3.705 \(\int \frac {\sqrt {c+d x}}{x^4 \sqrt {a+b x}} \, dx\)

Optimal. Leaf size=191 \[ -\frac {\sqrt {a+b x} \sqrt {c+d x} (5 b c-3 a d) (a d+3 b c)}{24 a^3 c^2 x}+\frac {\sqrt {a+b x} \sqrt {c+d x} (5 b c-a d)}{12 a^2 c x^2}+\frac {(b c-a d) \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{7/2} c^{5/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 a x^3} \]

[Out]

1/8*(-a*d+b*c)*(a^2*d^2+2*a*b*c*d+5*b^2*c^2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(7/2)/c^(5
/2)-1/3*(b*x+a)^(1/2)*(d*x+c)^(1/2)/x^3/a+1/12*(-a*d+5*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^2/c/x^2-1/24*(-3*a*d
+5*b*c)*(a*d+3*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^3/c^2/x

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Rubi [A]  time = 0.14, antiderivative size = 191, 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} \[ \frac {(b c-a d) \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{7/2} c^{5/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} (5 b c-3 a d) (a d+3 b c)}{24 a^3 c^2 x}+\frac {\sqrt {a+b x} \sqrt {c+d x} (5 b c-a d)}{12 a^2 c x^2}-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a + b*x]*Sqrt[c + d*x])/(3*a*x^3) + ((5*b*c - a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(12*a^2*c*x^2) - ((5*b*
c - 3*a*d)*(3*b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(24*a^3*c^2*x) + ((b*c - a*d)*(5*b^2*c^2 + 2*a*b*c*d + a
^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(8*a^(7/2)*c^(5/2))

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 {c+d x}}{x^4 \sqrt {a+b x}} \, dx &=-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 a x^3}+\frac {\int \frac {\frac {1}{2} (-5 b c+a d)-2 b d x}{x^3 \sqrt {a+b x} \sqrt {c+d x}} \, dx}{3 a}\\ &=-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 a x^3}+\frac {(5 b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{12 a^2 c x^2}-\frac {\int \frac {-\frac {1}{4} (5 b c-3 a d) (3 b c+a d)-\frac {1}{2} b d (5 b c-a d) x}{x^2 \sqrt {a+b x} \sqrt {c+d x}} \, dx}{6 a^2 c}\\ &=-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 a x^3}+\frac {(5 b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{12 a^2 c x^2}-\frac {(5 b c-3 a d) (3 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{24 a^3 c^2 x}+\frac {\int -\frac {3 (b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right )}{8 x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{6 a^3 c^2}\\ &=-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 a x^3}+\frac {(5 b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{12 a^2 c x^2}-\frac {(5 b c-3 a d) (3 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{24 a^3 c^2 x}-\frac {\left ((b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right )\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 a^3 c^2}\\ &=-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 a x^3}+\frac {(5 b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{12 a^2 c x^2}-\frac {(5 b c-3 a d) (3 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{24 a^3 c^2 x}-\frac {\left ((b c-a d) \left (5 b^2 c^2+2 a b c d+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^3 c^2}\\ &=-\frac {\sqrt {a+b x} \sqrt {c+d x}}{3 a x^3}+\frac {(5 b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{12 a^2 c x^2}-\frac {(5 b c-3 a d) (3 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{24 a^3 c^2 x}+\frac {(b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{7/2} c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 160, normalized size = 0.84 \[ \frac {\sqrt {a+b x} \sqrt {c+d x} \left (a^2 \left (-8 c^2-2 c d x+3 d^2 x^2\right )+2 a b c x (5 c+2 d x)-15 b^2 c^2 x^2\right )}{24 a^3 c^2 x^3}-\frac {\left (a^3 d^3+a^2 b c d^2+3 a b^2 c^2 d-5 b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{7/2} c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*Sqrt[c + d*x]*(-15*b^2*c^2*x^2 + 2*a*b*c*x*(5*c + 2*d*x) + a^2*(-8*c^2 - 2*c*d*x + 3*d^2*x^2)))
/(24*a^3*c^2*x^3) - ((-5*b^3*c^3 + 3*a*b^2*c^2*d + a^2*b*c*d^2 + a^3*d^3)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqr
t[a]*Sqrt[c + d*x])])/(8*a^(7/2)*c^(5/2))

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fricas [A]  time = 1.90, size = 440, normalized size = 2.30 \[ \left [-\frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a c} x^{3} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (8 \, a^{3} c^{3} + {\left (15 \, a b^{2} c^{3} - 4 \, a^{2} b c^{2} d - 3 \, a^{3} c d^{2}\right )} x^{2} - 2 \, {\left (5 \, a^{2} b c^{3} - a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, a^{4} c^{3} x^{3}}, -\frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-a c} x^{3} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) + 2 \, {\left (8 \, a^{3} c^{3} + {\left (15 \, a b^{2} c^{3} - 4 \, a^{2} b c^{2} d - 3 \, a^{3} c d^{2}\right )} x^{2} - 2 \, {\left (5 \, a^{2} b c^{3} - a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, a^{4} c^{3} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - 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 + (15*a*b^2*c^3 - 4*a^2*b*c^2*d - 3*a^3*c*d^2)*x^2 - 2*(5*a^2*b*c^3 - a^3*c^2*d)*x)*sqrt
(b*x + a)*sqrt(d*x + c))/(a^4*c^3*x^3), -1/48*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - 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 + (15*a*b^2*c^3 - 4*a^2*b*c^2*d - 3*a^3*c*d^2)*x^2 - 2*(5*a^2*b*c^3 - a^3*c^
2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^4*c^3*x^3)]

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giac [B]  time = 30.25, size = 2140, normalized size = 11.20 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(5*sqrt(b*d)*b^6*c^3 - 3*sqrt(b*d)*a*b^5*c^2*d - sqrt(b*d)*a^2*b^4*c*d^2 - sqrt(b*d)*a^3*b^3*d^3)*arct
an(-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^3*b*c^2) - 2*(15*sqrt(b*d)*b^16*c^8 - 94*sqrt(b*d)*a*b^15*c^7*d + 246*sqrt(b*d)*a^2*b^14*c
^6*d^2 - 342*sqrt(b*d)*a^3*b^13*c^5*d^3 + 260*sqrt(b*d)*a^4*b^12*c^4*d^4 - 90*sqrt(b*d)*a^5*b^11*c^3*d^5 - 6*s
qrt(b*d)*a^6*b^10*c^2*d^6 + 14*sqrt(b*d)*a^7*b^9*c*d^7 - 3*sqrt(b*d)*a^8*b^8*d^8 - 75*sqrt(b*d)*(sqrt(b*d)*sqr
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^14*c^7 + 225*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^13*c^6*d - 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^12*c^5*d^2 - 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^3*b^11*c^4*d^3 + 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^4*
b^10*c^3*d^4 - 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^5*b^9*c^2*d^5
- 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^6*b^8*c*d^6 + 15*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^7*d^7 + 150*sqrt(b*d)*(sqrt(b*d)*sqrt
(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^12*c^6 - 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*b^11*c^5*d - 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^2*b^10*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^9*c^3*d^3 + 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^4*b^8
*c^2*d^4 - 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^5*b^7*c*d^5 - 30*s
qrt(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^6*d^6 - 150*sqrt(b*d)*(sqrt(b
*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^10*c^5 - 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*b^9*c^4*d - 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^2*b^8*c^3*d^2 - 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^3*b^7*c^2*d^3 + 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^4
*b^6*c*d^4 + 30*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^5*d^5 + 75*s
qrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*b^8*c^4 + 30*sqrt(b*d)*(sqrt(b*d)*s
qrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a*b^7*c^3*d - 60*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sq
rt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^2*b^6*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^5*c*d^3 - 15*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^4*d^4 - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^10*b^6*c^3
+ 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*b^5*c^2*d + 3*sqrt(b*d)*(s
qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^10*a^2*b^4*c*d^2 + 3*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^3*d^3)/((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*a^3*c^
2))*abs(b)/b^3

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maple [B]  time = 0.03, size = 408, normalized size = 2.14 \[ -\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (3 a^{3} d^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+3 a^{2} b c \,d^{2} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+9 a \,b^{2} c^{2} d \,x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-15 b^{3} c^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} d^{2} x^{2}-8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a b c d \,x^{2}+30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, b^{2} c^{2} x^{2}+4 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} c d x -20 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a b \,c^{2} x +16 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} c^{2}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{3} c^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(d*x+c)^(1/2)*(b*x+a)^(1/2)/a^3/c^2*(3*a^3*d^3*x^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)+3*a^2*b*c*d^2*x^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)+9*a*b^2*c^2*d*x^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)-15*b^3*c^3*x^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)-6*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*d^2*x^2-8*((b*x+a)*(d*x+c))^(1/2)*
(a*c)^(1/2)*a*b*c*d*x^2+30*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*b^2*c^2*x^2+4*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/
2)*a^2*c*d*x-20*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a*b*c^2*x+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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/2)/x^4/(b*x+a)^(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 = 91.00, size = 1574, normalized size = 8.24 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((((a + b*x)^(1/2) - a^(1/2))^4*((5*a^4*b^2*d^4)/64 - (13*b^6*c^4)/64 + (a^3*b^3*c*d^3)/2 + (7*a^2*b^4*c^2*d^2
)/16 - (23*a*b^5*c^3*d)/16))/(a^4*c^3*d^3*((c + d*x)^(1/2) - c^(1/2))^4) - b^6/(192*a^2*c*d^3) - (((a + b*x)^(
1/2) - a^(1/2))^5*((17*b^6*c^5)/64 - (a^5*b*d^5)/64 + (7*a^4*b^2*c*d^4)/16 - (47*a^2*b^4*c^3*d^2)/32 + (37*a^3
*b^3*c^2*d^3)/32 - (3*a*b^5*c^4*d)/2))/(a^(9/2)*c^(7/2)*d^3*((c + d*x)^(1/2) - c^(1/2))^5) + (((a + b*x)^(1/2)
- a^(1/2))^2*((a^2*b^4*d^2)/64 - (5*b^6*c^2)/64 + (3*a*b^5*c*d)/32))/(a^3*c^2*d^3*((c + d*x)^(1/2) - c^(1/2))
^2) + (((a + b*x)^(1/2) - a^(1/2))^7*((3*a^5*d^5)/64 - (27*b^5*c^5)/64 + (69*a^2*b^3*c^3*d^2)/64 - (49*a^3*b^2
*c^2*d^3)/64 + (45*a*b^4*c^4*d)/64 - (13*a^4*b*c*d^4)/64))/(a^(9/2)*c^(7/2)*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^(5/2)*c^(3/2)*d^3*((c + d*x)^(1/2) - c^(1/2)))
+ (((a + b*x)^(1/2) - a^(1/2))^3*((73*b^6*c^3)/192 - (17*a^3*b^3*d^3)/192 - (9*a^2*b^4*c*d^2)/32 + (3*a*b^5*c
^2*d)/32))/(a^(7/2)*c^(5/2)*d^3*((c + d*x)^(1/2) - c^(1/2))^3) + (((a + b*x)^(1/2) - a^(1/2))^6*((37*b^6*c^6)/
192 - (5*a^6*d^6)/192 - (5*a^2*b^4*c^4*d^2)/2 + (5*a^3*b^3*c^3*d^3)/32 + a^4*b^2*c^2*d^4 + (a*b^5*c^5*d)/8 + (
a^5*b*c*d^5)/16))/(a^5*c^4*d^3*((c + d*x)^(1/2) - c^(1/2))^6) + (((a + b*x)^(1/2) - a^(1/2))^8*((15*b^4*c^4)/6
4 - (a^4*d^4)/64 + (13*a^2*b^2*c^2*d^2)/64 - (19*a*b^3*c^3*d)/32 + (3*a^3*b*c*d^3)/32))/(a^4*c^3*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)) + ((d^3/(32*a^(3/2)*c^(3/2)) - (d^2*(a*d +
b*c))/(32*a^(5/2)*c^(3/2)))*((a + b*x)^(1/2) - a^(1/2))^2)/((c + d*x)^(1/2) - c^(1/2))^2 + (((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^3*c^2) - (4*a*d^7 - 3*b*c*d^6)/(64*a^2*c^2*d^4) + (4*
(d^3/(16*a^(3/2)*c^(3/2)) - (d^2*(a*d + b*c))/(16*a^(5/2)*c^(3/2)))*(a*d + b*c))/(a^(1/2)*c^(1/2)*d)))/((c + d
*x)^(1/2) - c^(1/2)) + (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^(7/2)*c^(1/2)*d^3 - 5*a^
(1/2)*b^3*c^(7/2) + 3*a^(3/2)*b^2*c^(5/2)*d + a^(5/2)*b*c^(3/2)*d^2))/(16*a^4*c^3) - (log(((a + b*x)^(1/2) - a
^(1/2))/((c + d*x)^(1/2) - c^(1/2)))*(a^(7/2)*c^(1/2)*d^3 - 5*a^(1/2)*b^3*c^(7/2) + 3*a^(3/2)*b^2*c^(5/2)*d +
a^(5/2)*b*c^(3/2)*d^2))/(16*a^4*c^3) - (d^3*((a + b*x)^(1/2) - a^(1/2))^3)/(192*a^2*c*((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 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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