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

Optimal. Leaf size=165 \[ -\frac {2 c \sqrt {a+b x} \left (2 d x \left (-3 a^2 d^2-3 a b c d+2 b^2 c^2\right )+c (b c-3 a d) (a d+3 b c)\right )}{3 b d^2 (c+d x)^{3/2} (b c-a d)^3}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} d^{5/2}}+\frac {2 a x^2}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)} \]

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Rubi [A]  time = 0.12, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {98, 145, 63, 217, 206} \begin {gather*} -\frac {2 c \sqrt {a+b x} \left (2 d x \left (-3 a^2 d^2-3 a b c d+2 b^2 c^2\right )+c (b c-3 a d) (a d+3 b c)\right )}{3 b d^2 (c+d x)^{3/2} (b c-a d)^3}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} d^{5/2}}+\frac {2 a x^2}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 145

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g +
e*h) + d*e*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(
f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*(b*c - a*d)^2*(m + 1)*(m
+ 2)), x] + Dist[(f*h)/b^2 - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)
) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)), Int[(a + b*x)^(m +
2)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] &&  !L
tQ[n, -2]))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

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

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Mathematica [A]  time = 5.31, size = 160, normalized size = 0.97 \begin {gather*} \frac {2}{3} \sqrt {a+b x} \sqrt {c+d x} \left (\frac {3 a^3}{b (a+b x) (b c-a d)^3}+\frac {c^3}{d^2 (c+d x)^2 (b c-a d)^2}+\frac {c^2 (4 b c-9 a d)}{d^2 (c+d x) (a d-b c)^3}\right )+\frac {\log \left (2 \sqrt {b} \sqrt {d} \sqrt {a+b x} \sqrt {c+d x}+a d+b c+2 b d x\right )}{b^{3/2} d^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.24, size = 150, normalized size = 0.91 \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{3/2} d^{5/2}}-\frac {2 (a+b x)^{3/2} \left (-\frac {3 a^3 d^2 (c+d x)^2}{(a+b x)^2}+\frac {3 b^2 c^3 (c+d x)}{a+b x}-\frac {9 a b c^2 d (c+d x)}{a+b x}+b c^3 d\right )}{3 b d^2 (c+d x)^{3/2} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 3.77, size = 1326, normalized size = 8.04

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*
b^2*c*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^2 +
(b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x)*sqrt(b*d)*log(8*b^2*d^2*x^2 +
b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a
*b*d^2)*x) - 4*(3*a*b^3*c^4*d - 8*a^2*b^2*c^3*d^2 - 3*a^3*b*c^2*d^3 + (4*b^4*c^3*d^2 - 9*a*b^3*c^2*d^3 - 3*a^3
*b*d^5)*x^2 + (3*b^4*c^4*d - 4*a*b^3*c^3*d^2 - 9*a^2*b^2*c^2*d^3 - 6*a^3*b*c*d^4)*x)*sqrt(b*x + a)*sqrt(d*x +
c))/(a*b^5*c^5*d^3 - 3*a^2*b^4*c^4*d^4 + 3*a^3*b^3*c^3*d^5 - a^4*b^2*c^2*d^6 + (b^6*c^3*d^5 - 3*a*b^5*c^2*d^6
+ 3*a^2*b^4*c*d^7 - a^3*b^3*d^8)*x^3 + (2*b^6*c^4*d^4 - 5*a*b^5*c^3*d^5 + 3*a^2*b^4*c^2*d^6 + a^3*b^3*c*d^7 -
a^4*b^2*d^8)*x^2 + (b^6*c^5*d^3 - a*b^5*c^4*d^4 - 3*a^2*b^4*c^3*d^5 + 5*a^3*b^3*c^2*d^6 - 2*a^4*b^2*c*d^7)*x),
-1/3*(3*(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2
*b^2*c*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^2
+ (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x)*sqrt(-b*d)*arctan(1/2*(2*b*d*
x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) + 2*(3*
a*b^3*c^4*d - 8*a^2*b^2*c^3*d^2 - 3*a^3*b*c^2*d^3 + (4*b^4*c^3*d^2 - 9*a*b^3*c^2*d^3 - 3*a^3*b*d^5)*x^2 + (3*b
^4*c^4*d - 4*a*b^3*c^3*d^2 - 9*a^2*b^2*c^2*d^3 - 6*a^3*b*c*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b^5*c^5*d^3
- 3*a^2*b^4*c^4*d^4 + 3*a^3*b^3*c^3*d^5 - a^4*b^2*c^2*d^6 + (b^6*c^3*d^5 - 3*a*b^5*c^2*d^6 + 3*a^2*b^4*c*d^7
- a^3*b^3*d^8)*x^3 + (2*b^6*c^4*d^4 - 5*a*b^5*c^3*d^5 + 3*a^2*b^4*c^2*d^6 + a^3*b^3*c*d^7 - a^4*b^2*d^8)*x^2 +
(b^6*c^5*d^3 - a*b^5*c^4*d^4 - 3*a^2*b^4*c^3*d^5 + 5*a^3*b^3*c^2*d^6 - 2*a^4*b^2*c*d^7)*x)]

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giac [B]  time = 2.73, size = 478, normalized size = 2.90 \begin {gather*} \frac {4 \, \sqrt {b d} a^{3}}{{\left (b^{2} c^{2} {\left | b \right |} - 2 \, a b c d {\left | b \right |} + a^{2} d^{2} {\left | b \right |}\right )} {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}} - \frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (4 \, b^{8} c^{5} d^{2} - 17 \, a b^{7} c^{4} d^{3} + 22 \, a^{2} b^{6} c^{3} d^{4} - 9 \, a^{3} b^{5} c^{2} d^{5}\right )} {\left (b x + a\right )}}{b^{7} c^{5} d^{3} {\left | b \right |} - 5 \, a b^{6} c^{4} d^{4} {\left | b \right |} + 10 \, a^{2} b^{5} c^{3} d^{5} {\left | b \right |} - 10 \, a^{3} b^{4} c^{2} d^{6} {\left | b \right |} + 5 \, a^{4} b^{3} c d^{7} {\left | b \right |} - a^{5} b^{2} d^{8} {\left | b \right |}} + \frac {3 \, {\left (b^{9} c^{6} d - 6 \, a b^{8} c^{5} d^{2} + 12 \, a^{2} b^{7} c^{4} d^{3} - 10 \, a^{3} b^{6} c^{3} d^{4} + 3 \, a^{4} b^{5} c^{2} d^{5}\right )}}{b^{7} c^{5} d^{3} {\left | b \right |} - 5 \, a b^{6} c^{4} d^{4} {\left | b \right |} + 10 \, a^{2} b^{5} c^{3} d^{5} {\left | b \right |} - 10 \, a^{3} b^{4} c^{2} d^{6} {\left | b \right |} + 5 \, a^{4} b^{3} c d^{7} {\left | b \right |} - a^{5} b^{2} d^{8} {\left | b \right |}}\right )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {\log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{\sqrt {b d} d^{2} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*sqrt(b*d)*a^3/((b^2*c^2*abs(b) - 2*a*b*c*d*abs(b) + a^2*d^2*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)) - 2/3*sqrt(b*x + a)*((4*b^8*c^5*d^2 - 17*a*b^7*c^4*d^3 + 22*a^2*b
^6*c^3*d^4 - 9*a^3*b^5*c^2*d^5)*(b*x + a)/(b^7*c^5*d^3*abs(b) - 5*a*b^6*c^4*d^4*abs(b) + 10*a^2*b^5*c^3*d^5*ab
s(b) - 10*a^3*b^4*c^2*d^6*abs(b) + 5*a^4*b^3*c*d^7*abs(b) - a^5*b^2*d^8*abs(b)) + 3*(b^9*c^6*d - 6*a*b^8*c^5*d
^2 + 12*a^2*b^7*c^4*d^3 - 10*a^3*b^6*c^3*d^4 + 3*a^4*b^5*c^2*d^5)/(b^7*c^5*d^3*abs(b) - 5*a*b^6*c^4*d^4*abs(b)
+ 10*a^2*b^5*c^3*d^5*abs(b) - 10*a^3*b^4*c^2*d^6*abs(b) + 5*a^4*b^3*c*d^7*abs(b) - a^5*b^2*d^8*abs(b)))/(b^2*
c + (b*x + a)*b*d - a*b*d)^(3/2) - log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqr
t(b*d)*d^2*abs(b))

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maple [B]  time = 0.03, size = 1289, normalized size = 7.81

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-6*x^2*a^3*d^4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+3*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b
*d)^(1/2))/(b*d)^(1/2))*a^4*c^2*d^3-3*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/
2))*a*b^3*c^5+3*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*a^4*d^5-3*ln(1
/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x*b^4*c^5+8*x^2*b^3*c^3*d*((b*x+a)*(d*
x+c))^(1/2)*(b*d)^(1/2)-12*x*a^3*c*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-16*a^2*b*c^3*d*((b*x+a)*(d*x+c))^(1
/2)*(b*d)^(1/2)-9*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^3*a^2*b^2*c*d^
4+9*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^3*a*b^3*c^2*d^3-3*ln(1/2*(2*
b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*a^3*b*c*d^4-9*ln(1/2*(2*b*d*x+a*d+b*c+2*
((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*a^2*b^2*c^2*d^3+15*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d
*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*a*b^3*c^3*d^2-15*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*
(b*d)^(1/2))/(b*d)^(1/2))*x*a^3*b*c^2*d^3+9*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*
d)^(1/2))*x*a^2*b^2*c^3*d^2+3*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x*a*
b^3*c^4*d-18*x^2*a*b^2*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-18*x*a^2*b*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*
(b*d)^(1/2)-8*x*a*b^2*c^3*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+6*x*b^3*c^4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2
)-6*a^3*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+6*a*b^2*c^4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+3*ln(1/2*(
2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^3*a^3*b*d^5-3*ln(1/2*(2*b*d*x+a*d+b*c+2*
((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^3*b^4*c^3*d^2-6*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c)
)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*b^4*c^4*d+6*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2
))/(b*d)^(1/2))*x*a^4*c*d^4-9*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^3*
b*c^3*d^2+9*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^2*b^2*c^4*d)/(b*d)^(
1/2)/(a*d-b*c)^3/((b*x+a)*(d*x+c))^(1/2)/d^2/b/(d*x+c)^(3/2)/(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(x^3/(b*x+a)^(3/2)/(d*x+c)^(5/2),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 {x^3}{{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(b*x+a)**(3/2)/(d*x+c)**(5/2),x)

[Out]

Integral(x**3/((a + b*x)**(3/2)*(c + d*x)**(5/2)), x)

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