3.6.58 \(\int \frac {\sqrt {a+b x}}{x^4 \sqrt {c+d x}} \, dx\)
Optimal. Leaf size=190 \[ \frac {\sqrt {a+b x} \sqrt {c+d x} (3 b c-5 a d) (3 a d+b c)}{24 a^2 c^3 x}-\frac {(b c-a d) \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} c^{7/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d)}{12 a c^2 x^2}-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 c x^3} \]
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Rubi [A] time = 0.14, antiderivative size = 190, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used =
{99, 151, 12, 93, 208} \begin {gather*} -\frac {(b c-a d) \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} c^{7/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (3 b c-5 a d) (3 a d+b c)}{24 a^2 c^3 x}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d)}{12 a c^2 x^2}-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 c x^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + b*x]/(x^4*Sqrt[c + d*x]),x]
[Out]
-(Sqrt[a + b*x]*Sqrt[c + d*x])/(3*c*x^3) - ((b*c - 5*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(12*a*c^2*x^2) + ((3*b*
c - 5*a*d)*(b*c + 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(24*a^2*c^3*x) - ((b*c - a*d)*(b^2*c^2 + 2*a*b*c*d + 5*a
^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(8*a^(5/2)*c^(7/2))
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 99
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 + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])
Rule 151
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 + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]
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 {\sqrt {a+b x}}{x^4 \sqrt {c+d x}} \, dx &=-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 c x^3}+\frac {\int \frac {\frac {1}{2} (b c-5 a d)-2 b d x}{x^3 \sqrt {a+b x} \sqrt {c+d x}} \, dx}{3 c}\\ &=-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 c x^3}-\frac {(b c-5 a d) \sqrt {a+b x} \sqrt {c+d x}}{12 a c^2 x^2}-\frac {\int \frac {\frac {1}{4} (3 b c-5 a d) (b c+3 a d)+\frac {1}{2} b d (b c-5 a d) x}{x^2 \sqrt {a+b x} \sqrt {c+d x}} \, dx}{6 a c^2}\\ &=-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 c x^3}-\frac {(b c-5 a d) \sqrt {a+b x} \sqrt {c+d x}}{12 a c^2 x^2}+\frac {(3 b c-5 a d) (b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{24 a^2 c^3 x}+\frac {\int \frac {3 (b c-a d) \left (b^2 c^2+2 a b c d+5 a^2 d^2\right )}{8 x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{6 a^2 c^3}\\ &=-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 c x^3}-\frac {(b c-5 a d) \sqrt {a+b x} \sqrt {c+d x}}{12 a c^2 x^2}+\frac {(3 b c-5 a d) (b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{24 a^2 c^3 x}+\frac {\left ((b c-a d) \left (b^2 c^2+2 a b c d+5 a^2 d^2\right )\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 a^2 c^3}\\ &=-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 c x^3}-\frac {(b c-5 a d) \sqrt {a+b x} \sqrt {c+d x}}{12 a c^2 x^2}+\frac {(3 b c-5 a d) (b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{24 a^2 c^3 x}+\frac {\left ((b c-a d) \left (b^2 c^2+2 a b c d+5 a^2 d^2\right )\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 c^3}\\ &=-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 c x^3}-\frac {(b c-5 a d) \sqrt {a+b x} \sqrt {c+d x}}{12 a c^2 x^2}+\frac {(3 b c-5 a d) (b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{24 a^2 c^3 x}-\frac {(b c-a d) \left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} c^{7/2}}\\ \end {align*}
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Mathematica [A] time = 0.16, size = 158, normalized size = 0.83 \begin {gather*} \frac {\sqrt {a+b x} \sqrt {c+d x} \left (a^2 \left (-8 c^2+10 c d x-15 d^2 x^2\right )-2 a b c x (c-2 d x)+3 b^2 c^2 x^2\right )}{24 a^2 c^3 x^3}-\frac {\left (-5 a^3 d^3+3 a^2 b c d^2+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} c^{7/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + b*x]/(x^4*Sqrt[c + d*x]),x]
[Out]
(Sqrt[a + b*x]*Sqrt[c + d*x]*(3*b^2*c^2*x^2 - 2*a*b*c*x*(c - 2*d*x) + a^2*(-8*c^2 + 10*c*d*x - 15*d^2*x^2)))/(
24*a^2*c^3*x^3) - ((b^3*c^3 + a*b^2*c^2*d + 3*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)*c^(7/2))
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IntegrateAlgebraic [F] time = 180.38, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}
Verification is not applicable to the result.
[In]
IntegrateAlgebraic[Sqrt[a + b*x]/(x^4*Sqrt[c + d*x]),x]
[Out]
$Aborted
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fricas [A] time = 3.18, size = 436, normalized size = 2.29 \begin {gather*} \left [-\frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {a c} x^{3} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} + 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (8 \, a^{3} c^{3} - {\left (3 \, a b^{2} c^{3} + 4 \, a^{2} b c^{2} d - 15 \, a^{3} c d^{2}\right )} x^{2} + 2 \, {\left (a^{2} b c^{3} - 5 \, a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, a^{3} c^{4} x^{3}}, \frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {-a c} x^{3} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left (8 \, a^{3} c^{3} - {\left (3 \, a b^{2} c^{3} + 4 \, a^{2} b c^{2} d - 15 \, a^{3} c d^{2}\right )} x^{2} + 2 \, {\left (a^{2} b c^{3} - 5 \, a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, a^{3} c^{4} x^{3}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(1/2)/x^4/(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
[-1/96*(3*(b^3*c^3 + a*b^2*c^2*d + 3*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 + 4*a^2*b*c^2*d - 15*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 - 5*a^3*c^2*d)*x)*sqrt
(b*x + a)*sqrt(d*x + c))/(a^3*c^4*x^3), 1/48*(3*(b^3*c^3 + a*b^2*c^2*d + 3*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 + 4*a^2*b*c^2*d - 15*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 - 5*a^3*c^2
*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*c^4*x^3)]
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giac [B] time = 30.46, size = 2134, normalized size = 11.23
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(1/2)/x^4/(d*x+c)^(1/2),x, algorithm="giac")
[Out]
-1/24*b*(3*(sqrt(b*d)*b^4*c^3 + sqrt(b*d)*a*b^3*c^2*d + 3*sqrt(b*d)*a^2*b^2*c*d^2 - 5*sqrt(b*d)*a^3*b*d^3)*arc
tan(-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*c^3) - 2*(3*sqrt(b*d)*b^14*c^8 - 14*sqrt(b*d)*a*b^13*c^7*d + 6*sqrt(b*d)*a^2*b^12*c^6
*d^2 + 90*sqrt(b*d)*a^3*b^11*c^5*d^3 - 260*sqrt(b*d)*a^4*b^10*c^4*d^4 + 342*sqrt(b*d)*a^5*b^9*c^3*d^5 - 246*sq
rt(b*d)*a^6*b^8*c^2*d^6 + 94*sqrt(b*d)*a^7*b^7*c*d^7 - 15*sqrt(b*d)*a^8*b^6*d^8 - 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 + 21*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 + 141*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 - 399*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 + 291*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 + 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^5*b^7*c^2*d^5 - 2
25*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 + 75*sqrt(b*d)*(s
qrt(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 + 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 + 24*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 - 306*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 + 240*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 + 42*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 + 120*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^5*b^5*c*d^5 - 150*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 - 30*sqrt(b*d)*(sqrt(b*d)*sq
rt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^8*c^5 - 58*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 + 108*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 + 36*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 + 50*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 + 150*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 + 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 + 30*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 + 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^2*b^4*c^2*d^2 - 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a
)*b*d - a*b*d))^8*a^3*b^3*c*d^3 - 75*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 - 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 - 3*sq
rt(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 - 9*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 + 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^3*b*d^3)/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*d)*sq
rt(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*a^2*c^3))/abs(
b)
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maple [B] time = 0.03, size = 408, normalized size = 2.15 \begin {gather*} \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (15 a^{3} d^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-9 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 {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-3 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 {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-3 b^{3} c^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} d^{2} x^{2}+8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a b c d \,x^{2}+6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, b^{2} c^{2} x^{2}+20 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} c d x -4 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a b \,c^{2} x -16 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} c^{2}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} c^{3} x^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x+a)^(1/2)/x^4/(d*x+c)^(1/2),x)
[Out]
1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^2/c^3*(15*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x
^3*a^3*d^3-9*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a^2*b*c*d^2-3*ln((a*d*x+b*c*x
+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a*b^2*c^2*d-3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a
)*(d*x+c))^(1/2))/x)*x^3*b^3*c^3-30*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x^2*a^2*d^2+8*((b*x+a)*(d*x+c))^(1/2)*
(a*c)^(1/2)*x^2*a*b*c*d+6*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x^2*b^2*c^2+20*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/
2)*x*a^2*c*d-4*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x*a*b*c^2-16*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*c^2)/(
(b*x+a)*(d*x+c))^(1/2)/x^3/(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)^(1/2)/x^4/(d*x+c)^(1/2),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 [B] time = 91.08, size = 1570, normalized size = 8.26
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*x)^(1/2)/(x^4*(c + d*x)^(1/2)),x)
[Out]
((((a + b*x)^(1/2) - a^(1/2))^4*((5*b^6*c^4)/64 - (13*a^4*b^2*d^4)/64 - (23*a^3*b^3*c*d^3)/16 + (7*a^2*b^4*c^2
*d^2)/16 + (a*b^5*c^3*d)/2))/(a^3*c^4*d^3*((c + d*x)^(1/2) - c^(1/2))^4) - b^6/(192*a*c^2*d^3) + (((a + b*x)^(
1/2) - a^(1/2))^5*((b^6*c^5)/64 - (17*a^5*b*d^5)/64 + (3*a^4*b^2*c*d^4)/2 - (37*a^2*b^4*c^3*d^2)/32 + (47*a^3*
b^3*c^2*d^3)/32 - (7*a*b^5*c^4*d)/16))/(a^(7/2)*c^(9/2)*d^3*((c + d*x)^(1/2) - c^(1/2))^5) + (((a + b*x)^(1/2)
- a^(1/2))^2*((b^6*c^2)/64 - (5*a^2*b^4*d^2)/64 + (3*a*b^5*c*d)/32))/(a^2*c^3*d^3*((c + d*x)^(1/2) - c^(1/2))
^2) - (((a + b*x)^(1/2) - a^(1/2))^7*((27*a^5*d^5)/64 - (3*b^5*c^5)/64 + (49*a^2*b^3*c^3*d^2)/64 - (69*a^3*b^2
*c^2*d^3)/64 + (13*a*b^4*c^4*d)/64 - (45*a^4*b*c*d^4)/64))/(a^(7/2)*c^(9/2)*d^2*((c + d*x)^(1/2) - c^(1/2))^7)
+ (((b^6*c)/64 - (a*b^5*d)/64)*((a + b*x)^(1/2) - a^(1/2)))/(a^(3/2)*c^(5/2)*d^3*((c + d*x)^(1/2) - c^(1/2)))
- (((a + b*x)^(1/2) - a^(1/2))^3*((17*b^6*c^3)/192 - (73*a^3*b^3*d^3)/192 - (3*a^2*b^4*c*d^2)/32 + (9*a*b^5*c
^2*d)/32))/(a^(5/2)*c^(7/2)*d^3*((c + d*x)^(1/2) - c^(1/2))^3) + (((a + b*x)^(1/2) - a^(1/2))^6*((37*a^6*d^6)/
192 - (5*b^6*c^6)/192 + a^2*b^4*c^4*d^2 + (5*a^3*b^3*c^3*d^3)/32 - (5*a^4*b^2*c^2*d^4)/2 + (a*b^5*c^5*d)/16 +
(a^5*b*c*d^5)/8))/(a^4*c^5*d^3*((c + d*x)^(1/2) - c^(1/2))^6) + (((a + b*x)^(1/2) - a^(1/2))^8*((15*a^4*d^4)/6
4 - (b^4*c^4)/64 + (13*a^2*b^2*c^2*d^2)/64 + (3*a*b^3*c^3*d)/32 - (19*a^3*b*c*d^3)/32))/(a^3*c^4*d*((c + d*x)^
(1/2) - c^(1/2))^8))/(((a + b*x)^(1/2) - a^(1/2))^9/((c + d*x)^(1/2) - c^(1/2))^9 + (b^3*((a + b*x)^(1/2) - a^
(1/2))^3)/(d^3*((c + d*x)^(1/2) - c^(1/2))^3) + (((a + b*x)^(1/2) - a^(1/2))^7*(3*a^2*d^2 + 3*b^2*c^2 + 9*a*b*
c*d))/(a*c*d^2*((c + d*x)^(1/2) - c^(1/2))^7) - (((a + b*x)^(1/2) - a^(1/2))^6*(a^3*d^3 + b^3*c^3 + 9*a*b^2*c^
2*d + 9*a^2*b*c*d^2))/(a^(3/2)*c^(3/2)*d^3*((c + d*x)^(1/2) - c^(1/2))^6) - ((3*a*d + 3*b*c)*((a + b*x)^(1/2)
- a^(1/2))^8)/(a^(1/2)*c^(1/2)*d*((c + d*x)^(1/2) - c^(1/2))^8) - ((3*b^3*c + 3*a*b^2*d)*((a + b*x)^(1/2) - a^
(1/2))^4)/(a^(1/2)*c^(1/2)*d^3*((c + d*x)^(1/2) - c^(1/2))^4) + (((a + b*x)^(1/2) - a^(1/2))^5*(3*b^3*c^2 + 3*
a^2*b*d^2 + 9*a*b^2*c*d))/(a*c*d^3*((c + d*x)^(1/2) - c^(1/2))^5)) + (((a + b*x)^(1/2) - a^(1/2))*((d*(3*a^2*d
^2 + 3*b^2*c^2 + 8*a*b*c*d))/(32*a^2*c^3) + (4*(a*d + b*c)*((b*d^2)/(16*a^(3/2)*c^(3/2)) - (d^2*(a*d + b*c))/(
16*a^(3/2)*c^(5/2))))/(a^(1/2)*c^(1/2)*d) + (b*d*(3*a*d - 4*b*c))/(64*a^2*c^2)))/((c + d*x)^(1/2) - c^(1/2)) +
(((b*d^2)/(32*a^(3/2)*c^(3/2)) - (d^2*(a*d + b*c))/(32*a^(3/2)*c^(5/2)))*((a + b*x)^(1/2) - a^(1/2))^2)/((c +
d*x)^(1/2) - c^(1/2))^2 + (log(((c^(1/2)*(a + b*x)^(1/2) - a^(1/2)*(c + d*x)^(1/2))*(b*c^(1/2) - (a^(1/2)*d*(
(a + b*x)^(1/2) - a^(1/2)))/((c + d*x)^(1/2) - c^(1/2))))/((c + d*x)^(1/2) - c^(1/2)))*(a^(1/2)*b^3*c^(7/2) -
5*a^(7/2)*c^(1/2)*d^3 + a^(3/2)*b^2*c^(5/2)*d + 3*a^(5/2)*b*c^(3/2)*d^2))/(16*a^3*c^4) - (log(((a + b*x)^(1/2)
- a^(1/2))/((c + d*x)^(1/2) - c^(1/2)))*(a^(1/2)*b^3*c^(7/2) - 5*a^(7/2)*c^(1/2)*d^3 + a^(3/2)*b^2*c^(5/2)*d
+ 3*a^(5/2)*b*c^(3/2)*d^2))/(16*a^3*c^4) - (d^3*((a + b*x)^(1/2) - a^(1/2))^3)/(192*a*c^2*((c + d*x)^(1/2) - c
^(1/2))^3)
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)**(1/2)/x**4/(d*x+c)**(1/2),x)
[Out]
Timed out
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