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

Optimal. Leaf size=163 \[ \frac{c^{3/2} (3 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{5/2}}-\frac{\sqrt{c+d x} (3 b c-2 a d) (b c-a d)}{a^2 b \sqrt{a+b x}}+\frac{2 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{3/2}}-\frac{c (c+d x)^{3/2}}{a x \sqrt{a+b x}} \]

[Out]

-(((3*b*c - 2*a*d)*(b*c - a*d)*Sqrt[c + d*x])/(a^2*b*Sqrt[a + b*x])) - (c*(c + d*x)^(3/2))/(a*x*Sqrt[a + b*x])
+ (c^(3/2)*(3*b*c - 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/a^(5/2) + (2*d^(5/2)*Arc
Tanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/b^(3/2)

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Rubi [A]  time = 0.153911, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {98, 150, 157, 63, 217, 206, 93, 208} \[ \frac{c^{3/2} (3 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{5/2}}-\frac{\sqrt{c+d x} (3 b c-2 a d) (b c-a d)}{a^2 b \sqrt{a+b x}}+\frac{2 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{3/2}}-\frac{c (c+d x)^{3/2}}{a x \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((3*b*c - 2*a*d)*(b*c - a*d)*Sqrt[c + d*x])/(a^2*b*Sqrt[a + b*x])) - (c*(c + d*x)^(3/2))/(a*x*Sqrt[a + b*x])
+ (c^(3/2)*(3*b*c - 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/a^(5/2) + (2*d^(5/2)*Arc
Tanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/b^(3/2)

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 150

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] && IntegersQ[2*m, 2*n, 2*p]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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^2 (a+b x)^{3/2}} \, dx &=-\frac{c (c+d x)^{3/2}}{a x \sqrt{a+b x}}-\frac{\int \frac{\sqrt{c+d x} \left (\frac{1}{2} c (3 b c-5 a d)-a d^2 x\right )}{x (a+b x)^{3/2}} \, dx}{a}\\ &=-\frac{(3 b c-2 a d) (b c-a d) \sqrt{c+d x}}{a^2 b \sqrt{a+b x}}-\frac{c (c+d x)^{3/2}}{a x \sqrt{a+b x}}+\frac{2 \int \frac{-\frac{1}{4} b c^2 (3 b c-5 a d)+\frac{1}{2} a^2 d^3 x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{a^2 b}\\ &=-\frac{(3 b c-2 a d) (b c-a d) \sqrt{c+d x}}{a^2 b \sqrt{a+b x}}-\frac{c (c+d x)^{3/2}}{a x \sqrt{a+b x}}+\frac{d^3 \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{b}-\frac{\left (c^2 (3 b c-5 a d)\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 a^2}\\ &=-\frac{(3 b c-2 a d) (b c-a d) \sqrt{c+d x}}{a^2 b \sqrt{a+b x}}-\frac{c (c+d x)^{3/2}}{a x \sqrt{a+b x}}+\frac{\left (2 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{b^2}-\frac{\left (c^2 (3 b c-5 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{a^2}\\ &=-\frac{(3 b c-2 a d) (b c-a d) \sqrt{c+d x}}{a^2 b \sqrt{a+b x}}-\frac{c (c+d x)^{3/2}}{a x \sqrt{a+b x}}+\frac{c^{3/2} (3 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{5/2}}+\frac{\left (2 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{b^2}\\ &=-\frac{(3 b c-2 a d) (b c-a d) \sqrt{c+d x}}{a^2 b \sqrt{a+b x}}-\frac{c (c+d x)^{3/2}}{a x \sqrt{a+b x}}+\frac{c^{3/2} (3 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{5/2}}+\frac{2 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{3/2}}\\ \end{align*}

Mathematica [C]  time = 3.28075, size = 934, normalized size = 5.73 \[ -\frac{\sqrt{c+d x} \left (-3 d^{7/2} x \sqrt{a+b x} (c+d x)^2 \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right ) a^{7/2}-2 d^3 \sqrt{b c-a d} x (c+d x)^2 \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};\frac{d (a+b x)}{a d-b c}\right ) a^{7/2}+3 d^3 \sqrt{b c-a d} x (c+d x)^2 \sqrt{\frac{b (c+d x)}{b c-a d}} a^{7/2}+\frac{d^3 (b c-a d)^{5/2} x^2 \left (\frac{b (c+d x)}{b c-a d}\right )^{5/2} a^{5/2}}{b}-\frac{7 c d^2 (b c-a d)^{5/2} x \left (\frac{b (c+d x)}{b c-a d}\right )^{5/2} a^{5/2}}{b}+8 b c d^{5/2} x \sqrt{a+b x} (c+d x)^2 \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right ) a^{5/2}+8 b c d^2 \sqrt{b c-a d} x (c+d x)^2 \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};\frac{d (a+b x)}{a d-b c}\right ) a^{5/2}+2 c^3 (b c-a d)^{5/2} \left (\frac{b (c+d x)}{b c-a d}\right )^{5/2} a^{3/2}-3 c d^2 (b c-a d)^{5/2} x^2 \left (\frac{b (c+d x)}{b c-a d}\right )^{5/2} a^{3/2}-2 c^2 d (b c-a d)^{5/2} x \left (\frac{b (c+d x)}{b c-a d}\right )^{5/2} a^{3/2}-9 b^2 c^2 d^{3/2} x \sqrt{a+b x} (c+d x)^2 \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right ) a^{3/2}-6 b^2 c^2 d \sqrt{b c-a d} x (c+d x)^2 \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};\frac{d (a+b x)}{a d-b c}\right ) a^{3/2}+10 b c^{5/2} d (b c-a d)^{3/2} x \sqrt{a+b x} \sqrt{c+d x} \left (\frac{b (c+d x)}{b c-a d}\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right ) a+6 b c^3 (b c-a d)^{5/2} x \left (\frac{b (c+d x)}{b c-a d}\right )^{5/2} \sqrt{a}-6 b^3 c^{7/2} \sqrt{b c-a d} x \sqrt{a+b x} (c+d x)^{3/2} \sqrt{\frac{b (c+d x)}{b c-a d}} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )\right )}{2 a^{5/2} c (b c-a d)^{5/2} x \sqrt{a+b x} \left (\frac{b (c+d x)}{b c-a d}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c + d*x]*(3*a^(7/2)*d^3*Sqrt[b*c - a*d]*x*(c + d*x)^2*Sqrt[(b*(c + d*x))/(b*c - a*d)] + 2*a^(3/2)*c^3*(
b*c - a*d)^(5/2)*((b*(c + d*x))/(b*c - a*d))^(5/2) + 6*Sqrt[a]*b*c^3*(b*c - a*d)^(5/2)*x*((b*(c + d*x))/(b*c -
a*d))^(5/2) - 2*a^(3/2)*c^2*d*(b*c - a*d)^(5/2)*x*((b*(c + d*x))/(b*c - a*d))^(5/2) - (7*a^(5/2)*c*d^2*(b*c -
a*d)^(5/2)*x*((b*(c + d*x))/(b*c - a*d))^(5/2))/b - 3*a^(3/2)*c*d^2*(b*c - a*d)^(5/2)*x^2*((b*(c + d*x))/(b*c
- a*d))^(5/2) + (a^(5/2)*d^3*(b*c - a*d)^(5/2)*x^2*((b*(c + d*x))/(b*c - a*d))^(5/2))/b - 9*a^(3/2)*b^2*c^2*d
^(3/2)*x*Sqrt[a + b*x]*(c + d*x)^2*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]] + 8*a^(5/2)*b*c*d^(5/2)*x*
Sqrt[a + b*x]*(c + d*x)^2*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]] - 3*a^(7/2)*d^(7/2)*x*Sqrt[a + b*x]
*(c + d*x)^2*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]] - 6*b^3*c^(7/2)*Sqrt[b*c - a*d]*x*Sqrt[a + b*x]*
(c + d*x)^(3/2)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] + 10*
a*b*c^(5/2)*d*(b*c - a*d)^(3/2)*x*Sqrt[a + b*x]*Sqrt[c + d*x]*((b*(c + d*x))/(b*c - a*d))^(3/2)*ArcTanh[(Sqrt[
c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] - 6*a^(3/2)*b^2*c^2*d*Sqrt[b*c - a*d]*x*(c + d*x)^2*Hypergeometric2
F1[-3/2, -1/2, 1/2, (d*(a + b*x))/(-(b*c) + a*d)] + 8*a^(5/2)*b*c*d^2*Sqrt[b*c - a*d]*x*(c + d*x)^2*Hypergeome
tric2F1[-3/2, -1/2, 1/2, (d*(a + b*x))/(-(b*c) + a*d)] - 2*a^(7/2)*d^3*Sqrt[b*c - a*d]*x*(c + d*x)^2*Hypergeom
etric2F1[-3/2, -1/2, 1/2, (d*(a + b*x))/(-(b*c) + a*d)]))/(2*a^(5/2)*c*(b*c - a*d)^(5/2)*x*Sqrt[a + b*x]*((b*(
c + d*x))/(b*c - a*d))^(5/2))

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Maple [B]  time = 0.021, size = 502, normalized size = 3.1 \begin{align*} -{\frac{1}{2\,{a}^{2}xb}\sqrt{dx+c} \left ( 5\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}a{b}^{2}{c}^{2}d\sqrt{bd}-3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{b}^{3}{c}^{3}\sqrt{bd}-2\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}{a}^{2}b{d}^{3}\sqrt{ac}+5\,\sqrt{bd}\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) x{a}^{2}b{c}^{2}d-3\,\sqrt{bd}\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) xa{b}^{2}{c}^{3}-2\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) \sqrt{ac}x{a}^{3}{d}^{3}+4\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}x{a}^{2}{d}^{2}-8\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}xabcd+6\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}x{b}^{2}{c}^{2}+2\,ab{c}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)/x^2/(b*x+a)^(3/2),x)

[Out]

-1/2*(d*x+c)^(1/2)*(5*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a*b^2*c^2*d*(b*d)^(1
/2)-3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*b^3*c^3*(b*d)^(1/2)-2*ln(1/2*(2*b*d*
x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*a^2*b*d^3*(a*c)^(1/2)+5*(b*d)^(1/2)*ln((a*d*
x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x*a^2*b*c^2*d-3*(b*d)^(1/2)*ln((a*d*x+b*c*x+2*(a*c)^(1
/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x*a*b^2*c^3-2*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+
b*c)/(b*d)^(1/2))*(a*c)^(1/2)*x*a^3*d^3+4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*x*a^2*d^2-8*((b*x+a)
*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*x*a*b*c*d+6*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*x*b^2*c^2+
2*a*b*c^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2))/a^2/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(a*c)^(1/2)
/x/(b*x+a)^(1/2)/b

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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^2/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 13.904, size = 2691, normalized size = 16.51 \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^2/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(2*(a^2*b*d^2*x^2 + a^3*d^2*x)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b^2*d*x
+ b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) - ((3*b^3*c^2 - 5*a*b^2*c*d
)*x^2 + (3*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a
^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(a*b*c^2 +
(3*b^2*c^2 - 4*a*b*c*d + 2*a^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*b^2*x^2 + a^3*b*x), -1/4*(4*(a^2*b*d
^2*x^2 + a^3*d^2*x)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-d/b)/(b*d^2*
x^2 + a*c*d + (b*c*d + a*d^2)*x)) + ((3*b^3*c^2 - 5*a*b^2*c*d)*x^2 + (3*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(c/a)*
log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x
+ c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(a*b*c^2 + (3*b^2*c^2 - 4*a*b*c*d + 2*a^2*d^2)*x)*sqrt(b*x
+ a)*sqrt(d*x + c))/(a^2*b^2*x^2 + a^3*b*x), -1/2*(((3*b^3*c^2 - 5*a*b^2*c*d)*x^2 + (3*a*b^2*c^2 - 5*a^2*b*c*d
)*x)*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 +
(b*c^2 + a*c*d)*x)) - (a^2*b*d^2*x^2 + a^3*d^2*x)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2
+ 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) + 2*(a*b*c^2
+ (3*b^2*c^2 - 4*a*b*c*d + 2*a^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*b^2*x^2 + a^3*b*x), -1/2*(((3*b^3*
c^2 - 5*a*b^2*c*d)*x^2 + (3*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x
+ a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) + 2*(a^2*b*d^2*x^2 + a^3*d^2*x)*sqrt(-
d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d
^2)*x)) + 2*(a*b*c^2 + (3*b^2*c^2 - 4*a*b*c*d + 2*a^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*b^2*x^2 + a^3*
b*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((d*x+c)**(5/2)/x**2/(b*x+a)**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 3.64328, size = 1407, normalized size = 8.63 \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^2/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

-sqrt(b*d)*d^2*abs(b)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/b^3 + (3*sqrt(b*d
)*b*c^3*abs(b) - 5*sqrt(b*d)*a*c^2*d*abs(b))*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*
c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*a^2*b) - 2*(3*sqrt(b*d)*b^7*c^5*abs(b) - 13
*sqrt(b*d)*a*b^6*c^4*d*abs(b) + 23*sqrt(b*d)*a^2*b^5*c^3*d^2*abs(b) - 21*sqrt(b*d)*a^3*b^4*c^2*d^3*abs(b) + 10
*sqrt(b*d)*a^4*b^3*c*d^4*abs(b) - 2*sqrt(b*d)*a^5*b^2*d^5*abs(b) - 6*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt
(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^5*c^4*abs(b) + 10*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x
+ a)*b*d - a*b*d))^2*a*b^4*c^3*d*abs(b) - 8*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d -
a*b*d))^2*a^3*b^2*c*d^3*abs(b) + 4*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2
*a^4*b*d^4*abs(b) + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^3*c^3*abs(
b) - 5*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^2*c^2*d*abs(b) + 6*sqrt
(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b*c*d^2*abs(b) - 2*sqrt(b*d)*(sqrt
(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*d^3*abs(b))/((b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2
*b^4*c*d^2 - a^3*b^3*d^3 - 3*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^4*c^2 + 2*(sq
rt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^3*c*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2
*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^2*d^2 + 3*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))
^4*b^2*c + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b*d - (sqrt(b*d)*sqrt(b*x + a)
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6)*a^2*b^2)