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

Optimal. Leaf size=259 \[ -\frac{\sqrt{a+b x} \sqrt{c+d x} \left (-19 a^2 d^2-14 a b c d+b^2 c^2\right )}{8 d}-\frac{\left (-45 a^2 b c d^2-5 a^3 d^3-15 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 \sqrt{b} d^{3/2}}-a^{3/2} \sqrt{c} (3 a d+5 b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{x}+\frac{4}{3} b (a+b x)^{3/2} (c+d x)^{3/2}+\frac{b \sqrt{a+b x} (c+d x)^{3/2} (7 a d+b c)}{4 d} \]

[Out]

-((b^2*c^2 - 14*a*b*c*d - 19*a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*d) + (b*(b*c + 7*a*d)*Sqrt[a + b*x]*(c +
d*x)^(3/2))/(4*d) + (4*b*(a + b*x)^(3/2)*(c + d*x)^(3/2))/3 - ((a + b*x)^(5/2)*(c + d*x)^(3/2))/x - a^(3/2)*S
qrt[c]*(5*b*c + 3*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] - ((b^3*c^3 - 15*a*b^2*c^2*d -
45*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*Sqrt[b]*d^(3/2))

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Rubi [A]  time = 0.280158, antiderivative size = 259, 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} \[ -\frac{\sqrt{a+b x} \sqrt{c+d x} \left (-19 a^2 d^2-14 a b c d+b^2 c^2\right )}{8 d}-\frac{\left (-45 a^2 b c d^2-5 a^3 d^3-15 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 \sqrt{b} d^{3/2}}-a^{3/2} \sqrt{c} (3 a d+5 b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{x}+\frac{4}{3} b (a+b x)^{3/2} (c+d x)^{3/2}+\frac{b \sqrt{a+b x} (c+d x)^{3/2} (7 a d+b c)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2*c^2 - 14*a*b*c*d - 19*a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*d) + (b*(b*c + 7*a*d)*Sqrt[a + b*x]*(c +
d*x)^(3/2))/(4*d) + (4*b*(a + b*x)^(3/2)*(c + d*x)^(3/2))/3 - ((a + b*x)^(5/2)*(c + d*x)^(3/2))/x - a^(3/2)*S
qrt[c]*(5*b*c + 3*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] - ((b^3*c^3 - 15*a*b^2*c^2*d -
45*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*Sqrt[b]*d^(3/2))

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)^{5/2} (c+d x)^{3/2}}{x^2} \, dx &=-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{x}+\int \frac{(a+b x)^{3/2} \sqrt{c+d x} \left (\frac{1}{2} (5 b c+3 a d)+4 b d x\right )}{x} \, dx\\ &=\frac{4}{3} b (a+b x)^{3/2} (c+d x)^{3/2}-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{x}+\frac{\int \frac{\sqrt{a+b x} \sqrt{c+d x} \left (\frac{3}{2} a d (5 b c+3 a d)+\frac{3}{2} b d (b c+7 a d) x\right )}{x} \, dx}{3 d}\\ &=\frac{b (b c+7 a d) \sqrt{a+b x} (c+d x)^{3/2}}{4 d}+\frac{4}{3} b (a+b x)^{3/2} (c+d x)^{3/2}-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{x}+\frac{\int \frac{\sqrt{c+d x} \left (3 a^2 d^2 (5 b c+3 a d)-\frac{3}{4} b d \left (b^2 c^2-14 a b c d-19 a^2 d^2\right ) x\right )}{x \sqrt{a+b x}} \, dx}{6 d^2}\\ &=-\frac{\left (b^2 c^2-14 a b c d-19 a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 d}+\frac{b (b c+7 a d) \sqrt{a+b x} (c+d x)^{3/2}}{4 d}+\frac{4}{3} b (a+b x)^{3/2} (c+d x)^{3/2}-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{x}+\frac{\int \frac{3 a^2 b c d^2 (5 b c+3 a d)-\frac{3}{8} b d \left (b^3 c^3-15 a b^2 c^2 d-45 a^2 b c d^2-5 a^3 d^3\right ) x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{6 b d^2}\\ &=-\frac{\left (b^2 c^2-14 a b c d-19 a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 d}+\frac{b (b c+7 a d) \sqrt{a+b x} (c+d x)^{3/2}}{4 d}+\frac{4}{3} b (a+b x)^{3/2} (c+d x)^{3/2}-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{x}+\frac{1}{2} \left (a^2 c (5 b c+3 a d)\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx-\frac{\left (b^3 c^3-15 a b^2 c^2 d-45 a^2 b c d^2-5 a^3 d^3\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{16 d}\\ &=-\frac{\left (b^2 c^2-14 a b c d-19 a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 d}+\frac{b (b c+7 a d) \sqrt{a+b x} (c+d x)^{3/2}}{4 d}+\frac{4}{3} b (a+b x)^{3/2} (c+d x)^{3/2}-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{x}+\left (a^2 c (5 b c+3 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 (b^3 c^3-15 a b^2 c^2 d-45 a^2 b c d^2-5 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 d}\\ &=-\frac{\left (b^2 c^2-14 a b c d-19 a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 d}+\frac{b (b c+7 a d) \sqrt{a+b x} (c+d x)^{3/2}}{4 d}+\frac{4}{3} b (a+b x)^{3/2} (c+d x)^{3/2}-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{x}-a^{3/2} \sqrt{c} (5 b c+3 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )-\frac{\left (b^3 c^3-15 a b^2 c^2 d-45 a^2 b c d^2-5 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 d}\\ &=-\frac{\left (b^2 c^2-14 a b c d-19 a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 d}+\frac{b (b c+7 a d) \sqrt{a+b x} (c+d x)^{3/2}}{4 d}+\frac{4}{3} b (a+b x)^{3/2} (c+d x)^{3/2}-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{x}-a^{3/2} \sqrt{c} (5 b c+3 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )-\frac{\left (b^3 c^3-15 a b^2 c^2 d-45 a^2 b c d^2-5 a^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 \sqrt{b} d^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.42406, size = 261, normalized size = 1.01 \[ \frac{\frac{\sqrt{d} \left (\sqrt{a+b x} (c+d x) \left (3 a^2 d (11 d x-8 c)+2 a b d x (34 c+13 d x)+b^2 x \left (3 c^2+14 c d x+8 d^2 x^2\right )\right )-24 a^{3/2} \sqrt{c} d x \sqrt{c+d x} (3 a d+5 b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )\right )}{x}-\frac{3 \sqrt{b c-a d} \left (-45 a^2 b c d^2-5 a^3 d^3-15 a b^2 c^2 d+b^3 c^3\right ) \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{b}}{24 d^{3/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*Sqrt[b*c - a*d]*(b^3*c^3 - 15*a*b^2*c^2*d - 45*a^2*b*c*d^2 - 5*a^3*d^3)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*A
rcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/b + (Sqrt[d]*(Sqrt[a + b*x]*(c + d*x)*(3*a^2*d*(-8*c + 11*d*x
) + 2*a*b*d*x*(34*c + 13*d*x) + b^2*x*(3*c^2 + 14*c*d*x + 8*d^2*x^2)) - 24*a^(3/2)*Sqrt[c]*d*(5*b*c + 3*a*d)*x
*Sqrt[c + d*x]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/x)/(24*d^(3/2)*Sqrt[c + d*x])

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Maple [B]  time = 0.017, size = 696, normalized size = 2.7 \begin{align*}{\frac{1}{48\,dx}\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}+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{a}^{3}{d}^{3}+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}x{a}^{2}bc{d}^{2}+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}xa{b}^{2}{c}^{2}d-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{b}^{3}{c}^{3}-72\,\sqrt{bd}\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ) x{a}^{3}c{d}^{2}-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+52\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}{x}^{2}ab{d}^{2}+28\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}{x}^{2}{b}^{2}cd+66\,\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+6\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}x{b}^{2}{c}^{2}-48\,{a}^{2}cd\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd} \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)^(5/2)*(d*x+c)^(3/2)/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)+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*a^3*d^3+135*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+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*b^2*c^2*d
-3*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^3-7
2*(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^3*c*d^2-120*(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+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*a*b*d^2+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*b^2*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*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+6*(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^2-4
8*a^2*c*d*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/d/(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)^(5/2)*(d*x+c)^(3/2)/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 47.4538, size = 3015, normalized size = 11.64 \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)^(5/2)*(d*x+c)^(3/2)/x^2,x, algorithm="fricas")

[Out]

[-1/96*(3*(b^3*c^3 - 15*a*b^2*c^2*d - 45*a^2*b*c*d^2 - 5*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*(5*a*b^2*c*d^2 + 3*a^2*b*d^3)*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^2*b*c*d^2 + 2*(7*b^3*c*d^2 + 13*a*b^2*d^3)*x^2 + (3*b^3*c^2*d + 68*a*b^2*c*d^2 + 33*a^2*b*d^3)*x)*sqrt(b*x
+ a)*sqrt(d*x + c))/(b*d^2*x), 1/48*(3*(b^3*c^3 - 15*a*b^2*c^2*d - 45*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(-b*d)*x*ar
ctan(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*(5*a*b^2*c*d^2 + 3*a^2*b*d^3)*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^2*b*c*d^2 + 2*(7*b^3*c*d^2 + 13*a*b^2*d^3)*x^2 + (3*b^3*c^2*d + 68*a*b^2*c*d^2 + 33*a^2*b*d^3)*x)
*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^2*x), 1/96*(48*(5*a*b^2*c*d^2 + 3*a^2*b*d^3)*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*
(b^3*c^3 - 15*a*b^2*c^2*d - 45*a^2*b*c*d^2 - 5*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^2*b*c*d^2 + 2*(7*b^3*c*d^2 + 13*a*b^2*d^3)*x^2 + (3*b^3*c^2*d + 68*a*b^2*c*d^2 + 33*a^2*b*d^3)*
x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^2*x), 1/48*(24*(5*a*b^2*c*d^2 + 3*a^2*b*d^3)*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*(b^3*c^3 - 15*a*b^2*c^2*d - 45*a^2*b*c*d^2 - 5*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^2*
b*c*d^2 + 2*(7*b^3*c*d^2 + 13*a*b^2*d^3)*x^2 + (3*b^3*c^2*d + 68*a*b^2*c*d^2 + 33*a^2*b*d^3)*x)*sqrt(b*x + a)*
sqrt(d*x + c))/(b*d^2*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.41589, size = 907, normalized size = 3.5 \begin{align*} \frac{2 \, \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}{\left (2 \,{\left (b x + a\right )}{\left (\frac{4 \,{\left (b x + a\right )} d{\left | b \right |}}{b} + \frac{7 \, b c d^{4}{\left | b \right |} + 5 \, a d^{5}{\left | b \right |}}{b d^{4}}\right )} + \frac{3 \,{\left (b^{2} c^{2} d^{3}{\left | b \right |} + 18 \, a b c d^{4}{\left | b \right |} + 5 \, a^{2} d^{5}{\left | b \right |}\right )}}{b d^{4}}\right )} \sqrt{b x + a} - \frac{48 \,{\left (5 \, \sqrt{b d} a^{2} b^{2} c^{2}{\left | b \right |} + 3 \, \sqrt{b d} a^{3} b c d{\left | b \right |}\right )} \arctan \left (-\frac{b^{2} c + a b d -{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt{-a b c d} b}\right )}{\sqrt{-a b c d} b} - \frac{96 \,{\left (\sqrt{b d} a^{2} b^{4} c^{3}{\left | b \right |} - 2 \, \sqrt{b d} a^{3} b^{3} c^{2} d{\left | b \right |} + \sqrt{b d} a^{4} b^{2} c d^{2}{\left | b \right |} - \sqrt{b d}{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{2} c^{2}{\left | b \right |} - \sqrt{b d}{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} a^{3} b c d{\left | b \right |}\right )}}{b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \,{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \,{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} a b d +{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{4}} + \frac{3 \,{\left (\sqrt{b d} b^{3} c^{3}{\left | b \right |} - 15 \, \sqrt{b d} a b^{2} c^{2} d{\left | b \right |} - 45 \, \sqrt{b d} a^{2} b c d^{2}{\left | b \right |} - 5 \, \sqrt{b d} a^{3} d^{3}{\left | b \right |}\right )} \log \left ({\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{b d^{2}}}{48 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(3/2)/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*abs(b)/b + (7*b*c*d^4*abs(b) + 5*a*d^5
*abs(b))/(b*d^4)) + 3*(b^2*c^2*d^3*abs(b) + 18*a*b*c*d^4*abs(b) + 5*a^2*d^5*abs(b))/(b*d^4))*sqrt(b*x + a) - 4
8*(5*sqrt(b*d)*a^2*b^2*c^2*abs(b) + 3*sqrt(b*d)*a^3*b*c*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^2
*b^4*c^3*abs(b) - 2*sqrt(b*d)*a^3*b^3*c^2*d*abs(b) + sqrt(b*d)*a^4*b^2*c*d^2*abs(b) - sqrt(b*d)*(sqrt(b*d)*sqr
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^2*c^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^3*b*c*d*abs(b))/(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*d)*s
qrt(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*(sqrt(b*d)*
b^3*c^3*abs(b) - 15*sqrt(b*d)*a*b^2*c^2*d*abs(b) - 45*sqrt(b*d)*a^2*b*c*d^2*abs(b) - 5*sqrt(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*d^2))/b