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

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

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Rubi [A]  time = 0.28, antiderivative size = 257, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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) \end {gather*}

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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 2.50, size = 253, normalized size = 0.98 \begin {gather*} \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 (-a^3 d^3+15 a^2 b c d^2+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} \end {gather*}

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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IntegrateAlgebraic [B]  time = 0.80, size = 585, normalized size = 2.28 \begin {gather*} \left (-5 a^{3/2} c^{3/2} d-3 \sqrt {a} b c^{5/2}\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )+\frac {\left (-a^3 d^3+15 a^2 b c d^2+45 a b^2 c^2 d+5 b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 b^{3/2} \sqrt {d}}-\frac {\sqrt {c+d x} \left (\frac {3 a^4 b^2 d^3 (c+d x)^3}{(a+b x)^3}-\frac {3 a^4 d^5 (c+d x)}{a+b x}-\frac {8 a^4 b d^4 (c+d x)^2}{(a+b x)^2}-\frac {45 a^3 b^3 c d^2 (c+d x)^3}{(a+b x)^3}+\frac {117 a^3 b^2 c d^3 (c+d x)^2}{(a+b x)^2}-\frac {43 a^3 b c d^4 (c+d x)}{a+b x}+3 a^3 c d^5-\frac {15 a^2 b^4 c^2 d (c+d x)^3}{(a+b x)^3}+\frac {45 a^2 b^3 c^2 d^2 (c+d x)^2}{(a+b x)^2}-\frac {153 a^2 b^2 c^2 d^3 (c+d x)}{a+b x}+75 a^2 b c^2 d^4-\frac {33 b^5 c^4 (c+d x)^2}{(a+b x)^2}+\frac {57 a b^5 c^3 (c+d x)^3}{(a+b x)^3}+\frac {40 b^4 c^4 d (c+d x)}{a+b x}-\frac {121 a b^4 c^3 d (c+d x)^2}{(a+b x)^2}+\frac {159 a b^3 c^3 d^2 (c+d x)}{a+b x}-63 a b^2 c^3 d^3-15 b^3 c^4 d^2\right )}{24 b \sqrt {a+b x} \left (c-\frac {a (c+d x)}{a+b x}\right ) \left (\frac {b (c+d x)}{a+b x}-d\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(Sqrt[c + d*x]*(-15*b^3*c^4*d^2 - 63*a*b^2*c^3*d^3 + 75*a^2*b*c^2*d^4 + 3*a^3*c*d^5 + (40*b^4*c^4*d*(c +
d*x))/(a + b*x) + (159*a*b^3*c^3*d^2*(c + d*x))/(a + b*x) - (153*a^2*b^2*c^2*d^3*(c + d*x))/(a + b*x) - (43*a
^3*b*c*d^4*(c + d*x))/(a + b*x) - (3*a^4*d^5*(c + d*x))/(a + b*x) - (33*b^5*c^4*(c + d*x)^2)/(a + b*x)^2 - (12
1*a*b^4*c^3*d*(c + d*x)^2)/(a + b*x)^2 + (45*a^2*b^3*c^2*d^2*(c + d*x)^2)/(a + b*x)^2 + (117*a^3*b^2*c*d^3*(c
+ d*x)^2)/(a + b*x)^2 - (8*a^4*b*d^4*(c + d*x)^2)/(a + b*x)^2 + (57*a*b^5*c^3*(c + d*x)^3)/(a + b*x)^3 - (15*a
^2*b^4*c^2*d*(c + d*x)^3)/(a + b*x)^3 - (45*a^3*b^3*c*d^2*(c + d*x)^3)/(a + b*x)^3 + (3*a^4*b^2*d^3*(c + d*x)^
3)/(a + b*x)^3))/(b*Sqrt[a + b*x]*(c - (a*(c + d*x))/(a + b*x))*(-d + (b*(c + d*x))/(a + b*x))^3) + (-3*Sqrt[a
]*b*c^(5/2) - 5*a^(3/2)*c^(3/2)*d)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*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[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(8*b^(3/2)*
Sqrt[d])

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fricas [A]  time = 21.23, size = 1337, normalized size = 5.20

result too large to display

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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giac [B]  time = 5.63, size = 686, normalized size = 2.67 \begin {gather*} \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^{2} {\left | b \right |}}{b^{2}} + \frac {13 \, b^{3} c d^{5} {\left | b \right |} - a b^{2} d^{6} {\left | b \right |}}{b^{4} d^{4}}\right )} + \frac {3 \, {\left (11 \, b^{4} c^{2} d^{4} {\left | b \right |} + 14 \, a b^{3} c d^{5} {\left | b \right |} - a^{2} b^{2} d^{6} {\left | b \right |}\right )}}{b^{4} d^{4}}\right )} \sqrt {b x + a} - \frac {48 \, {\left (3 \, \sqrt {b d} a b^{2} c^{3} {\left | b \right |} + 5 \, \sqrt {b d} a^{2} b c^{2} 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 b^{4} c^{4} {\left | b \right |} - 2 \, \sqrt {b d} a^{2} b^{3} c^{3} d {\left | b \right |} + \sqrt {b d} a^{3} b^{2} c^{2} 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 b^{2} c^{3} {\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 c^{2} 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 (5 \, \sqrt {b d} b^{3} c^{3} {\left | b \right |} + 45 \, \sqrt {b d} a b^{2} c^{2} d {\left | b \right |} + 15 \, \sqrt {b d} a^{2} b c d^{2} {\left | b \right |} - \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^{2} d}}{48 \, b} \end {gather*}

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

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maple [B]  time = 0.02, size = 696, normalized size = 2.71 \begin {gather*} -\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (3 \sqrt {a c}\, a^{3} d^{3} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+120 \sqrt {b d}\, a^{2} b \,c^{2} d x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-45 \sqrt {a c}\, a^{2} b c \,d^{2} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+72 \sqrt {b d}\, a \,b^{2} c^{3} x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-135 \sqrt {a c}\, a \,b^{2} c^{2} d x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-15 \sqrt {a c}\, b^{3} c^{3} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-16 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{2} d^{2} x^{3}-28 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a b \,d^{2} x^{2}-52 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{2} c d \,x^{2}-6 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} d^{2} x -136 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a b c d x -66 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{2} c^{2} x +48 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a b \,c^{2}\right )}{48 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b x} \end {gather*}

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+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*(a*c)^(1/2)*x*a^3*d^3-45*ln
(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*(a*c)^(1/2)*x*a^2*b*c*d^2-13
5*ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*(a*c)^(1/2)*x*a*b^2*c^2*
d-15*ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/(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+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2))/x)*x*a^2*b*c^2*d+72*(b*d
)^(1/2)*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2))/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)/x/(a*c)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{5/2}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {5}{2}}}{x^{2}}\, dx \end {gather*}

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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