\(\int x^3 (c+d x)^{3/2} (c^2-d^2 x^2)^{3/2} \, dx\) [671]

Optimal result

Integrand size = 29, antiderivative size = 241 \[ \int x^3 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {4864 c^6 \left (c^2-d^2 x^2\right )^{5/2}}{255255 d^4 (c+d x)^{5/2}}-\frac {1216 c^5 \left (c^2-d^2 x^2\right )^{5/2}}{51051 d^4 (c+d x)^{3/2}}-\frac {152 c^4 \left (c^2-d^2 x^2\right )^{5/2}}{7293 d^4 \sqrt {c+d x}}-\frac {38 c^3 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{2431 d^4}-\frac {30 c^2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{221 d^4}+\frac {18 c (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{5/2}}{85 d^4}-\frac {2 (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{5/2}}{17 d^4} \] Output:

-4864/255255*c^6*(-d^2*x^2+c^2)^(5/2)/d^4/(d*x+c)^(5/2)-1216/51051*c^5*(-d 
^2*x^2+c^2)^(5/2)/d^4/(d*x+c)^(3/2)-152/7293*c^4*(-d^2*x^2+c^2)^(5/2)/d^4/ 
(d*x+c)^(1/2)-38/2431*c^3*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(5/2)/d^4-30/221*c^ 
2*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(5/2)/d^4+18/85*c*(d*x+c)^(5/2)*(-d^2*x^2+c 
^2)^(5/2)/d^4-2/17*(d*x+c)^(7/2)*(-d^2*x^2+c^2)^(5/2)/d^4
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.39 \[ \int x^3 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (15440 c^6+38600 c^5 d x+67550 c^4 d^2 x^2+101325 c^3 d^3 x^3+107415 c^2 d^4 x^4+63063 c d^5 x^5+15015 d^6 x^6\right )}{255255 d^4 (c+d x)^{5/2}} \] Input:

Integrate[x^3*(c + d*x)^(3/2)*(c^2 - d^2*x^2)^(3/2),x]
 

Output:

(-2*(c^2 - d^2*x^2)^(5/2)*(15440*c^6 + 38600*c^5*d*x + 67550*c^4*d^2*x^2 + 
 101325*c^3*d^3*x^3 + 107415*c^2*d^4*x^4 + 63063*c*d^5*x^5 + 15015*d^6*x^6 
))/(255255*d^4*(c + d*x)^(5/2))
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {581, 27, 2170, 27, 672, 459, 459, 459, 458}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^3 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx\)

\(\Big \downarrow \) 581

\(\displaystyle \frac {2 \int \frac {1}{2} (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \left (7 c^3-3 d x c^2-27 d^2 x^2 c\right )dx}{17 d^3}-\frac {2 (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{5/2}}{17 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \left (7 c^3-3 d x c^2-27 d^2 x^2 c\right )dx}{17 d^3}-\frac {2 (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{5/2}}{17 d^4}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {\frac {18 c (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{5/2}}{5 d}-\frac {2 \int \frac {15}{2} c^2 d^4 (2 c-15 d x) (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}dx}{15 d^4}}{17 d^3}-\frac {2 (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{5/2}}{17 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {18 c (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{5/2}}{5 d}-c^2 \int (2 c-15 d x) (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}dx}{17 d^3}-\frac {2 (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{5/2}}{17 d^4}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {\frac {18 c (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{5/2}}{5 d}-c^2 \left (\frac {30 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d}-\frac {19}{13} c \int (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}dx\right )}{17 d^3}-\frac {2 (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{5/2}}{17 d^4}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {\frac {18 c (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{5/2}}{5 d}-c^2 \left (\frac {30 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d}-\frac {19}{13} c \left (\frac {12}{11} c \int \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}dx-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d}\right )\right )}{17 d^3}-\frac {2 (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{5/2}}{17 d^4}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {\frac {18 c (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{5/2}}{5 d}-c^2 \left (\frac {30 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d}-\frac {19}{13} c \left (\frac {12}{11} c \left (\frac {8}{9} c \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{\sqrt {c+d x}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{9 d \sqrt {c+d x}}\right )-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d}\right )\right )}{17 d^3}-\frac {2 (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{5/2}}{17 d^4}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {\frac {18 c (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{5/2}}{5 d}-c^2 \left (\frac {30 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d}-\frac {19}{13} c \left (\frac {12}{11} c \left (\frac {8}{9} c \left (\frac {4}{7} c \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{3/2}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{7 d (c+d x)^{3/2}}\right )-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{9 d \sqrt {c+d x}}\right )-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d}\right )\right )}{17 d^3}-\frac {2 (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{5/2}}{17 d^4}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {\frac {18 c (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{5/2}}{5 d}-c^2 \left (\frac {30 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d}-\frac {19}{13} c \left (\frac {12}{11} c \left (\frac {8}{9} c \left (-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{7 d (c+d x)^{3/2}}-\frac {8 c \left (c^2-d^2 x^2\right )^{5/2}}{35 d (c+d x)^{5/2}}\right )-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{9 d \sqrt {c+d x}}\right )-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d}\right )\right )}{17 d^3}-\frac {2 (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{5/2}}{17 d^4}\)

Input:

Int[x^3*(c + d*x)^(3/2)*(c^2 - d^2*x^2)^(3/2),x]
 

Output:

(-2*(c + d*x)^(7/2)*(c^2 - d^2*x^2)^(5/2))/(17*d^4) + ((18*c*(c + d*x)^(5/ 
2)*(c^2 - d^2*x^2)^(5/2))/(5*d) - c^2*((30*(c + d*x)^(3/2)*(c^2 - d^2*x^2) 
^(5/2))/(13*d) - (19*c*((-2*Sqrt[c + d*x]*(c^2 - d^2*x^2)^(5/2))/(11*d) + 
(12*c*((-2*(c^2 - d^2*x^2)^(5/2))/(9*d*Sqrt[c + d*x]) + (8*c*((-8*c*(c^2 - 
 d^2*x^2)^(5/2))/(35*d*(c + d*x)^(5/2)) - (2*(c^2 - d^2*x^2)^(5/2))/(7*d*( 
c + d*x)^(3/2))))/9))/11))/13))/(17*d^3)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 458

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] /; FreeQ[{a, b, c 
, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + p, 0]
 

rule 459

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* 
(Simplify[n + p]/(n + 2*p + 1))   Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && IGtQ[Simplif 
y[n + p], 0]
 

rule 581

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 
2*p + 1))), x] + Simp[1/(d^m*(m + n + 2*p + 1))   Int[(c + d*x)^n*(a + b*x^ 
2)^p*ExpandToSum[d^m*(m + n + 2*p + 1)*x^m - (m + n + 2*p + 1)*(c + d*x)^m 
+ c*(c + d*x)^(m - 2)*(c*(m + n - 1) + c*(m + n + 2*p + 1) + 2*d*(m + n + p 
)*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] & 
& IGtQ[m, 1] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[2*p] || ILtQ[m + n, 0] 
)
 

rule 672

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), 
 x] + Simp[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2))   Int[(d + e*x 
)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^ 
2 + a*e^2, 0] && NeQ[m + 2*p + 2, 0]
 

rule 2170

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + 
 p + q)*(d + e*x)^(q - 2)*(a*e - b*d*x), x], x], x] /; NeQ[m + q + 2*p + 1, 
 0]] /; FreeQ[{a, b, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 
0] &&  !IGtQ[m, 0]
 

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.40

method	result	size
gosper	\(-\frac {2 \left (-d x +c \right ) \left (15015 d^{6} x^{6}+63063 c \,d^{5} x^{5}+107415 c^{2} d^{4} x^{4}+101325 d^{3} c^{3} x^{3}+67550 c^{4} d^{2} x^{2}+38600 c^{5} d x +15440 c^{6}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{255255 d^{4} \left (d x +c \right )^{\frac {3}{2}}}\)	\(96\)
orering	\(-\frac {2 \left (-d x +c \right ) \left (15015 d^{6} x^{6}+63063 c \,d^{5} x^{5}+107415 c^{2} d^{4} x^{4}+101325 d^{3} c^{3} x^{3}+67550 c^{4} d^{2} x^{2}+38600 c^{5} d x +15440 c^{6}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{255255 d^{4} \left (d x +c \right )^{\frac {3}{2}}}\)	\(96\)
default	\(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (-d x +c \right )^{2} \left (15015 d^{6} x^{6}+63063 c \,d^{5} x^{5}+107415 c^{2} d^{4} x^{4}+101325 d^{3} c^{3} x^{3}+67550 c^{4} d^{2} x^{2}+38600 c^{5} d x +15440 c^{6}\right )}{255255 \sqrt {d x +c}\, d^{4}}\)	\(98\)
risch	\(-\frac {2 \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}\, \left (15015 d^{8} x^{8}+33033 c \,d^{7} x^{7}-3696 c^{2} d^{6} x^{6}-50442 c^{3} d^{5} x^{5}-27685 c^{4} x^{4} d^{4}+4825 c^{5} d^{3} x^{3}+5790 c^{6} d^{2} x^{2}+7720 c^{7} d x +15440 c^{8}\right ) \sqrt {-d x +c}}{255255 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{4}}\)	\(142\)

Input:

int(x^3*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-2/255255*(-d*x+c)*(15015*d^6*x^6+63063*c*d^5*x^5+107415*c^2*d^4*x^4+10132 
5*c^3*d^3*x^3+67550*c^4*d^2*x^2+38600*c^5*d*x+15440*c^6)*(-d^2*x^2+c^2)^(3 
/2)/d^4/(d*x+c)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.50 \[ \int x^3 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {2 \, {\left (15015 \, d^{8} x^{8} + 33033 \, c d^{7} x^{7} - 3696 \, c^{2} d^{6} x^{6} - 50442 \, c^{3} d^{5} x^{5} - 27685 \, c^{4} d^{4} x^{4} + 4825 \, c^{5} d^{3} x^{3} + 5790 \, c^{6} d^{2} x^{2} + 7720 \, c^{7} d x + 15440 \, c^{8}\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{255255 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

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

Output:

-2/255255*(15015*d^8*x^8 + 33033*c*d^7*x^7 - 3696*c^2*d^6*x^6 - 50442*c^3* 
d^5*x^5 - 27685*c^4*d^4*x^4 + 4825*c^5*d^3*x^3 + 5790*c^6*d^2*x^2 + 7720*c 
^7*d*x + 15440*c^8)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)/(d^5*x + c*d^4)
 

Sympy [F]

\[ \int x^3 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx=\int x^{3} \left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {3}{2}}\, dx \] Input:

integrate(x**3*(d*x+c)**(3/2)*(-d**2*x**2+c**2)**(3/2),x)
 

Output:

Integral(x**3*(-(-c + d*x)*(c + d*x))**(3/2)*(c + d*x)**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.47 \[ \int x^3 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {2 \, {\left (15015 \, d^{8} x^{8} + 33033 \, c d^{7} x^{7} - 3696 \, c^{2} d^{6} x^{6} - 50442 \, c^{3} d^{5} x^{5} - 27685 \, c^{4} d^{4} x^{4} + 4825 \, c^{5} d^{3} x^{3} + 5790 \, c^{6} d^{2} x^{2} + 7720 \, c^{7} d x + 15440 \, c^{8}\right )} {\left (d x + c\right )} \sqrt {-d x + c}}{255255 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

integrate(x^3*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(3/2),x, algorithm="maxima")
 

Output:

-2/255255*(15015*d^8*x^8 + 33033*c*d^7*x^7 - 3696*c^2*d^6*x^6 - 50442*c^3* 
d^5*x^5 - 27685*c^4*d^4*x^4 + 4825*c^5*d^3*x^3 + 5790*c^6*d^2*x^2 + 7720*c 
^7*d*x + 15440*c^8)*(d*x + c)*sqrt(-d*x + c)/(d^5*x + c*d^4)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 754 vs. \(2 (199) = 398\).

Time = 0.13 (sec) , antiderivative size = 754, normalized size of antiderivative = 3.13 \[ \int x^3 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

-2/765765*(45045*(d*x - c)^8*sqrt(-d*x + c) + 408408*(d*x - c)^7*sqrt(-d*x 
 + c)*c + 1649340*(d*x - c)^6*sqrt(-d*x + c)*c^2 + 3898440*(d*x - c)^5*sqr 
t(-d*x + c)*c^3 + 5955950*(d*x - c)^4*sqrt(-d*x + c)*c^4 + 6126120*(d*x - 
c)^3*sqrt(-d*x + c)*c^5 + 4288284*(d*x - c)^2*sqrt(-d*x + c)*c^6 - 2042040 
*(-d*x + c)^(3/2)*c^7 + 765765*sqrt(-d*x + c)*c^8 + 21879*(5*(d*x - c)^3*s 
qrt(-d*x + c) + 21*(d*x - c)^2*sqrt(-d*x + c)*c - 35*(-d*x + c)^(3/2)*c^2 
+ 35*sqrt(-d*x + c)*c^3)*c^5 + 2431*(35*(d*x - c)^4*sqrt(-d*x + c) + 180*( 
d*x - c)^3*sqrt(-d*x + c)*c + 378*(d*x - c)^2*sqrt(-d*x + c)*c^2 - 420*(-d 
*x + c)^(3/2)*c^3 + 315*sqrt(-d*x + c)*c^4)*c^4 - 2210*(63*(d*x - c)^5*sqr 
t(-d*x + c) + 385*(d*x - c)^4*sqrt(-d*x + c)*c + 990*(d*x - c)^3*sqrt(-d*x 
 + c)*c^2 + 1386*(d*x - c)^2*sqrt(-d*x + c)*c^3 - 1155*(-d*x + c)^(3/2)*c^ 
4 + 693*sqrt(-d*x + c)*c^5)*c^3 - 510*(231*(d*x - c)^6*sqrt(-d*x + c) + 16 
38*(d*x - c)^5*sqrt(-d*x + c)*c + 5005*(d*x - c)^4*sqrt(-d*x + c)*c^2 + 85 
80*(d*x - c)^3*sqrt(-d*x + c)*c^3 + 9009*(d*x - c)^2*sqrt(-d*x + c)*c^4 - 
6006*(-d*x + c)^(3/2)*c^5 + 3003*sqrt(-d*x + c)*c^6)*c^2 + 119*(429*(d*x - 
 c)^7*sqrt(-d*x + c) + 3465*(d*x - c)^6*sqrt(-d*x + c)*c + 12285*(d*x - c) 
^5*sqrt(-d*x + c)*c^2 + 25025*(d*x - c)^4*sqrt(-d*x + c)*c^3 + 32175*(d*x 
- c)^3*sqrt(-d*x + c)*c^4 + 27027*(d*x - c)^2*sqrt(-d*x + c)*c^5 - 15015*( 
-d*x + c)^(3/2)*c^6 + 6435*sqrt(-d*x + c)*c^7)*c)/d^4
 

Mupad [B] (verification not implemented)

Time = 7.19 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.72 \[ \int x^3 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {6176\,c^8\,\sqrt {c+d\,x}}{51051\,d^5}-\frac {4804\,c^3\,x^5\,\sqrt {c+d\,x}}{12155}+\frac {2\,d^3\,x^8\,\sqrt {c+d\,x}}{17}-\frac {1582\,c^4\,x^4\,\sqrt {c+d\,x}}{7293\,d}+\frac {1930\,c^5\,x^3\,\sqrt {c+d\,x}}{51051\,d^2}+\frac {772\,c^6\,x^2\,\sqrt {c+d\,x}}{17017\,d^3}-\frac {32\,c^2\,d\,x^6\,\sqrt {c+d\,x}}{1105}+\frac {22\,c\,d^2\,x^7\,\sqrt {c+d\,x}}{85}+\frac {3088\,c^7\,x\,\sqrt {c+d\,x}}{51051\,d^4}\right )}{x+\frac {c}{d}} \] Input:

int(x^3*(c^2 - d^2*x^2)^(3/2)*(c + d*x)^(3/2),x)
 

Output:

-((c^2 - d^2*x^2)^(1/2)*((6176*c^8*(c + d*x)^(1/2))/(51051*d^5) - (4804*c^ 
3*x^5*(c + d*x)^(1/2))/12155 + (2*d^3*x^8*(c + d*x)^(1/2))/17 - (1582*c^4* 
x^4*(c + d*x)^(1/2))/(7293*d) + (1930*c^5*x^3*(c + d*x)^(1/2))/(51051*d^2) 
 + (772*c^6*x^2*(c + d*x)^(1/2))/(17017*d^3) - (32*c^2*d*x^6*(c + d*x)^(1/ 
2))/1105 + (22*c*d^2*x^7*(c + d*x)^(1/2))/85 + (3088*c^7*x*(c + d*x)^(1/2) 
)/(51051*d^4)))/(x + c/d)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.40 \[ \int x^3 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx=\frac {2 \sqrt {-d x +c}\, \left (-15015 d^{8} x^{8}-33033 c \,d^{7} x^{7}+3696 c^{2} d^{6} x^{6}+50442 c^{3} d^{5} x^{5}+27685 c^{4} d^{4} x^{4}-4825 c^{5} d^{3} x^{3}-5790 c^{6} d^{2} x^{2}-7720 c^{7} d x -15440 c^{8}\right )}{255255 d^{4}} \] Input:

int(x^3*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(3/2),x)
 

Output:

(2*sqrt(c - d*x)*( - 15440*c**8 - 7720*c**7*d*x - 5790*c**6*d**2*x**2 - 48 
25*c**5*d**3*x**3 + 27685*c**4*d**4*x**4 + 50442*c**3*d**5*x**5 + 3696*c** 
2*d**6*x**6 - 33033*c*d**7*x**7 - 15015*d**8*x**8))/(255255*d**4)