/2}}{b (a+b x)}-\frac{(c+d x)^{5/2} \left (20 b^2 c^2+180 a b c d-693 a^2 d^2-5 b d (10 b c-99 a d) x\right )}{315 b^4 d^2}+\frac{\left (a^2 (6 b c-11 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^6 d}\\ &=\frac{a^2 (6 b c-11 a d) (b c-a d) \sqrt{c+d x}}{b^6}+\frac{a^2 (6 b c-11 a d) (c+d x)^{3/2}}{3 b^5}+\frac{11 x^2 (c+d x)^{5/2}}{9 b^2}-\frac{x^3 (c+d x)^{5/2}}{b (a+b x)}-\frac{(c+d x)^{5/2} \left (20 b^2 c^2+180 a b c d-693 a^2 d^2-5 b d (10 b c-99 a d) x\right )}{315 b^4 d^2}-\frac{a^2 (6 b c-11 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^{13/2}}\\ \end{align*}

Mathematica [A]  time = 0.652182, size = 246, normalized size = 1.12 \[ \frac{21 a^2 d^2 (a+b x) (6 b c-11 a d) \left (\sqrt{b} \sqrt{c+d x} \left (15 a^2 d^2-5 a b d (7 c+d x)+b^2 \left (23 c^2+11 c d x+3 d^2 x^2\right )\right )-15 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )\right )+5 b^{7/2} (c+d x)^{7/2} \left (2 a^2 b d (11 d x-16 c)+99 a^3 d^2-2 a b^2 c (2 c+9 d x)-4 b^3 c^2 x\right )+70 b^{11/2} d x^2 (c+d x)^{7/2} (b c-a d)}{315 b^{13/2} d^2 (a+b x) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(70*b^(11/2)*d*(b*c - a*d)*x^2*(c + d*x)^(7/2) + 5*b^(7/2)*(c + d*x)^(7/2)*(99*a^3*d^2 - 4*b^3*c^2*x - 2*a*b^2
*c*(2*c + 9*d*x) + 2*a^2*b*d*(-16*c + 11*d*x)) + 21*a^2*d^2*(6*b*c - 11*a*d)*(a + b*x)*(Sqrt[b]*Sqrt[c + d*x]*
(15*a^2*d^2 - 5*a*b*d*(7*c + d*x) + b^2*(23*c^2 + 11*c*d*x + 3*d^2*x^2)) - 15*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[
b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]]))/(315*b^(13/2)*d^2*(b*c - a*d)*(a + b*x))

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Maple [B]  time = 0.018, size = 415, normalized size = 1.9 \begin{align*}{\frac{2}{9\,{b}^{2}{d}^{2}} \left ( dx+c \right ) ^{{\frac{9}{2}}}}-{\frac{4\,a}{7\,d{b}^{3}} \left ( dx+c \right ) ^{{\frac{7}{2}}}}-{\frac{2\,c}{7\,{b}^{2}{d}^{2}} \left ( dx+c \right ) ^{{\frac{7}{2}}}}+{\frac{6\,{a}^{2}}{5\,{b}^{4}} \left ( dx+c \right ) ^{{\frac{5}{2}}}}-{\frac{8\,d{a}^{3}}{3\,{b}^{5}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+2\,{\frac{ \left ( dx+c \right ) ^{3/2}{a}^{2}c}{{b}^{4}}}+10\,{\frac{{d}^{2}{a}^{4}\sqrt{dx+c}}{{b}^{6}}}-16\,{\frac{d{a}^{3}c\sqrt{dx+c}}{{b}^{5}}}+6\,{\frac{{a}^{2}{c}^{2}\sqrt{dx+c}}{{b}^{4}}}+{\frac{{d}^{3}{a}^{5}}{{b}^{6} \left ( bdx+ad \right ) }\sqrt{dx+c}}-2\,{\frac{{d}^{2}{a}^{4}\sqrt{dx+c}c}{{b}^{5} \left ( bdx+ad \right ) }}+{\frac{d{a}^{3}{c}^{2}}{{b}^{4} \left ( bdx+ad \right ) }\sqrt{dx+c}}-11\,{\frac{{d}^{3}{a}^{5}}{{b}^{6}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+28\,{\frac{{d}^{2}{a}^{4}c}{{b}^{5}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-23\,{\frac{d{a}^{3}{c}^{2}}{{b}^{4}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+6\,{\frac{{a}^{2}{c}^{3}}{{b}^{3}\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^3*(d*x+c)^(5/2)/(b*x+a)^2,x)

[Out]

2/9/d^2/b^2*(d*x+c)^(9/2)-4/7/d/b^3*(d*x+c)^(7/2)*a-2/7/d^2/b^2*(d*x+c)^(7/2)*c+6/5/b^4*a^2*(d*x+c)^(5/2)-8/3*
d/b^5*(d*x+c)^(3/2)*a^3+2/b^4*(d*x+c)^(3/2)*a^2*c+10*d^2/b^6*a^4*(d*x+c)^(1/2)-16*d/b^5*a^3*c*(d*x+c)^(1/2)+6/
b^4*a^2*c^2*(d*x+c)^(1/2)+d^3*a^5/b^6*(d*x+c)^(1/2)/(b*d*x+a*d)-2*d^2*a^4/b^5*(d*x+c)^(1/2)/(b*d*x+a*d)*c+d*a^
3/b^4*(d*x+c)^(1/2)/(b*d*x+a*d)*c^2-11*d^3*a^5/b^6/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1
/2))+28*d^2*a^4/b^5/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*c-23*d*a^3/b^4/((a*d-b*c)*
b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*c^2+6*a^2/b^3/((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^3*(d*x+c)^(5/2)/(b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.78741, size = 1721, normalized size = 7.82 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/630*(315*(6*a^3*b^2*c^2*d^2 - 17*a^4*b*c*d^3 + 11*a^5*d^4 + (6*a^2*b^3*c^2*d^2 - 17*a^3*b^2*c*d^3 + 11*a^4*
b*d^4)*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
*(70*b^5*d^4*x^5 - 20*a*b^4*c^4 - 180*a^2*b^3*c^3*d + 3213*a^3*b^2*c^2*d^2 - 6510*a^4*b*c*d^3 + 3465*a^5*d^4 +
10*(19*b^5*c*d^3 - 11*a*b^4*d^4)*x^4 + 2*(75*b^5*c^2*d^2 - 175*a*b^4*c*d^3 + 99*a^2*b^3*d^4)*x^3 + 2*(5*b^5*c
^3*d - 195*a*b^4*c^2*d^2 + 423*a^2*b^3*c*d^3 - 231*a^3*b^2*d^4)*x^2 - 2*(10*b^5*c^4 + 85*a*b^4*c^3*d - 1179*a^
2*b^3*c^2*d^2 + 2247*a^3*b^2*c*d^3 - 1155*a^4*b*d^4)*x)*sqrt(d*x + c))/(b^7*d^2*x + a*b^6*d^2), -1/315*(315*(6
*a^3*b^2*c^2*d^2 - 17*a^4*b*c*d^3 + 11*a^5*d^4 + (6*a^2*b^3*c^2*d^2 - 17*a^3*b^2*c*d^3 + 11*a^4*b*d^4)*x)*sqrt
(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (70*b^5*d^4*x^5 - 20*a*b^4*c^4 -
180*a^2*b^3*c^3*d + 3213*a^3*b^2*c^2*d^2 - 6510*a^4*b*c*d^3 + 3465*a^5*d^4 + 10*(19*b^5*c*d^3 - 11*a*b^4*d^4)*
x^4 + 2*(75*b^5*c^2*d^2 - 175*a*b^4*c*d^3 + 99*a^2*b^3*d^4)*x^3 + 2*(5*b^5*c^3*d - 195*a*b^4*c^2*d^2 + 423*a^2
*b^3*c*d^3 - 231*a^3*b^2*d^4)*x^2 - 2*(10*b^5*c^4 + 85*a*b^4*c^3*d - 1179*a^2*b^3*c^2*d^2 + 2247*a^3*b^2*c*d^3
- 1155*a^4*b*d^4)*x)*sqrt(d*x + c))/(b^7*d^2*x + a*b^6*d^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(x**3*(d*x+c)**(5/2)/(b*x+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.21316, size = 436, normalized size = 1.98 \begin{align*} \frac{{\left (6 \, a^{2} b^{3} c^{3} - 23 \, a^{3} b^{2} c^{2} d + 28 \, a^{4} b c d^{2} - 11 \, a^{5} 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^{6}} + \frac{\sqrt{d x + c} a^{3} b^{2} c^{2} d - 2 \, \sqrt{d x + c} a^{4} b c d^{2} + \sqrt{d x + c} a^{5} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b^{6}} + \frac{2 \,{\left (35 \,{\left (d x + c\right )}^{\frac{9}{2}} b^{16} d^{16} - 45 \,{\left (d x + c\right )}^{\frac{7}{2}} b^{16} c d^{16} - 90 \,{\left (d x + c\right )}^{\frac{7}{2}} a b^{15} d^{17} + 189 \,{\left (d x + c\right )}^{\frac{5}{2}} a^{2} b^{14} d^{18} + 315 \,{\left (d x + c\right )}^{\frac{3}{2}} a^{2} b^{14} c d^{18} + 945 \, \sqrt{d x + c} a^{2} b^{14} c^{2} d^{18} - 420 \,{\left (d x + c\right )}^{\frac{3}{2}} a^{3} b^{13} d^{19} - 2520 \, \sqrt{d x + c} a^{3} b^{13} c d^{19} + 1575 \, \sqrt{d x + c} a^{4} b^{12} d^{20}\right )}}{315 \, b^{18} d^{18}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*a^2*b^3*c^3 - 23*a^3*b^2*c^2*d + 28*a^4*b*c*d^2 - 11*a^5*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/
(sqrt(-b^2*c + a*b*d)*b^6) + (sqrt(d*x + c)*a^3*b^2*c^2*d - 2*sqrt(d*x + c)*a^4*b*c*d^2 + sqrt(d*x + c)*a^5*d^
3)/(((d*x + c)*b - b*c + a*d)*b^6) + 2/315*(35*(d*x + c)^(9/2)*b^16*d^16 - 45*(d*x + c)^(7/2)*b^16*c*d^16 - 90
*(d*x + c)^(7/2)*a*b^15*d^17 + 189*(d*x + c)^(5/2)*a^2*b^14*d^18 + 315*(d*x + c)^(3/2)*a^2*b^14*c*d^18 + 945*s
qrt(d*x + c)*a^2*b^14*c^2*d^18 - 420*(d*x + c)^(3/2)*a^3*b^13*d^19 - 2520*sqrt(d*x + c)*a^3*b^13*c*d^19 + 1575
*sqrt(d*x + c)*a^4*b^12*d^20)/(b^18*d^18)