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

Optimal. Leaf size=219 \[ \frac{\sqrt{c+d x} \left (4 a^2 d^2-17 a b c d+12 b^2 c^2\right )}{4 a^3 (a+b x)}-\frac{\sqrt{c} \left (15 a^2 d^2-40 a b c d+24 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{4 a^4}+\frac{c \sqrt{c+d x} (6 b c-7 a d)}{4 a^2 x (a+b x)}+\frac{(6 b c-a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^4 \sqrt{b}}-\frac{c (c+d x)^{3/2}}{2 a x^2 (a+b x)} \]

[Out]

((12*b^2*c^2 - 17*a*b*c*d + 4*a^2*d^2)*Sqrt[c + d*x])/(4*a^3*(a + b*x)) + (c*(6*b*c - 7*a*d)*Sqrt[c + d*x])/(4
*a^2*x*(a + b*x)) - (c*(c + d*x)^(3/2))/(2*a*x^2*(a + b*x)) - (Sqrt[c]*(24*b^2*c^2 - 40*a*b*c*d + 15*a^2*d^2)*
ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(4*a^4) + ((b*c - a*d)^(3/2)*(6*b*c - a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqr
t[b*c - a*d]])/(a^4*Sqrt[b])

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Rubi [A]  time = 0.239316, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {98, 149, 151, 156, 63, 208} \[ \frac{\sqrt{c+d x} \left (4 a^2 d^2-17 a b c d+12 b^2 c^2\right )}{4 a^3 (a+b x)}-\frac{\sqrt{c} \left (15 a^2 d^2-40 a b c d+24 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{4 a^4}+\frac{c \sqrt{c+d x} (6 b c-7 a d)}{4 a^2 x (a+b x)}+\frac{(6 b c-a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^4 \sqrt{b}}-\frac{c (c+d x)^{3/2}}{2 a x^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((12*b^2*c^2 - 17*a*b*c*d + 4*a^2*d^2)*Sqrt[c + d*x])/(4*a^3*(a + b*x)) + (c*(6*b*c - 7*a*d)*Sqrt[c + d*x])/(4
*a^2*x*(a + b*x)) - (c*(c + d*x)^(3/2))/(2*a*x^2*(a + b*x)) - (Sqrt[c]*(24*b^2*c^2 - 40*a*b*c*d + 15*a^2*d^2)*
ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(4*a^4) + ((b*c - a*d)^(3/2)*(6*b*c - a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqr
t[b*c - a*d]])/(a^4*Sqrt[b])

Rule 98

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

Rule 149

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

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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{(c+d x)^{5/2}}{x^3 (a+b x)^2} \, dx &=-\frac{c (c+d x)^{3/2}}{2 a x^2 (a+b x)}-\frac{\int \frac{\sqrt{c+d x} \left (\frac{1}{2} c (6 b c-7 a d)+\frac{1}{2} d (3 b c-4 a d) x\right )}{x^2 (a+b x)^2} \, dx}{2 a}\\ &=\frac{c (6 b c-7 a d) \sqrt{c+d x}}{4 a^2 x (a+b x)}-\frac{c (c+d x)^{3/2}}{2 a x^2 (a+b x)}-\frac{\int \frac{-\frac{1}{4} c \left (24 b^2 c^2-40 a b c d+15 a^2 d^2\right )-\frac{1}{4} d \left (18 b^2 c^2-27 a b c d+8 a^2 d^2\right ) x}{x (a+b x)^2 \sqrt{c+d x}} \, dx}{2 a^2}\\ &=\frac{\left (12 b^2 c^2-17 a b c d+4 a^2 d^2\right ) \sqrt{c+d x}}{4 a^3 (a+b x)}+\frac{c (6 b c-7 a d) \sqrt{c+d x}}{4 a^2 x (a+b x)}-\frac{c (c+d x)^{3/2}}{2 a x^2 (a+b x)}-\frac{\int \frac{-\frac{1}{4} c (b c-a d) \left (24 b^2 c^2-40 a b c d+15 a^2 d^2\right )-\frac{1}{4} d (b c-a d) \left (12 b^2 c^2-17 a b c d+4 a^2 d^2\right ) x}{x (a+b x) \sqrt{c+d x}} \, dx}{2 a^3 (b c-a d)}\\ &=\frac{\left (12 b^2 c^2-17 a b c d+4 a^2 d^2\right ) \sqrt{c+d x}}{4 a^3 (a+b x)}+\frac{c (6 b c-7 a d) \sqrt{c+d x}}{4 a^2 x (a+b x)}-\frac{c (c+d x)^{3/2}}{2 a x^2 (a+b x)}-\frac{\left ((b c-a d)^2 (6 b c-a d)\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{2 a^4}+\frac{\left (c \left (24 b^2 c^2-40 a b c d+15 a^2 d^2\right )\right ) \int \frac{1}{x \sqrt{c+d x}} \, dx}{8 a^4}\\ &=\frac{\left (12 b^2 c^2-17 a b c d+4 a^2 d^2\right ) \sqrt{c+d x}}{4 a^3 (a+b x)}+\frac{c (6 b c-7 a d) \sqrt{c+d x}}{4 a^2 x (a+b x)}-\frac{c (c+d x)^{3/2}}{2 a x^2 (a+b x)}-\frac{\left ((b c-a d)^2 (6 b c-a d)\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{a^4 d}+\frac{\left (c \left (24 b^2 c^2-40 a b c d+15 a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{4 a^4 d}\\ &=\frac{\left (12 b^2 c^2-17 a b c d+4 a^2 d^2\right ) \sqrt{c+d x}}{4 a^3 (a+b x)}+\frac{c (6 b c-7 a d) \sqrt{c+d x}}{4 a^2 x (a+b x)}-\frac{c (c+d x)^{3/2}}{2 a x^2 (a+b x)}-\frac{\sqrt{c} \left (24 b^2 c^2-40 a b c d+15 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{4 a^4}+\frac{(b c-a d)^{3/2} (6 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^4 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.291107, size = 192, normalized size = 0.88 \[ \frac{\frac{a \sqrt{c+d x} \left (a^2 \left (-2 c^2-9 c d x+4 d^2 x^2\right )+a b c x (6 c-17 d x)+12 b^2 c^2 x^2\right )}{x^2 (a+b x)}-\sqrt{c} \left (15 a^2 d^2-40 a b c d+24 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )+\frac{4 \sqrt{b c-a d} \left (a^2 d^2-7 a b c d+6 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{\sqrt{b}}}{4 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*Sqrt[c + d*x]*(12*b^2*c^2*x^2 + a*b*c*x*(6*c - 17*d*x) + a^2*(-2*c^2 - 9*c*d*x + 4*d^2*x^2)))/(x^2*(a + b*
x)) - Sqrt[c]*(24*b^2*c^2 - 40*a*b*c*d + 15*a^2*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]] + (4*Sqrt[b*c - a*d]*(6*b^
2*c^2 - 7*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/Sqrt[b])/(4*a^4)

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Maple [B]  time = 0.018, size = 403, normalized size = 1.8 \begin{align*} -{\frac{9\,c}{4\,{a}^{2}{x}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+2\,{\frac{{c}^{2} \left ( dx+c \right ) ^{3/2}b}{d{a}^{3}{x}^{2}}}+{\frac{7\,{c}^{2}}{4\,{a}^{2}{x}^{2}}\sqrt{dx+c}}-2\,{\frac{{c}^{3}\sqrt{dx+c}b}{d{a}^{3}{x}^{2}}}-{\frac{15\,{d}^{2}}{4\,{a}^{2}}\sqrt{c}{\it Artanh} \left ({\sqrt{dx+c}{\frac{1}{\sqrt{c}}}} \right ) }+10\,{\frac{d{c}^{3/2}b}{{a}^{3}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }-6\,{\frac{{c}^{5/2}{b}^{2}}{{a}^{4}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }+{\frac{{d}^{3}}{a \left ( bdx+ad \right ) }\sqrt{dx+c}}-2\,{\frac{{d}^{2}\sqrt{dx+c}bc}{{a}^{2} \left ( bdx+ad \right ) }}+{\frac{{b}^{2}d{c}^{2}}{{a}^{3} \left ( bdx+ad \right ) }\sqrt{dx+c}}+{\frac{{d}^{3}}{a}\arctan \left ({b\sqrt{dx+c}{\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}}-8\,{\frac{{d}^{2}bc}{{a}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+13\,{\frac{{b}^{2}d{c}^{2}}{{a}^{3}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-6\,{\frac{{b}^{3}{c}^{3}}{{a}^{4}\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((d*x+c)^(5/2)/x^3/(b*x+a)^2,x)

[Out]

-9/4*c/a^2/x^2*(d*x+c)^(3/2)+2/d*c^2/a^3/x^2*(d*x+c)^(3/2)*b+7/4*c^2/a^2/x^2*(d*x+c)^(1/2)-2/d*c^3/a^3/x^2*(d*
x+c)^(1/2)*b-15/4*d^2*c^(1/2)/a^2*arctanh((d*x+c)^(1/2)/c^(1/2))+10*d*c^(3/2)/a^3*arctanh((d*x+c)^(1/2)/c^(1/2
))*b-6*c^(5/2)/a^4*arctanh((d*x+c)^(1/2)/c^(1/2))*b^2+d^3/a*(d*x+c)^(1/2)/(b*d*x+a*d)-2*d^2/a^2*(d*x+c)^(1/2)/
(b*d*x+a*d)*b*c+d/a^3*(d*x+c)^(1/2)/(b*d*x+a*d)*b^2*c^2+d^3/a/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d
-b*c)*b)^(1/2))-8*d^2/a^2/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*b*c+13*d/a^3/((a*d-b
*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*b^2*c^2-6/a^4/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1
/2)/((a*d-b*c)*b)^(1/2))*b^3*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((d*x+c)^(5/2)/x^3/(b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.13861, size = 2514, normalized size = 11.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(4*((6*b^3*c^2 - 7*a*b^2*c*d + a^2*b*d^2)*x^3 + (6*a*b^2*c^2 - 7*a^2*b*c*d + a^3*d^2)*x^2)*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)) + ((24*b^3*c^2 - 40*a*b^2*c
*d + 15*a^2*b*d^2)*x^3 + (24*a*b^2*c^2 - 40*a^2*b*c*d + 15*a^3*d^2)*x^2)*sqrt(c)*log((d*x - 2*sqrt(d*x + c)*sq
rt(c) + 2*c)/x) - 2*(2*a^3*c^2 - (12*a*b^2*c^2 - 17*a^2*b*c*d + 4*a^3*d^2)*x^2 - 3*(2*a^2*b*c^2 - 3*a^3*c*d)*x
)*sqrt(d*x + c))/(a^4*b*x^3 + a^5*x^2), 1/8*(8*((6*b^3*c^2 - 7*a*b^2*c*d + a^2*b*d^2)*x^3 + (6*a*b^2*c^2 - 7*a
^2*b*c*d + a^3*d^2)*x^2)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + ((24
*b^3*c^2 - 40*a*b^2*c*d + 15*a^2*b*d^2)*x^3 + (24*a*b^2*c^2 - 40*a^2*b*c*d + 15*a^3*d^2)*x^2)*sqrt(c)*log((d*x
- 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(2*a^3*c^2 - (12*a*b^2*c^2 - 17*a^2*b*c*d + 4*a^3*d^2)*x^2 - 3*(2*a^2
*b*c^2 - 3*a^3*c*d)*x)*sqrt(d*x + c))/(a^4*b*x^3 + a^5*x^2), 1/4*(((24*b^3*c^2 - 40*a*b^2*c*d + 15*a^2*b*d^2)*
x^3 + (24*a*b^2*c^2 - 40*a^2*b*c*d + 15*a^3*d^2)*x^2)*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt(-c)/c) + 2*((6*b^3*c^
2 - 7*a*b^2*c*d + a^2*b*d^2)*x^3 + (6*a*b^2*c^2 - 7*a^2*b*c*d + a^3*d^2)*x^2)*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*a^3*c^2 - (12*a*b^2*c^2 - 17*a^2*b*c*d +
4*a^3*d^2)*x^2 - 3*(2*a^2*b*c^2 - 3*a^3*c*d)*x)*sqrt(d*x + c))/(a^4*b*x^3 + a^5*x^2), 1/4*(4*((6*b^3*c^2 - 7*
a*b^2*c*d + a^2*b*d^2)*x^3 + (6*a*b^2*c^2 - 7*a^2*b*c*d + a^3*d^2)*x^2)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x
+ c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + ((24*b^3*c^2 - 40*a*b^2*c*d + 15*a^2*b*d^2)*x^3 + (24*a*b^2*c^2 - 4
0*a^2*b*c*d + 15*a^3*d^2)*x^2)*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt(-c)/c) - (2*a^3*c^2 - (12*a*b^2*c^2 - 17*a^2
*b*c*d + 4*a^3*d^2)*x^2 - 3*(2*a^2*b*c^2 - 3*a^3*c*d)*x)*sqrt(d*x + c))/(a^4*b*x^3 + a^5*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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.2208, size = 358, normalized size = 1.63 \begin{align*} -\frac{{\left (6 \, b^{3} c^{3} - 13 \, a b^{2} c^{2} d + 8 \, a^{2} b c d^{2} - 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} a^{4}} + \frac{{\left (24 \, b^{2} c^{3} - 40 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} \arctan \left (\frac{\sqrt{d x + c}}{\sqrt{-c}}\right )}{4 \, a^{4} \sqrt{-c}} + \frac{\sqrt{d x + c} b^{2} c^{2} d - 2 \, \sqrt{d x + c} a b c d^{2} + \sqrt{d x + c} a^{2} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} a^{3}} + \frac{8 \,{\left (d x + c\right )}^{\frac{3}{2}} b c^{2} d - 8 \, \sqrt{d x + c} b c^{3} d - 9 \,{\left (d x + c\right )}^{\frac{3}{2}} a c d^{2} + 7 \, \sqrt{d x + c} a c^{2} d^{2}}{4 \, a^{3} d^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(6*b^3*c^3 - 13*a*b^2*c^2*d + 8*a^2*b*c*d^2 - a^3*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^
2*c + a*b*d)*a^4) + 1/4*(24*b^2*c^3 - 40*a*b*c^2*d + 15*a^2*c*d^2)*arctan(sqrt(d*x + c)/sqrt(-c))/(a^4*sqrt(-c
)) + (sqrt(d*x + c)*b^2*c^2*d - 2*sqrt(d*x + c)*a*b*c*d^2 + sqrt(d*x + c)*a^2*d^3)/(((d*x + c)*b - b*c + a*d)*
a^3) + 1/4*(8*(d*x + c)^(3/2)*b*c^2*d - 8*sqrt(d*x + c)*b*c^3*d - 9*(d*x + c)^(3/2)*a*c*d^2 + 7*sqrt(d*x + c)*
a*c^2*d^2)/(a^3*d^2*x^2)