3.572 \(\int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=227 \[ \frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{8 a^2 x}-\frac {\left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} \sqrt {c}}+2 \sqrt {b} d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{12 a x^2} \]

[Out]

2*d^(5/2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))*b^(1/2)-1/8*(5*a^3*d^3+15*a^2*b*c*d^2-5*a*b^2*c
^2*d+b^3*c^3)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(5/2)/c^(1/2)-1/12*(5*a*d+b*c)*(d*x+c)^(3
/2)*(b*x+a)^(1/2)/a/x^2-1/3*(d*x+c)^(5/2)*(b*x+a)^(1/2)/x^3+1/8*(-5*a*d+b*c)*(a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(
1/2)/a^2/x

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Rubi [A]  time = 0.19, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {97, 149, 157, 63, 217, 206, 93, 208} \[ -\frac {\left (15 a^2 b c d^2+5 a^3 d^3-5 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} \sqrt {c}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{8 a^2 x}+2 \sqrt {b} d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{12 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x]*(c + d*x)^(5/2))/x^4,x]

[Out]

((b*c - 5*a*d)*(b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*a^2*x) - ((b*c + 5*a*d)*Sqrt[a + b*x]*(c + d*x)^(3/
2))/(12*a*x^2) - (Sqrt[a + b*x]*(c + d*x)^(5/2))/(3*x^3) - ((b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*
d^3)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(8*a^(5/2)*Sqrt[c]) + 2*Sqrt[b]*d^(5/2)*ArcTanh
[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*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 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 149

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] && IntegerQ[m]

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

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Mathematica [A]  time = 2.26, size = 233, normalized size = 1.03 \[ -\frac {\sqrt {a+b x} \sqrt {c+d x} \left (a^2 \left (8 c^2+26 c d x+33 d^2 x^2\right )+2 a b c x (c+7 d x)-3 b^2 c^2 x^2\right )}{24 a^2 x^3}-\frac {\left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} \sqrt {c}}+\frac {2 d^{5/2} \sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x]*(c + d*x)^(5/2))/x^4,x]

[Out]

-1/24*(Sqrt[a + b*x]*Sqrt[c + d*x]*(-3*b^2*c^2*x^2 + 2*a*b*c*x*(c + 7*d*x) + a^2*(8*c^2 + 26*c*d*x + 33*d^2*x^
2)))/(a^2*x^3) + (2*d^(5/2)*Sqrt[b*c - a*d]*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sq
rt[b*c - a*d]])/Sqrt[c + d*x] - ((b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3)*ArcTanh[(Sqrt[c]*Sqrt[
a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(8*a^(5/2)*Sqrt[c])

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fricas [A]  time = 5.35, size = 1245, normalized size = 5.48 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^4,x, algorithm="fricas")

[Out]

[1/96*(48*sqrt(b*d)*a^3*c*d^2*x^3*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) + 3*(b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2
+ 5*a^3*d^3)*sqrt(a*c)*x^3*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)*s
qrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(8*a^3*c^3 - (3*a*b^2*c^3 - 14*a^2*b*
c^2*d - 33*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 + 13*a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*c*x^3), -1/96*(9
6*sqrt(-b*d)*a^3*c*d^2*x^3*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)) - 3*(b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(a*c)*x^3
*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*a^3*c^3 - (3*a*b^2*c^3 - 14*a^2*b*c^2*d - 33*a^3*c*d^2)*x^2 +
2*(a^2*b*c^3 + 13*a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*c*x^3), 1/48*(24*sqrt(b*d)*a^3*c*d^2*x^3*lo
g(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) + 3*(b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(-a*c)*x^3*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)) - 2*(8*a^3*c^3 - (3*a*b^2*c^3 - 14*a^2*b*c^2*d - 33*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 + 13*a^3*c^2*d)*x)*sqr
t(b*x + a)*sqrt(d*x + c))/(a^3*c*x^3), -1/48*(48*sqrt(-b*d)*a^3*c*d^2*x^3*arctan(1/2*(2*b*d*x + b*c + a*d)*sqr
t(-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)) - 3*(b^3*c^3 - 5*a*b^2*c^
2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(-a*c)*x^3*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*s
qrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) + 2*(8*a^3*c^3 - (3*a*b^2*c^3 - 14*a^2*b*c^2*d -
33*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 + 13*a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*c*x^3)]

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giac [B]  time = 98.29, size = 2265, normalized size = 9.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^4,x, algorithm="giac")

[Out]

-1/24*(24*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) + 3*(sqr
t(b*d)*b^4*c^3*abs(b) - 5*sqrt(b*d)*a*b^3*c^2*d*abs(b) + 15*sqrt(b*d)*a^2*b^2*c*d^2*abs(b) + 5*sqrt(b*d)*a^3*b
*d^3*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^14*c^8*abs(b) - 32*sqrt(b*d)*a*b^13*c^7*d*abs(b)
+ 96*sqrt(b*d)*a^2*b^12*c^6*d^2*abs(b) - 72*sqrt(b*d)*a^3*b^11*c^5*d^3*abs(b) - 170*sqrt(b*d)*a^4*b^10*c^4*d^4
*abs(b) + 432*sqrt(b*d)*a^5*b^9*c^3*d^5*abs(b) - 408*sqrt(b*d)*a^6*b^8*c^2*d^6*abs(b) + 184*sqrt(b*d)*a^7*b^7*
c*d^7*abs(b) - 33*sqrt(b*d)*a^8*b^6*d^8*abs(b) - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a
)*b*d - a*b*d))^2*b^12*c^7*abs(b) + 111*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^11*c^6*d*abs(b) - 111*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^10*c^5*d^2*abs(b) - 201*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^
9*c^4*d^3*abs(b) + 219*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^8*c^3
*d^4*abs(b) + 309*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^5*b^7*c^2*d^5*
abs(b) - 477*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^6*b^6*c*d^6*abs(b)
+ 165*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^7*b^5*d^7*abs(b) + 30*sqrt
(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^10*c^6*abs(b) - 156*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^9*c^5*d*abs(b) - 198*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^8*c^4*d^2*abs(b) + 312*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*b^7*c^3*d^3*abs(b) + 114*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a)
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^4*b^6*c^2*d^4*abs(b) + 228*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqr
t(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^5*b^5*c*d^5*abs(b) - 330*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c
+ (b*x + a)*b*d - a*b*d))^4*a^6*b^4*d^6*abs(b) - 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x +
a)*b*d - a*b*d))^6*b^8*c^5*abs(b) + 122*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*
d))^6*a*b^7*c^4*d*abs(b) + 396*sqrt(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^6*c^3*d^2*abs(b) + 252*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^3*b^5*
c^2*d^3*abs(b) + 338*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^4*b^4*c*d^4
*abs(b) + 330*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^5*b^3*d^5*abs(b) +
15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*b^6*c^4*abs(b) - 60*sqrt(b*d)*
(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a*b^5*c^3*d*abs(b) - 282*sqrt(b*d)*(sqrt(b*d
)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^2*b^4*c^2*d^2*abs(b) - 372*sqrt(b*d)*(sqrt(b*d)*sqr
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^3*b^3*c*d^3*abs(b) - 165*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x +
a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^4*b^2*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))^10*b^4*c^3*abs(b) + 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x
+ a)*b*d - a*b*d))^10*a*b^3*c^2*d*abs(b) + 99*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d
- a*b*d))^10*a^2*b^2*c*d^2*abs(b) + 33*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d
))^10*a^3*b*d^3*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 + (sqr
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)^3*a^2))/b

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maple [B]  time = 0.02, size = 601, normalized size = 2.65 \[ \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-15 \sqrt {b d}\, a^{3} d^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-45 \sqrt {b d}\, a^{2} b c \,d^{2} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+48 \sqrt {a c}\, a^{2} b \,d^{3} x^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+15 \sqrt {b d}\, a \,b^{2} c^{2} d \,x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-3 \sqrt {b d}\, b^{3} c^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-66 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} d^{2} x^{2}-28 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a b c d \,x^{2}+6 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{2} c^{2} x^{2}-52 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} c d x -4 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a b \,c^{2} x -16 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} c^{2}\right )}{48 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^4,x)

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^2*(48*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))*x^3*a^2*b*d^3*(a*c)^(1/2)-15*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^3*a^3*d^3*(b*d)^(1/2)-45*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^3*a
^2*b*c*d^2*(b*d)^(1/2)+15*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^3*a*b^2*c^
2*d*(b*d)^(1/2)-3*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^3*b^3*c^3*(b*d)^(1
/2)-66*(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^2*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*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^2*b^2*c^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*a^2*c*d-4*(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^2-16*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^2*c^2)/(b*d*x^2+a*d*x+b*c*
x+a*c)^(1/2)/x^3/(b*d)^(1/2)/(a*c)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^4,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 \[ \int \frac {\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{5/2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^(1/2)*(c + d*x)^(5/2))/x^4,x)

[Out]

int(((a + b*x)^(1/2)*(c + d*x)^(5/2))/x^4, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(5/2)*(b*x+a)**(1/2)/x**4,x)

[Out]

Timed out

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