3.21 \(\int \frac{\sqrt{a+b x} (e+f x)}{x (c+d x)^3} \, dx\)

Optimal. Leaf size=205 \[ -\frac{\left (-8 a^2 d^3 e+12 a b c d^2 e-b^2 c^2 (c f+3 d e)\right ) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{4 c^3 d^{3/2} (b c-a d)^{3/2}}-\frac{\sqrt{a+b x} \left (4 a d^2 e-b c (c f+3 d e)\right )}{4 c^2 d (c+d x) (b c-a d)}-\frac{2 \sqrt{a} e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{c^3}+\frac{\sqrt{a+b x} (d e-c f)}{2 c d (c+d x)^2} \]

[Out]

((d*e - c*f)*Sqrt[a + b*x])/(2*c*d*(c + d*x)^2) - ((4*a*d^2*e - b*c*(3*d*e + c*f))*Sqrt[a + b*x])/(4*c^2*d*(b*
c - a*d)*(c + d*x)) - ((12*a*b*c*d^2*e - 8*a^2*d^3*e - b^2*c^2*(3*d*e + c*f))*ArcTan[(Sqrt[d]*Sqrt[a + b*x])/S
qrt[b*c - a*d]])/(4*c^3*d^(3/2)*(b*c - a*d)^(3/2)) - (2*Sqrt[a]*e*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/c^3

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Rubi [A]  time = 0.279463, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {149, 151, 156, 63, 208, 205} \[ -\frac{\left (-8 a^2 d^3 e+12 a b c d^2 e-b^2 c^2 (c f+3 d e)\right ) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{4 c^3 d^{3/2} (b c-a d)^{3/2}}-\frac{\sqrt{a+b x} \left (4 a d^2 e-b c (c f+3 d e)\right )}{4 c^2 d (c+d x) (b c-a d)}-\frac{2 \sqrt{a} e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{c^3}+\frac{\sqrt{a+b x} (d e-c f)}{2 c d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x]*(e + f*x))/(x*(c + d*x)^3),x]

[Out]

((d*e - c*f)*Sqrt[a + b*x])/(2*c*d*(c + d*x)^2) - ((4*a*d^2*e - b*c*(3*d*e + c*f))*Sqrt[a + b*x])/(4*c^2*d*(b*
c - a*d)*(c + d*x)) - ((12*a*b*c*d^2*e - 8*a^2*d^3*e - b^2*c^2*(3*d*e + c*f))*ArcTan[(Sqrt[d]*Sqrt[a + b*x])/S
qrt[b*c - a*d]])/(4*c^3*d^(3/2)*(b*c - a*d)^(3/2)) - (2*Sqrt[a]*e*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/c^3

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]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sqrt{a+b x} (e+f x)}{x (c+d x)^3} \, dx &=\frac{(d e-c f) \sqrt{a+b x}}{2 c d (c+d x)^2}-\frac{\int \frac{-2 a d e-\frac{1}{2} b (3 d e+c f) x}{x \sqrt{a+b x} (c+d x)^2} \, dx}{2 c d}\\ &=\frac{(d e-c f) \sqrt{a+b x}}{2 c d (c+d x)^2}-\frac{\left (4 a d^2 e-b c (3 d e+c f)\right ) \sqrt{a+b x}}{4 c^2 d (b c-a d) (c+d x)}+\frac{\int \frac{2 a d (b c-a d) e-\frac{1}{4} b \left (4 a d^2 e-b c (3 d e+c f)\right ) x}{x \sqrt{a+b x} (c+d x)} \, dx}{2 c^2 d (b c-a d)}\\ &=\frac{(d e-c f) \sqrt{a+b x}}{2 c d (c+d x)^2}-\frac{\left (4 a d^2 e-b c (3 d e+c f)\right ) \sqrt{a+b x}}{4 c^2 d (b c-a d) (c+d x)}+\frac{(a e) \int \frac{1}{x \sqrt{a+b x}} \, dx}{c^3}-\frac{\left (12 a b c d^2 e-8 a^2 d^3 e-b^2 c^2 (3 d e+c f)\right ) \int \frac{1}{\sqrt{a+b x} (c+d x)} \, dx}{8 c^3 d (b c-a d)}\\ &=\frac{(d e-c f) \sqrt{a+b x}}{2 c d (c+d x)^2}-\frac{\left (4 a d^2 e-b c (3 d e+c f)\right ) \sqrt{a+b x}}{4 c^2 d (b c-a d) (c+d x)}+\frac{(2 a e) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{b c^3}-\frac{\left (12 a b c d^2 e-8 a^2 d^3 e-b^2 c^2 (3 d e+c f)\right ) \operatorname{Subst}\left (\int \frac{1}{c-\frac{a d}{b}+\frac{d x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{4 b c^3 d (b c-a d)}\\ &=\frac{(d e-c f) \sqrt{a+b x}}{2 c d (c+d x)^2}-\frac{\left (4 a d^2 e-b c (3 d e+c f)\right ) \sqrt{a+b x}}{4 c^2 d (b c-a d) (c+d x)}-\frac{\left (12 a b c d^2 e-8 a^2 d^3 e-b^2 c^2 (3 d e+c f)\right ) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{4 c^3 d^{3/2} (b c-a d)^{3/2}}-\frac{2 \sqrt{a} e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{c^3}\\ \end{align*}

Mathematica [A]  time = 0.55004, size = 259, normalized size = 1.26 \[ \frac{\frac{2 \left (\frac{\left (8 a^2 d^3 e-12 a b c d^2 e+b^2 c^2 (c f+3 d e)\right ) \left (\sqrt{d} \sqrt{a+b x}-\sqrt{b c-a d} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )\right )}{4 d^{3/2}}+2 e (b c-a d)^2 \left (\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )-\sqrt{a+b x}\right )\right )}{c^2 (b c-a d)}-\frac{(a+b x)^{3/2} \left (4 a d^2 e+b c (c f-5 d e)\right )}{2 c (c+d x) (b c-a d)}+\frac{(a+b x)^{3/2} (d e-c f)}{(c+d x)^2}}{2 c (a d-b c)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x]*(e + f*x))/(x*(c + d*x)^3),x]

[Out]

(((d*e - c*f)*(a + b*x)^(3/2))/(c + d*x)^2 - ((4*a*d^2*e + b*c*(-5*d*e + c*f))*(a + b*x)^(3/2))/(2*c*(b*c - a*
d)*(c + d*x)) + (2*(((-12*a*b*c*d^2*e + 8*a^2*d^3*e + b^2*c^2*(3*d*e + c*f))*(Sqrt[d]*Sqrt[a + b*x] - Sqrt[b*c
- a*d]*ArcTan[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]]))/(4*d^(3/2)) + 2*(b*c - a*d)^2*e*(-Sqrt[a + b*x] + Sq
rt[a]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])))/(c^2*(b*c - a*d)))/(2*c*(-(b*c) + a*d))

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Maple [A]  time = 0.014, size = 221, normalized size = 1.1 \begin{align*} 2\,{b}^{2} \left ( -{\frac{1}{{c}^{3}{b}^{2}} \left ({\frac{1}{ \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{2}} \left ( -1/8\,{\frac{bc \left ( 4\,a{d}^{2}e-b{c}^{2}f-3\,bcde \right ) \left ( bx+a \right ) ^{3/2}}{ad-bc}}+1/8\,{\frac{ \left ( 4\,a{d}^{2}e+b{c}^{2}f-5\,bcde \right ) bc\sqrt{bx+a}}{d}} \right ) }-1/8\,{\frac{8\,{a}^{2}{d}^{3}e-12\,abc{d}^{2}e+{c}^{3}{b}^{2}f+3\,{b}^{2}{c}^{2}de}{ \left ( ad-bc \right ) d\sqrt{ \left ( ad-bc \right ) d}}{\it Artanh} \left ({\frac{\sqrt{bx+a}d}{\sqrt{ \left ( ad-bc \right ) d}}} \right ) } \right ) }-{\frac{\sqrt{a}e}{{c}^{3}{b}^{2}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*(b*x+a)^(1/2)/x/(d*x+c)^3,x)

[Out]

2*b^2*(-1/b^2/c^3*((-1/8*b*c*(4*a*d^2*e-b*c^2*f-3*b*c*d*e)/(a*d-b*c)*(b*x+a)^(3/2)+1/8*(4*a*d^2*e+b*c^2*f-5*b*
c*d*e)*b*c/d*(b*x+a)^(1/2))/((b*x+a)*d-a*d+b*c)^2-1/8*(8*a^2*d^3*e-12*a*b*c*d^2*e+b^2*c^3*f+3*b^2*c^2*d*e)/(a*
d-b*c)/d/((a*d-b*c)*d)^(1/2)*arctanh((b*x+a)^(1/2)*d/((a*d-b*c)*d)^(1/2)))-a^(1/2)*e/b^2/c^3*arctanh((b*x+a)^(
1/2)/a^(1/2)))

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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((f*x+e)*(b*x+a)^(1/2)/x/(d*x+c)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.97522, size = 4551, normalized size = 22.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(b*x+a)^(1/2)/x/(d*x+c)^3,x, algorithm="fricas")

[Out]

[1/8*((b^2*c^5*f + (b^2*c^3*d^2*f + (3*b^2*c^2*d^3 - 12*a*b*c*d^4 + 8*a^2*d^5)*e)*x^2 + (3*b^2*c^4*d - 12*a*b*
c^3*d^2 + 8*a^2*c^2*d^3)*e + 2*(b^2*c^4*d*f + (3*b^2*c^3*d^2 - 12*a*b*c^2*d^3 + 8*a^2*c*d^4)*e)*x)*sqrt(-b*c*d
+ a*d^2)*log((b*d*x - b*c + 2*a*d + 2*sqrt(-b*c*d + a*d^2)*sqrt(b*x + a))/(d*x + c)) + 8*((b^2*c^2*d^4 - 2*a*
b*c*d^5 + a^2*d^6)*e*x^2 + 2*(b^2*c^3*d^3 - 2*a*b*c^2*d^4 + a^2*c*d^5)*e*x + (b^2*c^4*d^2 - 2*a*b*c^3*d^3 + a^
2*c^2*d^4)*e)*sqrt(a)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*((5*b^2*c^4*d^2 - 11*a*b*c^3*d^3 + 6*a^
2*c^2*d^4)*e - (b^2*c^5*d - 3*a*b*c^4*d^2 + 2*a^2*c^3*d^3)*f + ((3*b^2*c^3*d^3 - 7*a*b*c^2*d^4 + 4*a^2*c*d^5)*
e + (b^2*c^4*d^2 - a*b*c^3*d^3)*f)*x)*sqrt(b*x + a))/(b^2*c^7*d^2 - 2*a*b*c^6*d^3 + a^2*c^5*d^4 + (b^2*c^5*d^4
- 2*a*b*c^4*d^5 + a^2*c^3*d^6)*x^2 + 2*(b^2*c^6*d^3 - 2*a*b*c^5*d^4 + a^2*c^4*d^5)*x), 1/8*(16*((b^2*c^2*d^4
- 2*a*b*c*d^5 + a^2*d^6)*e*x^2 + 2*(b^2*c^3*d^3 - 2*a*b*c^2*d^4 + a^2*c*d^5)*e*x + (b^2*c^4*d^2 - 2*a*b*c^3*d^
3 + a^2*c^2*d^4)*e)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (b^2*c^5*f + (b^2*c^3*d^2*f + (3*b^2*c^2*d^3 -
12*a*b*c*d^4 + 8*a^2*d^5)*e)*x^2 + (3*b^2*c^4*d - 12*a*b*c^3*d^2 + 8*a^2*c^2*d^3)*e + 2*(b^2*c^4*d*f + (3*b^2
*c^3*d^2 - 12*a*b*c^2*d^3 + 8*a^2*c*d^4)*e)*x)*sqrt(-b*c*d + a*d^2)*log((b*d*x - b*c + 2*a*d + 2*sqrt(-b*c*d +
a*d^2)*sqrt(b*x + a))/(d*x + c)) + 2*((5*b^2*c^4*d^2 - 11*a*b*c^3*d^3 + 6*a^2*c^2*d^4)*e - (b^2*c^5*d - 3*a*b
*c^4*d^2 + 2*a^2*c^3*d^3)*f + ((3*b^2*c^3*d^3 - 7*a*b*c^2*d^4 + 4*a^2*c*d^5)*e + (b^2*c^4*d^2 - a*b*c^3*d^3)*f
)*x)*sqrt(b*x + a))/(b^2*c^7*d^2 - 2*a*b*c^6*d^3 + a^2*c^5*d^4 + (b^2*c^5*d^4 - 2*a*b*c^4*d^5 + a^2*c^3*d^6)*x
^2 + 2*(b^2*c^6*d^3 - 2*a*b*c^5*d^4 + a^2*c^4*d^5)*x), -1/4*((b^2*c^5*f + (b^2*c^3*d^2*f + (3*b^2*c^2*d^3 - 12
*a*b*c*d^4 + 8*a^2*d^5)*e)*x^2 + (3*b^2*c^4*d - 12*a*b*c^3*d^2 + 8*a^2*c^2*d^3)*e + 2*(b^2*c^4*d*f + (3*b^2*c^
3*d^2 - 12*a*b*c^2*d^3 + 8*a^2*c*d^4)*e)*x)*sqrt(b*c*d - a*d^2)*arctan(sqrt(b*c*d - a*d^2)*sqrt(b*x + a)/(b*d*
x + a*d)) - 4*((b^2*c^2*d^4 - 2*a*b*c*d^5 + a^2*d^6)*e*x^2 + 2*(b^2*c^3*d^3 - 2*a*b*c^2*d^4 + a^2*c*d^5)*e*x +
(b^2*c^4*d^2 - 2*a*b*c^3*d^3 + a^2*c^2*d^4)*e)*sqrt(a)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - ((5*b^2
*c^4*d^2 - 11*a*b*c^3*d^3 + 6*a^2*c^2*d^4)*e - (b^2*c^5*d - 3*a*b*c^4*d^2 + 2*a^2*c^3*d^3)*f + ((3*b^2*c^3*d^3
- 7*a*b*c^2*d^4 + 4*a^2*c*d^5)*e + (b^2*c^4*d^2 - a*b*c^3*d^3)*f)*x)*sqrt(b*x + a))/(b^2*c^7*d^2 - 2*a*b*c^6*
d^3 + a^2*c^5*d^4 + (b^2*c^5*d^4 - 2*a*b*c^4*d^5 + a^2*c^3*d^6)*x^2 + 2*(b^2*c^6*d^3 - 2*a*b*c^5*d^4 + a^2*c^4
*d^5)*x), -1/4*((b^2*c^5*f + (b^2*c^3*d^2*f + (3*b^2*c^2*d^3 - 12*a*b*c*d^4 + 8*a^2*d^5)*e)*x^2 + (3*b^2*c^4*d
- 12*a*b*c^3*d^2 + 8*a^2*c^2*d^3)*e + 2*(b^2*c^4*d*f + (3*b^2*c^3*d^2 - 12*a*b*c^2*d^3 + 8*a^2*c*d^4)*e)*x)*s
qrt(b*c*d - a*d^2)*arctan(sqrt(b*c*d - a*d^2)*sqrt(b*x + a)/(b*d*x + a*d)) - 8*((b^2*c^2*d^4 - 2*a*b*c*d^5 + a
^2*d^6)*e*x^2 + 2*(b^2*c^3*d^3 - 2*a*b*c^2*d^4 + a^2*c*d^5)*e*x + (b^2*c^4*d^2 - 2*a*b*c^3*d^3 + a^2*c^2*d^4)*
e)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) - ((5*b^2*c^4*d^2 - 11*a*b*c^3*d^3 + 6*a^2*c^2*d^4)*e - (b^2*c^5*
d - 3*a*b*c^4*d^2 + 2*a^2*c^3*d^3)*f + ((3*b^2*c^3*d^3 - 7*a*b*c^2*d^4 + 4*a^2*c*d^5)*e + (b^2*c^4*d^2 - a*b*c
^3*d^3)*f)*x)*sqrt(b*x + a))/(b^2*c^7*d^2 - 2*a*b*c^6*d^3 + a^2*c^5*d^4 + (b^2*c^5*d^4 - 2*a*b*c^4*d^5 + a^2*c
^3*d^6)*x^2 + 2*(b^2*c^6*d^3 - 2*a*b*c^5*d^4 + a^2*c^4*d^5)*x)]

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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((f*x+e)*(b*x+a)**(1/2)/x/(d*x+c)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.82932, size = 406, normalized size = 1.98 \begin{align*} \frac{{\left (b^{2} c^{3} f + 3 \, b^{2} c^{2} d e - 12 \, a b c d^{2} e + 8 \, a^{2} d^{3} e\right )} \arctan \left (\frac{\sqrt{b x + a} d}{\sqrt{b c d - a d^{2}}}\right )}{4 \,{\left (b c^{4} d - a c^{3} d^{2}\right )} \sqrt{b c d - a d^{2}}} + \frac{2 \, a \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right ) e}{\sqrt{-a} c^{3}} - \frac{\sqrt{b x + a} b^{3} c^{3} f -{\left (b x + a\right )}^{\frac{3}{2}} b^{2} c^{2} d f - \sqrt{b x + a} a b^{2} c^{2} d f - 5 \, \sqrt{b x + a} b^{3} c^{2} d e - 3 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{2} c d^{2} e + 9 \, \sqrt{b x + a} a b^{2} c d^{2} e + 4 \,{\left (b x + a\right )}^{\frac{3}{2}} a b d^{3} e - 4 \, \sqrt{b x + a} a^{2} b d^{3} e}{4 \,{\left (b c^{3} d - a c^{2} d^{2}\right )}{\left (b c +{\left (b x + a\right )} d - a d\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(b*x+a)^(1/2)/x/(d*x+c)^3,x, algorithm="giac")

[Out]

1/4*(b^2*c^3*f + 3*b^2*c^2*d*e - 12*a*b*c*d^2*e + 8*a^2*d^3*e)*arctan(sqrt(b*x + a)*d/sqrt(b*c*d - a*d^2))/((b
*c^4*d - a*c^3*d^2)*sqrt(b*c*d - a*d^2)) + 2*a*arctan(sqrt(b*x + a)/sqrt(-a))*e/(sqrt(-a)*c^3) - 1/4*(sqrt(b*x
+ a)*b^3*c^3*f - (b*x + a)^(3/2)*b^2*c^2*d*f - sqrt(b*x + a)*a*b^2*c^2*d*f - 5*sqrt(b*x + a)*b^3*c^2*d*e - 3*
(b*x + a)^(3/2)*b^2*c*d^2*e + 9*sqrt(b*x + a)*a*b^2*c*d^2*e + 4*(b*x + a)^(3/2)*a*b*d^3*e - 4*sqrt(b*x + a)*a^
2*b*d^3*e)/((b*c^3*d - a*c^2*d^2)*(b*c + (b*x + a)*d - a*d)^2)