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

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

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Rubi [A]  time = 0.28, antiderivative size = 259, 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 (-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} \end {gather*}

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 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)^{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*}

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Mathematica [A]  time = 1.41, size = 261, normalized size = 1.01 \begin {gather*} \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 (-5 a^3 d^3-45 a^2 b c d^2-15 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}} \end {gather*}

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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IntegrateAlgebraic [B]  time = 0.75, size = 585, normalized size = 2.26 \begin {gather*} \left (-5 a^{3/2} b c^{3/2}-3 a^{5/2} \sqrt {c} d\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (5 a^3 d^3+45 a^2 b c d^2+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}}-\frac {\sqrt {a+b x} \left (-15 a^4 b^2 d^3-\frac {33 a^4 d^5 (a+b x)^2}{(c+d x)^2}+\frac {40 a^4 b d^4 (a+b x)}{c+d x}-63 a^3 b^3 c d^2+\frac {159 a^3 b^2 c d^3 (a+b x)}{c+d x}+\frac {57 a^3 c d^5 (a+b x)^3}{(c+d x)^3}-\frac {121 a^3 b c d^4 (a+b x)^2}{(c+d x)^2}+75 a^2 b^4 c^2 d-\frac {153 a^2 b^3 c^2 d^2 (a+b x)}{c+d x}+\frac {45 a^2 b^2 c^2 d^3 (a+b x)^2}{(c+d x)^2}-\frac {15 a^2 b c^2 d^4 (a+b x)^3}{(c+d x)^3}-\frac {3 b^5 c^4 (a+b x)}{c+d x}+3 a b^5 c^3-\frac {8 b^4 c^4 d (a+b x)^2}{(c+d x)^2}-\frac {43 a b^4 c^3 d (a+b x)}{c+d x}+\frac {3 b^3 c^4 d^2 (a+b x)^3}{(c+d x)^3}+\frac {117 a b^3 c^3 d^2 (a+b x)^2}{(c+d x)^2}-\frac {45 a b^2 c^3 d^3 (a+b x)^3}{(c+d x)^3}\right )}{24 d \sqrt {c+d x} \left (a-\frac {c (a+b x)}{c+d x}\right ) \left (\frac {d (a+b x)}{c+d x}-b\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(Sqrt[a + b*x]*(3*a*b^5*c^3 + 75*a^2*b^4*c^2*d - 63*a^3*b^3*c*d^2 - 15*a^4*b^2*d^3 + (3*b^3*c^4*d^2*(a +
b*x)^3)/(c + d*x)^3 - (45*a*b^2*c^3*d^3*(a + b*x)^3)/(c + d*x)^3 - (15*a^2*b*c^2*d^4*(a + b*x)^3)/(c + d*x)^3
+ (57*a^3*c*d^5*(a + b*x)^3)/(c + d*x)^3 - (8*b^4*c^4*d*(a + b*x)^2)/(c + d*x)^2 + (117*a*b^3*c^3*d^2*(a + b*
x)^2)/(c + d*x)^2 + (45*a^2*b^2*c^2*d^3*(a + b*x)^2)/(c + d*x)^2 - (121*a^3*b*c*d^4*(a + b*x)^2)/(c + d*x)^2 -
(33*a^4*d^5*(a + b*x)^2)/(c + d*x)^2 - (3*b^5*c^4*(a + b*x))/(c + d*x) - (43*a*b^4*c^3*d*(a + b*x))/(c + d*x)
- (153*a^2*b^3*c^2*d^2*(a + b*x))/(c + d*x) + (159*a^3*b^2*c*d^3*(a + b*x))/(c + d*x) + (40*a^4*b*d^4*(a + b*
x))/(c + d*x)))/(d*Sqrt[c + d*x]*(a - (c*(a + b*x))/(c + d*x))*(-b + (d*(a + b*x))/(c + d*x))^3) + (-5*a^(3/2)
*b*c^(3/2) - 3*a^(5/2)*Sqrt[c]*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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fricas [A]  time = 12.73, size = 1333, normalized size = 5.15

result too large to display

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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giac [B]  time = 5.32, size = 672, normalized size = 2.59 \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 {\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 {gather*}

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

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maple [B]  time = 0.02, size = 696, normalized size = 2.69 \begin {gather*} \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-72 \sqrt {b d}\, a^{3} c \,d^{2} 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 )+15 \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 )+135 \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 )+45 \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 )-3 \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}+52 \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}+28 \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}+66 \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 +6 \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^{2} c d \right )}{48 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, d x} \end {gather*}

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*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*b^2*d^2*x^3+15*(a
*c)^(1/2)*a^3*d^3*x*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))+135*(a
*c)^(1/2)*a^2*b*c*d^2*x*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))+45
*(a*c)^(1/2)*a*b^2*c^2*d*x*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))
-3*(a*c)^(1/2)*b^3*c^3*x*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))-7
2*(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^3*c*d^2-120*(b*d)^(1
/2)*a^2*b*c^2*d*x*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)+52*(b*d*x^2+a*d*x+b*
c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a*b*d^2*x^2+28*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*
b^2*c*d*x^2+66*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^2*d^2*x+136*(b*d*x^2+a*d*x+b*c*x+a*c)
^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a*b*c*d*x+6*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*b^2*c^2*x-4
8*a^2*c*d*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2))/d/(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)^(5/2)*(d*x+c)^(3/2)/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 )}^{5/2}\,{\left (c+d\,x\right )}^{3/2}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((a + b*x)^(5/2)*(c + d*x)^(3/2))/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 {5}{2}} \left (c + d x\right )^{\frac {3}{2}}}{x^{2}}\, dx \end {gather*}

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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