3.8.49 \(\int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=230 \[ \frac {5 (b c-a d)^2 (7 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{9/2} \sqrt {c}}-\frac {5 \sqrt {c+d x} (b c-a d)^2 (7 b c-a d)}{8 a^4 c \sqrt {a+b x}}-\frac {5 (c+d x)^{3/2} (b c-a d) (7 b c-a d)}{24 a^3 c x \sqrt {a+b x}}+\frac {(c+d x)^{5/2} (7 b c-a d)}{12 a^2 c x^2 \sqrt {a+b x}}-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}} \]

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Rubi [A]  time = 0.11, antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {(c+d x)^{5/2} (7 b c-a d)}{12 a^2 c x^2 \sqrt {a+b x}}-\frac {5 (c+d x)^{3/2} (b c-a d) (7 b c-a d)}{24 a^3 c x \sqrt {a+b x}}-\frac {5 \sqrt {c+d x} (b c-a d)^2 (7 b c-a d)}{8 a^4 c \sqrt {a+b x}}+\frac {5 (b c-a d)^2 (7 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{9/2} \sqrt {c}}-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)/(x^4*(a + b*x)^(3/2)),x]

[Out]

(-5*(b*c - a*d)^2*(7*b*c - a*d)*Sqrt[c + d*x])/(8*a^4*c*Sqrt[a + b*x]) - (5*(b*c - a*d)*(7*b*c - a*d)*(c + d*x
)^(3/2))/(24*a^3*c*x*Sqrt[a + b*x]) + ((7*b*c - a*d)*(c + d*x)^(5/2))/(12*a^2*c*x^2*Sqrt[a + b*x]) - (c + d*x)
^(7/2)/(3*a*c*x^3*Sqrt[a + b*x]) + (5*(b*c - a*d)^2*(7*b*c - a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqr
t[c + d*x])])/(8*a^(9/2)*Sqrt[c])

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)^{5/2}}{x^4 (a+b x)^{3/2}} \, dx &=-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}}-\frac {\left (\frac {7 b c}{2}-\frac {a d}{2}\right ) \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{3/2}} \, dx}{3 a c}\\ &=\frac {(7 b c-a d) (c+d x)^{5/2}}{12 a^2 c x^2 \sqrt {a+b x}}-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}}+\frac {(5 (b c-a d) (7 b c-a d)) \int \frac {(c+d x)^{3/2}}{x^2 (a+b x)^{3/2}} \, dx}{24 a^2 c}\\ &=-\frac {5 (b c-a d) (7 b c-a d) (c+d x)^{3/2}}{24 a^3 c x \sqrt {a+b x}}+\frac {(7 b c-a d) (c+d x)^{5/2}}{12 a^2 c x^2 \sqrt {a+b x}}-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}}-\frac {\left (5 (b c-a d)^2 (7 b c-a d)\right ) \int \frac {\sqrt {c+d x}}{x (a+b x)^{3/2}} \, dx}{16 a^3 c}\\ &=-\frac {5 (b c-a d)^2 (7 b c-a d) \sqrt {c+d x}}{8 a^4 c \sqrt {a+b x}}-\frac {5 (b c-a d) (7 b c-a d) (c+d x)^{3/2}}{24 a^3 c x \sqrt {a+b x}}+\frac {(7 b c-a d) (c+d x)^{5/2}}{12 a^2 c x^2 \sqrt {a+b x}}-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}}-\frac {\left (5 (b c-a d)^2 (7 b c-a d)\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 a^4}\\ &=-\frac {5 (b c-a d)^2 (7 b c-a d) \sqrt {c+d x}}{8 a^4 c \sqrt {a+b x}}-\frac {5 (b c-a d) (7 b c-a d) (c+d x)^{3/2}}{24 a^3 c x \sqrt {a+b x}}+\frac {(7 b c-a d) (c+d x)^{5/2}}{12 a^2 c x^2 \sqrt {a+b x}}-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}}-\frac {\left (5 (b c-a d)^2 (7 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^4}\\ &=-\frac {5 (b c-a d)^2 (7 b c-a d) \sqrt {c+d x}}{8 a^4 c \sqrt {a+b x}}-\frac {5 (b c-a d) (7 b c-a d) (c+d x)^{3/2}}{24 a^3 c x \sqrt {a+b x}}+\frac {(7 b c-a d) (c+d x)^{5/2}}{12 a^2 c x^2 \sqrt {a+b x}}-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}}+\frac {5 (b c-a d)^2 (7 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{9/2} \sqrt {c}}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 168, normalized size = 0.73 \begin {gather*} \frac {\frac {1}{2} x (7 b c-a d) \left (2 a^{5/2} (c+d x)^{5/2}-5 x (b c-a d) \left (\sqrt {a} \sqrt {c+d x} (a (c-2 d x)+3 b c x)-3 \sqrt {c} x \sqrt {a+b x} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )\right )\right )-4 a^{7/2} (c+d x)^{7/2}}{12 a^{9/2} c x^3 \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/2)/(x^4*(a + b*x)^(3/2)),x]

[Out]

(-4*a^(7/2)*(c + d*x)^(7/2) + ((7*b*c - a*d)*x*(2*a^(5/2)*(c + d*x)^(5/2) - 5*(b*c - a*d)*x*(Sqrt[a]*Sqrt[c +
d*x]*(3*b*c*x + a*(c - 2*d*x)) - 3*Sqrt[c]*(b*c - a*d)*x*Sqrt[a + b*x]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a
]*Sqrt[c + d*x])])))/2)/(12*a^(9/2)*c*x^3*Sqrt[a + b*x])

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IntegrateAlgebraic [A]  time = 0.52, size = 231, normalized size = 1.00 \begin {gather*} -\frac {5 (a d-7 b c) (a d-b c)^2 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{8 a^{9/2} \sqrt {c}}-\frac {\sqrt {c+d x} (a d-b c)^2 \left (\frac {48 a^3 b (c+d x)^3}{(a+b x)^3}+\frac {33 a^3 d (c+d x)^2}{(a+b x)^2}-\frac {231 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 {280 a b c^2 (c+d x)}{a+b x}+15 a c^2 d-105 b c^3\right )}{24 a^4 \sqrt {a+b x} \left (\frac {a (c+d x)}{a+b x}-c\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + d*x)^(5/2)/(x^4*(a + b*x)^(3/2)),x]

[Out]

-1/24*((-(b*c) + a*d)^2*Sqrt[c + d*x]*(-105*b*c^3 + 15*a*c^2*d + (280*a*b*c^2*(c + d*x))/(a + b*x) - (40*a^2*c
*d*(c + d*x))/(a + b*x) - (231*a^2*b*c*(c + d*x)^2)/(a + b*x)^2 + (33*a^3*d*(c + d*x)^2)/(a + b*x)^2 + (48*a^3
*b*(c + d*x)^3)/(a + b*x)^3))/(a^4*Sqrt[a + b*x]*(-c + (a*(c + d*x))/(a + b*x))^3) - (5*(-7*b*c + a*d)*(-(b*c)
+ a*d)^2*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/(8*a^(9/2)*Sqrt[c])

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fricas [A]  time = 5.95, size = 636, normalized size = 2.77 \begin {gather*} \left [-\frac {15 \, {\left ({\left (7 \, b^{4} c^{3} - 15 \, a b^{3} c^{2} d + 9 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} x^{4} + {\left (7 \, a b^{3} c^{3} - 15 \, a^{2} b^{2} c^{2} d + 9 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} x^{3}\right )} \sqrt {a c} \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^{4} c^{3} + {\left (105 \, a b^{3} c^{3} - 190 \, a^{2} b^{2} c^{2} d + 81 \, a^{3} b c d^{2}\right )} x^{3} + {\left (35 \, a^{2} b^{2} c^{3} - 68 \, a^{3} b c^{2} d + 33 \, a^{4} c d^{2}\right )} x^{2} - 2 \, {\left (7 \, a^{3} b c^{3} - 13 \, a^{4} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, {\left (a^{5} b c x^{4} + a^{6} c x^{3}\right )}}, -\frac {15 \, {\left ({\left (7 \, b^{4} c^{3} - 15 \, a b^{3} c^{2} d + 9 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} x^{4} + {\left (7 \, a b^{3} c^{3} - 15 \, a^{2} b^{2} c^{2} d + 9 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} x^{3}\right )} \sqrt {-a c} \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^{4} c^{3} + {\left (105 \, a b^{3} c^{3} - 190 \, a^{2} b^{2} c^{2} d + 81 \, a^{3} b c d^{2}\right )} x^{3} + {\left (35 \, a^{2} b^{2} c^{3} - 68 \, a^{3} b c^{2} d + 33 \, a^{4} c d^{2}\right )} x^{2} - 2 \, {\left (7 \, a^{3} b c^{3} - 13 \, a^{4} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (a^{5} b c x^{4} + a^{6} c x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/x^4/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/96*(15*((7*b^4*c^3 - 15*a*b^3*c^2*d + 9*a^2*b^2*c*d^2 - a^3*b*d^3)*x^4 + (7*a*b^3*c^3 - 15*a^2*b^2*c^2*d +
9*a^3*b*c*d^2 - a^4*d^3)*x^3)*sqrt(a*c)*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^4*c^3 + (105*a*b^3*
c^3 - 190*a^2*b^2*c^2*d + 81*a^3*b*c*d^2)*x^3 + (35*a^2*b^2*c^3 - 68*a^3*b*c^2*d + 33*a^4*c*d^2)*x^2 - 2*(7*a^
3*b*c^3 - 13*a^4*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^5*b*c*x^4 + a^6*c*x^3), -1/48*(15*((7*b^4*c^3 - 15*
a*b^3*c^2*d + 9*a^2*b^2*c*d^2 - a^3*b*d^3)*x^4 + (7*a*b^3*c^3 - 15*a^2*b^2*c^2*d + 9*a^3*b*c*d^2 - a^4*d^3)*x^
3)*sqrt(-a*c)*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^4*c^3 + (105*a*b^3*c^3 - 190*a^2*b^2*c^2*d + 81*a^3*b*c*d^2)*x^3 + (35*a^2
*b^2*c^3 - 68*a^3*b*c^2*d + 33*a^4*c*d^2)*x^2 - 2*(7*a^3*b*c^3 - 13*a^4*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))
/(a^5*b*c*x^4 + a^6*c*x^3)]

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giac [B]  time = 85.14, size = 2324, normalized size = 10.10

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/x^4/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

5/8*(7*sqrt(b*d)*b^3*c^3*abs(b) - 15*sqrt(b*d)*a*b^2*c^2*d*abs(b) + 9*sqrt(b*d)*a^2*b*c*d^2*abs(b) - sqrt(b*d)
*a^3*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^4*b) - 4*(sqrt(b*d)*b^3*c^3*abs(b) - 3*sqrt(b*d)*a*b^2*c^2*d*abs(b) +
3*sqrt(b*d)*a^2*b*c*d^2*abs(b) - sqrt(b*d)*a^3*d^3*abs(b))/((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)*a^4) - 1/12*(57*sqrt(b*d)*b^13*c^8*abs(b) - 436*sqrt(b*d)*a*b^12*c^7*d*abs(
b) + 1452*sqrt(b*d)*a^2*b^11*c^6*d^2*abs(b) - 2748*sqrt(b*d)*a^3*b^10*c^5*d^3*abs(b) + 3230*sqrt(b*d)*a^4*b^9*
c^4*d^4*abs(b) - 2412*sqrt(b*d)*a^5*b^8*c^3*d^5*abs(b) + 1116*sqrt(b*d)*a^6*b^7*c^2*d^6*abs(b) - 292*sqrt(b*d)
*a^7*b^6*c*d^7*abs(b) + 33*sqrt(b*d)*a^8*b^5*d^8*abs(b) - 285*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c
+ (b*x + a)*b*d - a*b*d))^2*b^11*c^7*abs(b) + 1281*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^10*c^6*d*abs(b) - 1917*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^9*c^5*d^2*abs(b) + 393*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^8*c^4*d^3*abs(b) + 1953*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^7*c^3*d^4*abs(b) - 2277*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^6*c^2*d^5*abs(b) + 1017*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^5
*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^4*d^7*ab
s(b) + 570*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^9*c^6*abs(b) - 1308*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^8*c^5*d*abs(b) + 726*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^7*c^4*d^2*abs(b) - 456*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^6*c^3*d^3*abs(b) + 1446*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^5*c^2*d^4*abs(b) - 1308*sqrt(b*d)*(sqrt(b*d)*sq
rt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^5*b^4*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^3*d^6*abs(b) - 570*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sq
rt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^7*c^5*abs(b) + 598*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^6*c^4*d*abs(b) - 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^5*c^3*d^2*abs(b) - 204*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^4*c^2*d^3*abs(b) + 742*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^3*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^2*d^5*abs(b) + 285*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*b^5*c^4*abs
(b) - 216*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^4*c^3*d*abs(b) - 162
*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^3*c^2*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))^8*a^3*b^2*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*d^4*abs(b) - 57*sqrt(b*d)*(sqrt(b*d)*sq
rt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^10*b^3*c^3*abs(b) + 81*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^2*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*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*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) - sqr
t(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^4)

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maple [B]  time = 0.03, size = 704, normalized size = 3.06 \begin {gather*} -\frac {\sqrt {d x +c}\, \left (15 a^{3} b \,d^{3} x^{4} \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 )-135 a^{2} b^{2} c \,d^{2} x^{4} \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 )+225 a \,b^{3} c^{2} d \,x^{4} \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 )-105 b^{4} c^{3} x^{4} \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 a^{4} 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 )-135 a^{3} 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 )+225 a^{2} 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 )-105 a \,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 )+162 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} b \,d^{2} x^{3}-380 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,b^{2} c d \,x^{3}+210 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{3} c^{2} x^{3}+66 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{3} d^{2} x^{2}-136 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} b c d \,x^{2}+70 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,b^{2} c^{2} x^{2}+52 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{3} c d x -28 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} b \,c^{2} x +16 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{3} c^{2}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, \sqrt {b x +a}\, a^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)/x^4/(b*x+a)^(3/2),x)

[Out]

-1/48*(d*x+c)^(1/2)*(15*a^3*b*d^3*x^4*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)-135*a^2*
b^2*c*d^2*x^4*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)+225*a*b^3*c^2*d*x^4*ln((a*d*x+b*
c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)-105*b^4*c^3*x^4*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*a^4*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)-135*a^3*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)+225*a^2*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)-105*a*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)+162*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b*d^2*x^3-380*(a*c)^(1/2)*((b*x+a)*(d*x+c))
^(1/2)*a*b^2*c*d*x^3+210*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^3*c^2*x^3+66*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2
)*a^3*d^2*x^2-136*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b*c*d*x^2+70*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b
^2*c^2*x^2+52*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*c*d*x-28*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b*c^2*x
+16*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*c^2)/a^4/((b*x+a)*(d*x+c))^(1/2)/x^3/(a*c)^(1/2)/(b*x+a)^(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)^(5/2)/x^4/(b*x+a)^(3/2),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 = -0.00 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^{5/2}}{x^4\,{\left (a+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/2)/(x^4*(a + b*x)^(3/2)),x)

[Out]

int((c + d*x)^(5/2)/(x^4*(a + b*x)^(3/2)), 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)**(5/2)/x**4/(b*x+a)**(3/2),x)

[Out]

Timed out

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