3.7.7 \(\int \frac {(a+b x)^{3/2}}{x^4 \sqrt {c+d x}} \, dx\)

Optimal. Leaf size=180 \[ \frac {(b c-a d)^2 (5 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{3/2} c^{7/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d) (5 a d+b c)}{8 a c^3 x}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 a d+b c)}{12 a c^2 x^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 a c x^3} \]

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {96, 94, 93, 208} \begin {gather*} \frac {(b c-a d)^2 (5 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{3/2} c^{7/2}}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 a d+b c)}{12 a c^2 x^2}+\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d) (5 a d+b c)}{8 a c^3 x}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)/(x^4*Sqrt[c + d*x]),x]

[Out]

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

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 94

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[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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

________________________________________________________________________________________

Mathematica [A]  time = 0.18, size = 144, normalized size = 0.80 \begin {gather*} \frac {\frac {x (5 a d+b c) \left (3 x^2 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\sqrt {a} \sqrt {c} \sqrt {a+b x} \sqrt {c+d x} (2 a c-3 a d x+5 b c x)\right )}{\sqrt {a} c^{5/2}}-8 (a+b x)^{5/2} \sqrt {c+d x}}{24 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)/(x^4*Sqrt[c + d*x]),x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.31, size = 214, normalized size = 1.19 \begin {gather*} \frac {(a d-b c)^2 (5 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{8 a^{3/2} c^{7/2}}-\frac {\sqrt {c+d x} (a d-b c)^2 \left (\frac {15 a^3 d (c+d x)^2}{(a+b x)^2}+\frac {3 a^2 b c (c+d x)^2}{(a+b x)^2}-\frac {40 a^2 c d (c+d x)}{a+b x}-\frac {8 a b c^2 (c+d x)}{a+b x}+33 a c^2 d-3 b c^3\right )}{24 a c^3 \sqrt {a+b x} \left (\frac {a (c+d x)}{a+b x}-c\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(3/2)/(x^4*Sqrt[c + d*x]),x]

[Out]

-1/24*((-(b*c) + a*d)^2*Sqrt[c + d*x]*(-3*b*c^3 + 33*a*c^2*d - (8*a*b*c^2*(c + d*x))/(a + b*x) - (40*a^2*c*d*(
c + d*x))/(a + b*x) + (3*a^2*b*c*(c + d*x)^2)/(a + b*x)^2 + (15*a^3*d*(c + d*x)^2)/(a + b*x)^2))/(a*c^3*Sqrt[a
+ b*x]*(-c + (a*(c + d*x))/(a + b*x))^3) + ((-(b*c) + a*d)^2*(b*c + 5*a*d)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(S
qrt[c]*Sqrt[a + b*x])])/(8*a^(3/2)*c^(7/2))

________________________________________________________________________________________

fricas [A]  time = 3.41, size = 438, normalized size = 2.43 \begin {gather*} \left [\frac {3 \, {\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d - 9 \, 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} - 22 \, a^{2} b c^{2} d + 15 \, a^{3} c d^{2}\right )} x^{2} + 2 \, {\left (7 \, a^{2} b c^{3} - 5 \, a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, a^{2} c^{4} x^{3}}, -\frac {3 \, {\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d - 9 \, 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} - 22 \, a^{2} b c^{2} d + 15 \, a^{3} c d^{2}\right )} x^{2} + 2 \, {\left (7 \, a^{2} b c^{3} - 5 \, a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, a^{2} c^{4} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^4/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(3*(b^3*c^3 + 3*a*b^2*c^2*d - 9*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 - 22*a^2*b*c^2*d + 15*a^3*c*d^2)*x^2 + 2*(7*a^2*b*c^3 - 5*a^3*c^2*d)*x)*
sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^4*x^3), -1/48*(3*(b^3*c^3 + 3*a*b^2*c^2*d - 9*a^2*b*c*d^2 + 5*a^3*d^3)*sqr
t(-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 - 22*a^2*b*c^2*d + 15*a^3*c*d^2)*x^2 + 2*(7*a^2*b*c^3 -
5*a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^4*x^3)]

________________________________________________________________________________________

giac [B]  time = 31.37, size = 2135, normalized size = 11.86

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^4/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

1/24*b*(3*(sqrt(b*d)*b^4*c^3 + 3*sqrt(b*d)*a*b^3*c^2*d - 9*sqrt(b*d)*a^2*b^2*c*d^2 + 5*sqrt(b*d)*a^3*b*d^3)*ar
ctan(-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*b*c^3) - 2*(3*sqrt(b*d)*b^14*c^8 - 40*sqrt(b*d)*a*b^13*c^7*d + 192*sqrt(b*d)*a^2*b^12*c^
6*d^2 - 480*sqrt(b*d)*a^3*b^11*c^5*d^3 + 710*sqrt(b*d)*a^4*b^10*c^4*d^4 - 648*sqrt(b*d)*a^5*b^9*c^3*d^5 + 360*
sqrt(b*d)*a^6*b^8*c^2*d^6 - 112*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)*s
qrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^12*c^7 + 183*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 - 567*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 + 663*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 - 117*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 - 387*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
+ 315*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
)*(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 + 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 - 300*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 + 474*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 - 168*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 + 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 - 300*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)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^8*c^5 + 202*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 - 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^6*c^3*d^2 + 12*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 + 130*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 - 36*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 - 90*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - s
qrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^2*b^4*c^2*d^2 - 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^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
- 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^3*c^2*d + 27*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)*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*c^3)
)/abs(b)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 408, normalized size = 2.27 \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 )-27 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 )+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}+44 \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 -28 \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 \,c^{3} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)/x^4/(d*x+c)^(1/2),x)

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a/c^3*(15*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)-27*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)+3*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)-30*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*d^2*x^2+44*((b*x+a)*(d*x+c))^(1/2)*
(a*c)^(1/2)*a*b*c*d*x^2-6*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*b^2*c^2*x^2+20*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/
2)*a^2*c*d*x-28*((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)

________________________________________________________________________________________

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)/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?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{3/2}}{x^4\,\sqrt {c+d\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(3/2)/(x^4*(c + d*x)^(1/2)),x)

[Out]

int((a + b*x)^(3/2)/(x^4*(c + d*x)^(1/2)), x)

________________________________________________________________________________________

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)**(3/2)/x**4/(d*x+c)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________