3.462 \(\int \frac{(c+d x)^{5/2}}{x^4 (a+b x)} \, dx\)

Optimal. Leaf size=207 \[ -\frac{\sqrt{c+d x} \left (11 a^2 d^2-18 a b c d+8 b^2 c^2\right )}{8 a^3 x}+\frac{\left (30 a^2 b c d^2-5 a^3 d^3-40 a b^2 c^2 d+16 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{8 a^4 \sqrt{c}}+\frac{c \sqrt{c+d x} (2 b c-3 a d)}{4 a^2 x^2}-\frac{2 \sqrt{b} (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^4}-\frac{c (c+d x)^{3/2}}{3 a x^3} \]

[Out]

(c*(2*b*c - 3*a*d)*Sqrt[c + d*x])/(4*a^2*x^2) - ((8*b^2*c^2 - 18*a*b*c*d + 11*a^2*d^2)*Sqrt[c + d*x])/(8*a^3*x
) - (c*(c + d*x)^(3/2))/(3*a*x^3) + ((16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTanh[Sqrt[c
+ d*x]/Sqrt[c]])/(8*a^4*Sqrt[c]) - (2*Sqrt[b]*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*
d]])/a^4

________________________________________________________________________________________

Rubi [A]  time = 0.26857, antiderivative size = 207, 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} \[ -\frac{\sqrt{c+d x} \left (11 a^2 d^2-18 a b c d+8 b^2 c^2\right )}{8 a^3 x}+\frac{\left (30 a^2 b c d^2-5 a^3 d^3-40 a b^2 c^2 d+16 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{8 a^4 \sqrt{c}}+\frac{c \sqrt{c+d x} (2 b c-3 a d)}{4 a^2 x^2}-\frac{2 \sqrt{b} (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^4}-\frac{c (c+d x)^{3/2}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)/(x^4*(a + b*x)),x]

[Out]

(c*(2*b*c - 3*a*d)*Sqrt[c + d*x])/(4*a^2*x^2) - ((8*b^2*c^2 - 18*a*b*c*d + 11*a^2*d^2)*Sqrt[c + d*x])/(8*a^3*x
) - (c*(c + d*x)^(3/2))/(3*a*x^3) + ((16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTanh[Sqrt[c
+ d*x]/Sqrt[c]])/(8*a^4*Sqrt[c]) - (2*Sqrt[b]*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*
d]])/a^4

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^4 (a+b x)} \, dx &=-\frac{c (c+d x)^{3/2}}{3 a x^3}-\frac{\int \frac{\sqrt{c+d x} \left (\frac{3}{2} c (2 b c-3 a d)+\frac{3}{2} d (b c-2 a d) x\right )}{x^3 (a+b x)} \, dx}{3 a}\\ &=\frac{c (2 b c-3 a d) \sqrt{c+d x}}{4 a^2 x^2}-\frac{c (c+d x)^{3/2}}{3 a x^3}-\frac{\int \frac{-\frac{3}{4} c \left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right )-\frac{3}{4} d \left (6 b^2 c^2-13 a b c d+8 a^2 d^2\right ) x}{x^2 (a+b x) \sqrt{c+d x}} \, dx}{6 a^2}\\ &=\frac{c (2 b c-3 a d) \sqrt{c+d x}}{4 a^2 x^2}-\frac{\left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) \sqrt{c+d x}}{8 a^3 x}-\frac{c (c+d x)^{3/2}}{3 a x^3}+\frac{\int \frac{-\frac{3}{8} c \left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right )-\frac{3}{8} b c d \left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) x}{x (a+b x) \sqrt{c+d x}} \, dx}{6 a^3 c}\\ &=\frac{c (2 b c-3 a d) \sqrt{c+d x}}{4 a^2 x^2}-\frac{\left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) \sqrt{c+d x}}{8 a^3 x}-\frac{c (c+d x)^{3/2}}{3 a x^3}+\frac{\left (b (b c-a d)^3\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{a^4}-\frac{\left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right ) \int \frac{1}{x \sqrt{c+d x}} \, dx}{16 a^4}\\ &=\frac{c (2 b c-3 a d) \sqrt{c+d x}}{4 a^2 x^2}-\frac{\left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) \sqrt{c+d x}}{8 a^3 x}-\frac{c (c+d x)^{3/2}}{3 a x^3}+\frac{\left (2 b (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 )}{a^4 d}-\frac{\left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{8 a^4 d}\\ &=\frac{c (2 b c-3 a d) \sqrt{c+d x}}{4 a^2 x^2}-\frac{\left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) \sqrt{c+d x}}{8 a^3 x}-\frac{c (c+d x)^{3/2}}{3 a x^3}+\frac{\left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{8 a^4 \sqrt{c}}-\frac{2 \sqrt{b} (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^4}\\ \end{align*}

Mathematica [A]  time = 0.327246, size = 178, normalized size = 0.86 \[ -\frac{\frac{a \sqrt{c+d x} \left (a^2 \left (8 c^2+26 c d x+33 d^2 x^2\right )-6 a b c x (2 c+9 d x)+24 b^2 c^2 x^2\right )}{x^3}-\frac{3 \left (30 a^2 b c d^2-5 a^3 d^3-40 a b^2 c^2 d+16 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{\sqrt{c}}+48 \sqrt{b} (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{24 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/2)/(x^4*(a + b*x)),x]

[Out]

-((a*Sqrt[c + d*x]*(24*b^2*c^2*x^2 - 6*a*b*c*x*(2*c + 9*d*x) + a^2*(8*c^2 + 26*c*d*x + 33*d^2*x^2)))/x^3 - (3*
(16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/Sqrt[c] + 48*Sqrt[b
]*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(24*a^4)

________________________________________________________________________________________

Maple [B]  time = 0.014, size = 461, normalized size = 2.2 \begin{align*} -{\frac{11}{8\,a{x}^{3}} \left ( dx+c \right ) ^{{\frac{5}{2}}}}+{\frac{9\,bc}{4\,d{a}^{2}{x}^{3}} \left ( dx+c \right ) ^{{\frac{5}{2}}}}-{\frac{{b}^{2}{c}^{2}}{{d}^{2}{a}^{3}{x}^{3}} \left ( dx+c \right ) ^{{\frac{5}{2}}}}+{\frac{5\,c}{3\,a{x}^{3}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-4\,{\frac{ \left ( dx+c \right ) ^{3/2}b{c}^{2}}{d{a}^{2}{x}^{3}}}+2\,{\frac{ \left ( dx+c \right ) ^{3/2}{b}^{2}{c}^{3}}{{d}^{2}{a}^{3}{x}^{3}}}+{\frac{7\,{c}^{3}b}{4\,d{a}^{2}{x}^{3}}\sqrt{dx+c}}-{\frac{{b}^{2}{c}^{4}}{{d}^{2}{a}^{3}{x}^{3}}\sqrt{dx+c}}-{\frac{5\,{c}^{2}}{8\,a{x}^{3}}\sqrt{dx+c}}-{\frac{5\,{d}^{3}}{8\,a}{\it Artanh} \left ({\sqrt{dx+c}{\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c}}}}+{\frac{15\,{d}^{2}b}{4\,{a}^{2}}\sqrt{c}{\it Artanh} \left ({\sqrt{dx+c}{\frac{1}{\sqrt{c}}}} \right ) }-5\,{\frac{d{c}^{3/2}{b}^{2}}{{a}^{3}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }+2\,{\frac{{c}^{5/2}{b}^{3}}{{a}^{4}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }-2\,{\frac{{d}^{3}b}{a\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+6\,{\frac{{d}^{2}{b}^{2}c}{{a}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-6\,{\frac{d{b}^{3}{c}^{2}}{{a}^{3}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+2\,{\frac{{b}^{4}{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^4/(b*x+a),x)

[Out]

-11/8/a/x^3*(d*x+c)^(5/2)+9/4/d/a^2/x^3*(d*x+c)^(5/2)*c*b-1/d^2/a^3/x^3*(d*x+c)^(5/2)*b^2*c^2+5/3*c*(d*x+c)^(3
/2)/a/x^3-4/d/a^2/x^3*(d*x+c)^(3/2)*b*c^2+2/d^2/a^3/x^3*(d*x+c)^(3/2)*b^2*c^3+7/4/d/a^2/x^3*(d*x+c)^(1/2)*b*c^
3-1/d^2/a^3/x^3*(d*x+c)^(1/2)*b^2*c^4-5/8/a/x^3*(d*x+c)^(1/2)*c^2-5/8*d^3/a/c^(1/2)*arctanh((d*x+c)^(1/2)/c^(1
/2))+15/4*d^2/a^2*c^(1/2)*arctanh((d*x+c)^(1/2)/c^(1/2))*b-5*d/a^3*c^(3/2)*arctanh((d*x+c)^(1/2)/c^(1/2))*b^2+
2/a^4*c^(5/2)*arctanh((d*x+c)^(1/2)/c^(1/2))*b^3-2*d^3*b/a/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*
c)*b)^(1/2))+6*d^2*b^2/a^2/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*c-6*d*b^3/a^3/((a*d
-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*c^2+2*b^4/a^4/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^
(1/2)/((a*d-b*c)*b)^(1/2))*c^3

________________________________________________________________________________________

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^4/(b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 5.71013, size = 2075, normalized size = 10.02 \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^4/(b*x+a),x, algorithm="fricas")

[Out]

[1/48*(48*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(b^2*c - a*b*d)*x^3*log((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c
- a*b*d)*sqrt(d*x + c))/(b*x + a)) - 3*(16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(c)*x^3*
log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(8*a^3*c^3 + 3*(8*a*b^2*c^3 - 18*a^2*b*c^2*d + 11*a^3*c*d^2)*
x^2 - 2*(6*a^2*b*c^3 - 13*a^3*c^2*d)*x)*sqrt(d*x + c))/(a^4*c*x^3), 1/48*(96*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^
2)*sqrt(-b^2*c + a*b*d)*x^3*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - 3*(16*b^3*c^3 - 40*a*b^
2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(c)*x^3*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(8*a^3*c^3
+ 3*(8*a*b^2*c^3 - 18*a^2*b*c^2*d + 11*a^3*c*d^2)*x^2 - 2*(6*a^2*b*c^3 - 13*a^3*c^2*d)*x)*sqrt(d*x + c))/(a^4*
c*x^3), -1/24*(3*(16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(-c)*x^3*arctan(sqrt(d*x + c)*
sqrt(-c)/c) - 24*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(b^2*c - a*b*d)*x^3*log((b*d*x + 2*b*c - a*d - 2*sqrt
(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) + (8*a^3*c^3 + 3*(8*a*b^2*c^3 - 18*a^2*b*c^2*d + 11*a^3*c*d^2)*x^2 -
2*(6*a^2*b*c^3 - 13*a^3*c^2*d)*x)*sqrt(d*x + c))/(a^4*c*x^3), 1/24*(48*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sq
rt(-b^2*c + a*b*d)*x^3*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - 3*(16*b^3*c^3 - 40*a*b^2*c^2
*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(-c)*x^3*arctan(sqrt(d*x + c)*sqrt(-c)/c) - (8*a^3*c^3 + 3*(8*a*b^2*c^3 -
18*a^2*b*c^2*d + 11*a^3*c*d^2)*x^2 - 2*(6*a^2*b*c^3 - 13*a^3*c^2*d)*x)*sqrt(d*x + c))/(a^4*c*x^3)]

________________________________________________________________________________________

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**4/(b*x+a),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.17614, size = 405, normalized size = 1.96 \begin{align*} \frac{2 \,{\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b 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 (16 \, b^{3} c^{3} - 40 \, a b^{2} c^{2} d + 30 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \arctan \left (\frac{\sqrt{d x + c}}{\sqrt{-c}}\right )}{8 \, a^{4} \sqrt{-c}} - \frac{24 \,{\left (d x + c\right )}^{\frac{5}{2}} b^{2} c^{2} d - 48 \,{\left (d x + c\right )}^{\frac{3}{2}} b^{2} c^{3} d + 24 \, \sqrt{d x + c} b^{2} c^{4} d - 54 \,{\left (d x + c\right )}^{\frac{5}{2}} a b c d^{2} + 96 \,{\left (d x + c\right )}^{\frac{3}{2}} a b c^{2} d^{2} - 42 \, \sqrt{d x + c} a b c^{3} d^{2} + 33 \,{\left (d x + c\right )}^{\frac{5}{2}} a^{2} d^{3} - 40 \,{\left (d x + c\right )}^{\frac{3}{2}} a^{2} c d^{3} + 15 \, \sqrt{d x + c} a^{2} c^{2} d^{3}}{24 \, a^{3} d^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/x^4/(b*x+a),x, algorithm="giac")

[Out]

2*(b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-
b^2*c + a*b*d)*a^4) - 1/8*(16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*arctan(sqrt(d*x + c)/sqrt
(-c))/(a^4*sqrt(-c)) - 1/24*(24*(d*x + c)^(5/2)*b^2*c^2*d - 48*(d*x + c)^(3/2)*b^2*c^3*d + 24*sqrt(d*x + c)*b^
2*c^4*d - 54*(d*x + c)^(5/2)*a*b*c*d^2 + 96*(d*x + c)^(3/2)*a*b*c^2*d^2 - 42*sqrt(d*x + c)*a*b*c^3*d^2 + 33*(d
*x + c)^(5/2)*a^2*d^3 - 40*(d*x + c)^(3/2)*a^2*c*d^3 + 15*sqrt(d*x + c)*a^2*c^2*d^3)/(a^3*d^3*x^3)