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

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

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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 (a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{7/2} c^{3/2}}+\frac {\sqrt {a+b x} (c+d x)^{3/2} (a d+5 b c)}{12 a^2 c x^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d) (a d+5 b c)}{8 a^3 c x}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.19, size = 144, normalized size = 0.80 \begin {gather*} \frac {\frac {x (a d+5 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+5 a d x-3 b c x)\right )}{a^{5/2} \sqrt {c}}-8 \sqrt {a+b x} (c+d x)^{5/2}}{24 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.29, size = 213, normalized size = 1.18 \begin {gather*} \frac {(a d-b c)^2 (a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{7/2} c^{3/2}}-\frac {\sqrt {a+b x} (a d-b c)^2 \left (3 a^3 d+\frac {8 a^2 c d (a+b x)}{c+d x}-33 a^2 b c-\frac {15 b c^3 (a+b x)^2}{(c+d x)^2}+\frac {40 a b c^2 (a+b x)}{c+d x}-\frac {3 a c^2 d (a+b x)^2}{(c+d x)^2}\right )}{24 a^3 c \sqrt {c+d x} \left (a-\frac {c (a+b x)}{c+d x}\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 2.27, size = 438, normalized size = 2.43 \begin {gather*} \left [\frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, 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} - 22 \, a^{2} b c^{2} d + 3 \, a^{3} c d^{2}\right )} x^{2} - 2 \, {\left (5 \, a^{2} b c^{3} - 7 \, a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, a^{4} c^{2} x^{3}}, -\frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, 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} - 22 \, a^{2} b c^{2} d + 3 \, a^{3} c d^{2}\right )} x^{2} - 2 \, {\left (5 \, a^{2} b c^{3} - 7 \, a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, a^{4} c^{2} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 30.80, size = 2219, normalized size = 12.33

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(5*sqrt(b*d)*b^4*c^3*abs(b) - 9*sqrt(b*d)*a*b^3*c^2*d*abs(b) + 3*sqrt(b*d)*a^2*b^2*c*d^2*abs(b) + 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^3*b*c) - 2*(15*sqrt(b*d)*b^14*c^8*abs(b) - 112*sqrt(b*d)*a*b^1
3*c^7*d*abs(b) + 360*sqrt(b*d)*a^2*b^12*c^6*d^2*abs(b) - 648*sqrt(b*d)*a^3*b^11*c^5*d^3*abs(b) + 710*sqrt(b*d)
*a^4*b^10*c^4*d^4*abs(b) - 480*sqrt(b*d)*a^5*b^9*c^3*d^5*abs(b) + 192*sqrt(b*d)*a^6*b^8*c^2*d^6*abs(b) - 40*sq
rt(b*d)*a^7*b^7*c*d^7*abs(b) + 3*sqrt(b*d)*a^8*b^6*d^8*abs(b) - 75*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) + 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*b^11*c^6*d*abs(b) - 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^2*b^10*c^5*d^2*abs(b) - 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^3*b^9*c^4*d^3*abs(b) + 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^4*b^8*c^3*d^4*abs(b) - 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
^5*b^7*c^2*d^5*abs(b) + 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^6*b^
6*c*d^6*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*a^7*b^5*d^7*ab
s(b) + 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^10*c^6*abs(b) - 300*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*b^9*c^5*d*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^2*b^8*c^4*d^2*abs(b) - 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*abs(b) + 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^4*b^6*c^2*d^4*abs(b) - 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*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*a^6*b^4*d^6*abs(b) - 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^8*c^5*abs(b) + 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*b^7*c^4*d*abs(b) + 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^2*b^6*c^3*d^2*abs(b) - 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*abs(b) + 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^4*b^4*c*d^4*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*a^5*b^3*
d^5*abs(b) + 75*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) - 6
0*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) - 90*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) - 36*sqrt(b*d)*(sqr
t(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*abs(b) + 15*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^4*b^2*d^4*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*b^4*c^3*abs(b) + 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*b^3*c^2*d*abs(b) - 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*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*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 + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)^3*a^3*c))/b

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maple [B]  time = 0.03, size = 408, normalized size = 2.27 \begin {gather*} \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 )+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 )-27 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}+44 \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}-28 \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 \,x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(d*x+c)^(1/2)*(b*x+a)^(1/2)/a^3/c*(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)+9*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)-27*a*b^2*c^2*d*x^3*l
n((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+44*((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^2-28*((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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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