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

Optimal. Leaf size=257 \[ \frac{\sqrt{a+b x} \sqrt{c+d x} \left (a^2 d^2+26 a b c d+5 b^2 c^2\right )}{8 b}+\frac{\left (15 a^2 b c d^2-a^3 d^3+45 a b^2 c^2 d+5 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{3/2} \sqrt{d}}-\sqrt{a} c^{3/2} (5 a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{x}+\frac{4}{3} b \sqrt{a+b x} (c+d x)^{5/2}+\frac{1}{12} \sqrt{a+b x} (c+d x)^{3/2} (19 a d+5 b c) \]

[Out]

((5*b^2*c^2 + 26*a*b*c*d + a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b) + ((5*b*c + 19*a*d)*Sqrt[a + b*x]*(c +
d*x)^(3/2))/12 + (4*b*Sqrt[a + b*x]*(c + d*x)^(5/2))/3 - ((a + b*x)^(3/2)*(c + d*x)^(5/2))/x - Sqrt[a]*c^(3/2)
*(3*b*c + 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] + ((5*b^3*c^3 + 45*a*b^2*c^2*d + 15*
a^2*b*c*d^2 - a^3*d^3)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(3/2)*Sqrt[d])

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Rubi [A]  time = 0.284188, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {97, 154, 157, 63, 217, 206, 93, 208} \[ \frac{\sqrt{a+b x} \sqrt{c+d x} \left (a^2 d^2+26 a b c d+5 b^2 c^2\right )}{8 b}+\frac{\left (15 a^2 b c d^2-a^3 d^3+45 a b^2 c^2 d+5 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{3/2} \sqrt{d}}-\sqrt{a} c^{3/2} (5 a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{x}+\frac{4}{3} b \sqrt{a+b x} (c+d x)^{5/2}+\frac{1}{12} \sqrt{a+b x} (c+d x)^{3/2} (19 a d+5 b c) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*b^2*c^2 + 26*a*b*c*d + a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b) + ((5*b*c + 19*a*d)*Sqrt[a + b*x]*(c +
d*x)^(3/2))/12 + (4*b*Sqrt[a + b*x]*(c + d*x)^(5/2))/3 - ((a + b*x)^(3/2)*(c + d*x)^(5/2))/x - Sqrt[a]*c^(3/2)
*(3*b*c + 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] + ((5*b^3*c^3 + 45*a*b^2*c^2*d + 15*
a^2*b*c*d^2 - a^3*d^3)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(3/2)*Sqrt[d])

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 154

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] && 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{(a+b x)^{3/2} (c+d x)^{5/2}}{x^2} \, dx &=-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{x}+\int \frac{\sqrt{a+b x} (c+d x)^{3/2} \left (\frac{1}{2} (3 b c+5 a d)+4 b d x\right )}{x} \, dx\\ &=\frac{4}{3} b \sqrt{a+b x} (c+d x)^{5/2}-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{x}+\frac{\int \frac{(c+d x)^{3/2} \left (\frac{3}{2} a d (3 b c+5 a d)+\frac{1}{2} b d (5 b c+19 a d) x\right )}{x \sqrt{a+b x}} \, dx}{3 d}\\ &=\frac{1}{12} (5 b c+19 a d) \sqrt{a+b x} (c+d x)^{3/2}+\frac{4}{3} b \sqrt{a+b x} (c+d x)^{5/2}-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{x}+\frac{\int \frac{\sqrt{c+d x} \left (3 a b c d (3 b c+5 a d)+\frac{3}{4} b d \left (5 b^2 c^2+26 a b c d+a^2 d^2\right ) x\right )}{x \sqrt{a+b x}} \, dx}{6 b d}\\ &=\frac{\left (5 b^2 c^2+26 a b c d+a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 b}+\frac{1}{12} (5 b c+19 a d) \sqrt{a+b x} (c+d x)^{3/2}+\frac{4}{3} b \sqrt{a+b x} (c+d x)^{5/2}-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{x}+\frac{\int \frac{3 a b^2 c^2 d (3 b c+5 a d)+\frac{3}{8} b d \left (5 b^3 c^3+45 a b^2 c^2 d+15 a^2 b c d^2-a^3 d^3\right ) x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{6 b^2 d}\\ &=\frac{\left (5 b^2 c^2+26 a b c d+a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 b}+\frac{1}{12} (5 b c+19 a d) \sqrt{a+b x} (c+d x)^{3/2}+\frac{4}{3} b \sqrt{a+b x} (c+d x)^{5/2}-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{x}+\frac{1}{2} \left (a c^2 (3 b c+5 a d)\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx+\frac{\left (5 b^3 c^3+45 a b^2 c^2 d+15 a^2 b c d^2-a^3 d^3\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{16 b}\\ &=\frac{\left (5 b^2 c^2+26 a b c d+a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 b}+\frac{1}{12} (5 b c+19 a d) \sqrt{a+b x} (c+d x)^{3/2}+\frac{4}{3} b \sqrt{a+b x} (c+d x)^{5/2}-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{x}+\left (a 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 )+\frac{\left (5 b^3 c^3+45 a b^2 c^2 d+15 a^2 b c d^2-a^3 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 )}{8 b^2}\\ &=\frac{\left (5 b^2 c^2+26 a b c d+a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 b}+\frac{1}{12} (5 b c+19 a d) \sqrt{a+b x} (c+d x)^{3/2}+\frac{4}{3} b \sqrt{a+b x} (c+d x)^{5/2}-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{x}-\sqrt{a} 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 )+\frac{\left (5 b^3 c^3+45 a b^2 c^2 d+15 a^2 b c d^2-a^3 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 )}{8 b^2}\\ &=\frac{\left (5 b^2 c^2+26 a b c d+a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 b}+\frac{1}{12} (5 b c+19 a d) \sqrt{a+b x} (c+d x)^{3/2}+\frac{4}{3} b \sqrt{a+b x} (c+d x)^{5/2}-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{x}-\sqrt{a} 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 )+\frac{\left (5 b^3 c^3+45 a b^2 c^2 d+15 a^2 b c d^2-a^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{3/2} \sqrt{d}}\\ \end{align*}

Mathematica [A]  time = 2.18621, size = 253, normalized size = 0.98 \[ \frac{\frac{\sqrt{a+b x} \sqrt{c+d x} \left (3 a^2 d^2 x+2 a b \left (-12 c^2+34 c d x+7 d^2 x^2\right )+b^2 x \left (33 c^2+26 c d x+8 d^2 x^2\right )\right )}{x}+\frac{3 \sqrt{c+d x} \left (15 a^2 b c d^2-a^3 d^3+45 a b^2 c^2 d+5 b^3 c^3\right ) \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{\sqrt{d} \sqrt{b c-a d} \sqrt{\frac{b (c+d x)}{b c-a d}}}-24 \sqrt{a} b c^{3/2} (5 a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{24 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[a + b*x]*Sqrt[c + d*x]*(3*a^2*d^2*x + 2*a*b*(-12*c^2 + 34*c*d*x + 7*d^2*x^2) + b^2*x*(33*c^2 + 26*c*d*x
+ 8*d^2*x^2)))/x + (3*(5*b^3*c^3 + 45*a*b^2*c^2*d + 15*a^2*b*c*d^2 - a^3*d^3)*Sqrt[c + d*x]*ArcSinh[(Sqrt[d]*
Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(Sqrt[d]*Sqrt[b*c - a*d]*Sqrt[(b*(c + d*x))/(b*c - a*d)]) - 24*Sqrt[a]*b*c^(3
/2)*(3*b*c + 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(24*b)

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Maple [B]  time = 0.015, size = 696, normalized size = 2.7 \begin{align*} -{\frac{1}{48\,bx}\sqrt{bx+a}\sqrt{dx+c} \left ( -16\,{x}^{3}{b}^{2}{d}^{2}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}+3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) \sqrt{ac}x{a}^{3}{d}^{3}-45\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) \sqrt{ac}x{a}^{2}bc{d}^{2}-135\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) \sqrt{ac}xa{b}^{2}{c}^{2}d-15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) \sqrt{ac}x{b}^{3}{c}^{3}+120\,\sqrt{bd}\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ) x{a}^{2}b{c}^{2}d+72\,\sqrt{bd}\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ) xa{b}^{2}{c}^{3}-28\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}{x}^{2}ab{d}^{2}-52\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}{x}^{2}{b}^{2}cd-6\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}x{a}^{2}{d}^{2}-136\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}xabcd-66\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}x{b}^{2}{c}^{2}+48\,ab{c}^{2}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac} \right ){\frac{1}{\sqrt{d{x}^{2}b+adx+bcx+ac}}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(-16*x^3*b^2*d^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)+3*l
n(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*(a*c)^(1/2)*x*a^3*d^3-45*ln
(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*(a*c)^(1/2)*x*a^2*b*c*d^2-13
5*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*(a*c)^(1/2)*x*a*b^2*c^2*
d-15*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*(a*c)^(1/2)*x*b^3*c^3
+120*(b*d)^(1/2)*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x*a^2*b*c^2*d+72*(b*d
)^(1/2)*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x*a*b^2*c^3-28*(b*d*x^2+a*d*x+
b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*x^2*a*b*d^2-52*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2
)*x^2*b^2*c*d-6*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*x*a^2*d^2-136*(b*d*x^2+a*d*x+b*c*x+a*c
)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*x*a*b*c*d-66*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*x*b^2*c^2
+48*a*b*c^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2))/b/(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)/(b*d)^(
1/2)/(a*c)^(1/2)/x

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 50.3407, size = 3016, normalized size = 11.74 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(3*(5*b^3*c^3 + 45*a*b^2*c^2*d + 15*a^2*b*c*d^2 - a^3*d^3)*sqrt(b*d)*x*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*
a*b*c*d + a^2*d^2 - 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) -
24*(3*b^3*c^2*d + 5*a*b^2*c*d^2)*sqrt(a*c)*x*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a*c
+ (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(8*b^3*d^3*x^3 - 24
*a*b^2*c^2*d + 2*(13*b^3*c*d^2 + 7*a*b^2*d^3)*x^2 + (33*b^3*c^2*d + 68*a*b^2*c*d^2 + 3*a^2*b*d^3)*x)*sqrt(b*x
+ a)*sqrt(d*x + c))/(b^2*d*x), -1/48*(3*(5*b^3*c^3 + 45*a*b^2*c^2*d + 15*a^2*b*c*d^2 - a^3*d^3)*sqrt(-b*d)*x*a
rctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b
*d^2)*x)) - 12*(3*b^3*c^2*d + 5*a*b^2*c*d^2)*sqrt(a*c)*x*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2
- 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 2*(8*b^3*d
^3*x^3 - 24*a*b^2*c^2*d + 2*(13*b^3*c*d^2 + 7*a*b^2*d^3)*x^2 + (33*b^3*c^2*d + 68*a*b^2*c*d^2 + 3*a^2*b*d^3)*x
)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d*x), 1/96*(48*(3*b^3*c^2*d + 5*a*b^2*c*d^2)*sqrt(-a*c)*x*arctan(1/2*(2*a*
c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) - 3
*(5*b^3*c^3 + 45*a*b^2*c^2*d + 15*a^2*b*c*d^2 - a^3*d^3)*sqrt(b*d)*x*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d +
a^2*d^2 - 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(8*b^3
*d^3*x^3 - 24*a*b^2*c^2*d + 2*(13*b^3*c*d^2 + 7*a*b^2*d^3)*x^2 + (33*b^3*c^2*d + 68*a*b^2*c*d^2 + 3*a^2*b*d^3)
*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d*x), 1/48*(24*(3*b^3*c^2*d + 5*a*b^2*c*d^2)*sqrt(-a*c)*x*arctan(1/2*(2*
a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) -
3*(5*b^3*c^3 + 45*a*b^2*c^2*d + 15*a^2*b*c*d^2 - a^3*d^3)*sqrt(-b*d)*x*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(
-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) + 2*(8*b^3*d^3*x^3 - 24*a*b
^2*c^2*d + 2*(13*b^3*c*d^2 + 7*a*b^2*d^3)*x^2 + (33*b^3*c^2*d + 68*a*b^2*c*d^2 + 3*a^2*b*d^3)*x)*sqrt(b*x + a)
*sqrt(d*x + c))/(b^2*d*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{\frac{3}{2}} \left (c + d x\right )^{\frac{5}{2}}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 3.02074, size = 926, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*d^2*abs(b)/b^2 + (13*b^3*c*d^5*abs(b) -
a*b^2*d^6*abs(b))/(b^4*d^4)) + 3*(11*b^4*c^2*d^4*abs(b) + 14*a*b^3*c*d^5*abs(b) - a^2*b^2*d^6*abs(b))/(b^4*d^4
))*sqrt(b*x + a) - 48*(3*sqrt(b*d)*a*b^2*c^3*abs(b) + 5*sqrt(b*d)*a^2*b*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)*b)
- 96*(sqrt(b*d)*a*b^4*c^4*abs(b) - 2*sqrt(b*d)*a^2*b^3*c^3*d*abs(b) + sqrt(b*d)*a^3*b^2*c^2*d^2*abs(b) - 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^2*c^3*abs(b) - sqrt(b*d)*(sqrt(b*d
)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b*c^2*d*abs(b))/(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*
d^2 - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt(b*x + a) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)
)^4) - 3*(5*sqrt(b*d)*b^3*c^3*abs(b) + 45*sqrt(b*d)*a*b^2*c^2*d*abs(b) + 15*sqrt(b*d)*a^2*b*c*d^2*abs(b) - sqr
t(b*d)*a^3*d^3*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(b^2*d))/b