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

Optimal. Leaf size=226 \[ -\frac {5 (b c-7 a d) (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 \sqrt {a} c^{9/2}}+\frac {5 \sqrt {a+b x} (b c-7 a d) (b c-a d)^2}{8 a c^4 \sqrt {c+d x}}-\frac {5 (a+b x)^{3/2} (b c-7 a d) (b c-a d)}{24 a c^3 x \sqrt {c+d x}}-\frac {(a+b x)^{5/2} (b c-7 a d)}{12 a c^2 x^2 \sqrt {c+d x}}-\frac {(a+b x)^{7/2}}{3 a c x^3 \sqrt {c+d x}} \]

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Rubi [A]  time = 0.11, antiderivative size = 226, 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 {(a+b x)^{5/2} (b c-7 a d)}{12 a c^2 x^2 \sqrt {c+d x}}-\frac {5 (a+b x)^{3/2} (b c-7 a d) (b c-a d)}{24 a c^3 x \sqrt {c+d x}}+\frac {5 \sqrt {a+b x} (b c-7 a d) (b c-a d)^2}{8 a c^4 \sqrt {c+d x}}-\frac {5 (b c-7 a d) (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 \sqrt {a} c^{9/2}}-\frac {(a+b x)^{7/2}}{3 a c x^3 \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.30, size = 167, normalized size = 0.74 \begin {gather*} \frac {-\frac {1}{2} x (b c-7 a d) \left (2 c^{5/2} (a+b x)^{5/2}+5 x (b c-a d) \left (\sqrt {c} \sqrt {a+b x} (a (c+3 d x)-2 b c x)+3 \sqrt {a} x \sqrt {c+d 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 c^{7/2} (a+b x)^{7/2}}{12 a c^{9/2} x^3 \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.38, size = 232, normalized size = 1.03 \begin {gather*} \frac {5 (a d-b c)^2 (7 a d-b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 \sqrt {a} c^{9/2}}-\frac {\sqrt {a+b x} (b c-a d)^2 \left (-105 a^3 d+\frac {280 a^2 c d (a+b x)}{c+d x}+15 a^2 b c+\frac {33 b c^3 (a+b x)^2}{(c+d x)^2}+\frac {48 c^3 d (a+b x)^3}{(c+d x)^3}-\frac {40 a b c^2 (a+b x)}{c+d x}-\frac {231 a c^2 d (a+b x)^2}{(c+d x)^2}\right )}{24 c^4 \sqrt {c+d x} \left (\frac {c (a+b x)}{c+d x}-a\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 9.12, size = 632, normalized size = 2.80 \begin {gather*} \left [-\frac {15 \, {\left ({\left (b^{3} c^{3} d - 9 \, a b^{2} c^{2} d^{2} + 15 \, a^{2} b c d^{3} - 7 \, a^{3} d^{4}\right )} x^{4} + {\left (b^{3} c^{4} - 9 \, a b^{2} c^{3} d + 15 \, a^{2} b c^{2} d^{2} - 7 \, a^{3} c 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^{3} c^{4} + {\left (81 \, a b^{2} c^{3} d - 190 \, a^{2} b c^{2} d^{2} + 105 \, a^{3} c d^{3}\right )} x^{3} + {\left (33 \, a b^{2} c^{4} - 68 \, a^{2} b c^{3} d + 35 \, a^{3} c^{2} d^{2}\right )} x^{2} + 2 \, {\left (13 \, a^{2} b c^{4} - 7 \, a^{3} c^{3} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, {\left (a c^{5} d x^{4} + a c^{6} x^{3}\right )}}, \frac {15 \, {\left ({\left (b^{3} c^{3} d - 9 \, a b^{2} c^{2} d^{2} + 15 \, a^{2} b c d^{3} - 7 \, a^{3} d^{4}\right )} x^{4} + {\left (b^{3} c^{4} - 9 \, a b^{2} c^{3} d + 15 \, a^{2} b c^{2} d^{2} - 7 \, a^{3} c 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^{3} c^{4} + {\left (81 \, a b^{2} c^{3} d - 190 \, a^{2} b c^{2} d^{2} + 105 \, a^{3} c d^{3}\right )} x^{3} + {\left (33 \, a b^{2} c^{4} - 68 \, a^{2} b c^{3} d + 35 \, a^{3} c^{2} d^{2}\right )} x^{2} + 2 \, {\left (13 \, a^{2} b c^{4} - 7 \, a^{3} c^{3} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (a c^{5} d x^{4} + a c^{6} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 65.96, size = 2200, normalized size = 9.73

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b^4*c^2*d - 2*a*b^3*c*d^2 + a^2*b^2*d^3)*sqrt(b*x + a)/(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*c^4*abs(b)) -
5/8*(sqrt(b*d)*b^5*c^3 - 9*sqrt(b*d)*a*b^4*c^2*d + 15*sqrt(b*d)*a^2*b^3*c*d^2 - 7*sqrt(b*d)*a^3*b^2*d^3)*arcta
n(-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)*b*c^4*abs(b)) - 1/12*(33*sqrt(b*d)*b^15*c^8 - 292*sqrt(b*d)*a*b^14*c^7*d + 1116*sqrt(b*d)*a^2
*b^13*c^6*d^2 - 2412*sqrt(b*d)*a^3*b^12*c^5*d^3 + 3230*sqrt(b*d)*a^4*b^11*c^4*d^4 - 2748*sqrt(b*d)*a^5*b^10*c^
3*d^5 + 1452*sqrt(b*d)*a^6*b^9*c^2*d^6 - 436*sqrt(b*d)*a^7*b^8*c*d^7 + 57*sqrt(b*d)*a^8*b^7*d^8 - 165*sqrt(b*d
)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^13*c^7 + 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*b^12*c^6*d - 2277*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqr
t(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^11*c^5*d^2 + 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^3*b^10*c^4*d^3 + 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^4*b^9*c^3*d^4 - 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^5*b^8*c^2*d^5 + 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^6*b^7
*c*d^6 - 285*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^6*d^7 + 330*sqr
t(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^11*c^6 - 1308*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^10*c^5*d + 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^2*b^9*c^4*d^2 - 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^8*c^3*d^3 + 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^4*b^7*c^2*d^4 - 1308*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^6*c*d^5 + 570*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^5*
d^6 - 330*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^9*c^5 + 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*b^8*c^4*d - 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^2*b^7*c^3*d^2 - 108*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - s
qrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^3*b^6*c^2*d^3 + 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^4*b^5*c*d^4 - 570*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^4*d^5 + 165*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*b^7*
c^4 - 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*b^6*c^3*d - 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^5*c^2*d^2 - 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^3*b^4*c*d^3 + 285*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^3*d^4 - 33*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^10*b^5*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^4*c^2*d + 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^2*b^3*c
*d^2 - 57*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^2*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*(s
qrt(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*c^4*abs(b))

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maple [B]  time = 0.03, size = 704, normalized size = 3.12 \begin {gather*} \frac {\sqrt {b x +a}\, \left (105 a^{3} d^{4} 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^{2} b c \,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 \,b^{2} c^{2} 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 )-15 b^{3} c^{3} 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 a^{3} c \,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 )-225 a^{2} b \,c^{2} 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 )+135 a \,b^{2} c^{3} 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^{4} 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 )-210 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{3} x^{3}+380 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c \,d^{2} x^{3}-162 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2} d \,x^{3}-70 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c \,d^{2} x^{2}+136 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,c^{2} d \,x^{2}-66 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{3} x^{2}+28 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c^{2} d x -52 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,c^{3} x -16 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c^{3}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, \sqrt {d x +c}\, c^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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