\(\int x^3 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx\) [601]

Optimal result

Integrand size = 32, antiderivative size = 201 \[ \int x^3 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=-\frac {136 c^4 \left (b c^2-b d^2 x^2\right )^{3/2}}{3465 b d^4 (c+d x)^{3/2}}-\frac {34 c^3 \left (b c^2-b d^2 x^2\right )^{3/2}}{1155 b d^4 \sqrt {c+d x}}-\frac {62 c^2 \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}{231 b d^4}+\frac {34 c (c+d x)^{3/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{99 b d^4}-\frac {2 (c+d x)^{5/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b d^4} \] Output:

-136/3465*c^4*(-b*d^2*x^2+b*c^2)^(3/2)/b/d^4/(d*x+c)^(3/2)-34/1155*c^3*(-b 
*d^2*x^2+b*c^2)^(3/2)/b/d^4/(d*x+c)^(1/2)-62/231*c^2*(d*x+c)^(1/2)*(-b*d^2 
*x^2+b*c^2)^(3/2)/b/d^4+34/99*c*(d*x+c)^(3/2)*(-b*d^2*x^2+b*c^2)^(3/2)/b/d 
^4-2/11*(d*x+c)^(5/2)*(-b*d^2*x^2+b*c^2)^(3/2)/b/d^4
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.40 \[ \int x^3 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=-\frac {2 (c-d x) \sqrt {b \left (c^2-d^2 x^2\right )} \left (304 c^4+456 c^3 d x+570 c^2 d^2 x^2+665 c d^3 x^3+315 d^4 x^4\right )}{3465 d^4 \sqrt {c+d x}} \] Input:

Integrate[x^3*Sqrt[c + d*x]*Sqrt[b*c^2 - b*d^2*x^2],x]
 

Output:

(-2*(c - d*x)*Sqrt[b*(c^2 - d^2*x^2)]*(304*c^4 + 456*c^3*d*x + 570*c^2*d^2 
*x^2 + 665*c*d^3*x^3 + 315*d^4*x^4))/(3465*d^4*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {574, 581, 27, 2170, 27, 672, 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.

\(\displaystyle \int x^3 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx\)

\(\Big \downarrow \) 574

\(\displaystyle \frac {19}{11} c \int \frac {x^3 \sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}}dx-\frac {2 x^4 \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b (c+d x)^{3/2}}\)

\(\Big \downarrow \) 581

\(\displaystyle \frac {19}{11} c \left (\frac {2 \int \frac {3 \sqrt {b c^2-b d^2 x^2} \left (c^3-d x c^2-5 d^2 x^2 c\right )}{2 \sqrt {c+d x}}dx}{9 d^3}-\frac {2 (c+d x)^{3/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{9 b d^4}\right )-\frac {2 x^4 \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b (c+d x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {19}{11} c \left (\frac {\int \frac {\sqrt {b c^2-b d^2 x^2} \left (c^3-d x c^2-5 d^2 x^2 c\right )}{\sqrt {c+d x}}dx}{3 d^3}-\frac {2 (c+d x)^{3/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{9 b d^4}\right )-\frac {2 x^4 \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b (c+d x)^{3/2}}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {19}{11} c \left (\frac {\frac {10 c \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}{7 b d}-\frac {2 \int -\frac {b c^2 d^4 (2 c+23 d x) \sqrt {b c^2-b d^2 x^2}}{2 \sqrt {c+d x}}dx}{7 b d^4}}{3 d^3}-\frac {2 (c+d x)^{3/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{9 b d^4}\right )-\frac {2 x^4 \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b (c+d x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {19}{11} c \left (\frac {\frac {1}{7} c^2 \int \frac {(2 c+23 d x) \sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}}dx+\frac {10 c \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}{7 b d}}{3 d^3}-\frac {2 (c+d x)^{3/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{9 b d^4}\right )-\frac {2 x^4 \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b (c+d x)^{3/2}}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {19}{11} c \left (\frac {\frac {1}{7} c^2 \left (-\frac {13}{5} c \int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}}dx-\frac {46 \left (b c^2-b d^2 x^2\right )^{3/2}}{5 b d \sqrt {c+d x}}\right )+\frac {10 c \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}{7 b d}}{3 d^3}-\frac {2 (c+d x)^{3/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{9 b d^4}\right )-\frac {2 x^4 \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b (c+d x)^{3/2}}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {19}{11} c \left (\frac {\frac {1}{7} c^2 \left (\frac {26 c \left (b c^2-b d^2 x^2\right )^{3/2}}{15 b d (c+d x)^{3/2}}-\frac {46 \left (b c^2-b d^2 x^2\right )^{3/2}}{5 b d \sqrt {c+d x}}\right )+\frac {10 c \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}{7 b d}}{3 d^3}-\frac {2 (c+d x)^{3/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{9 b d^4}\right )-\frac {2 x^4 \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b (c+d x)^{3/2}}\)

Input:

Int[x^3*Sqrt[c + d*x]*Sqrt[b*c^2 - b*d^2*x^2],x]
 

Output:

(-2*x^4*(b*c^2 - b*d^2*x^2)^(3/2))/(11*b*(c + d*x)^(3/2)) + (19*c*((-2*(c 
+ d*x)^(3/2)*(b*c^2 - b*d^2*x^2)^(3/2))/(9*b*d^4) + ((10*c*Sqrt[c + d*x]*( 
b*c^2 - b*d^2*x^2)^(3/2))/(7*b*d) + (c^2*((26*c*(b*c^2 - b*d^2*x^2)^(3/2)) 
/(15*b*d*(c + d*x)^(3/2)) - (46*(b*c^2 - b*d^2*x^2)^(3/2))/(5*b*d*Sqrt[c + 
 d*x])))/7)/(3*d^3)))/11
 

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 574

Int[((e_.)*(x_))^(n_)*((c_) + (d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[d^2*(e*x)^(n + 1)*(c + d*x)^(m - 2)*((a + b*x^2)^(p + 1)/ 
(b*e*(n + p + 2))), x] + Simp[c*((2*n + p + 3)/(n + p + 2))   Int[(e*x)^n*( 
c + d*x)^(m - 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x 
] && EqQ[b*c^2 + a*d^2, 0] && EqQ[m + p - 1, 0] &&  !LtQ[n, -1] && IntegerQ 
[2*p]
 

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 = 0.22 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.38

method	result	size
default	\(-\frac {2 \sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, \left (-d x +c \right ) \left (315 d^{4} x^{4}+665 c \,d^{3} x^{3}+570 d^{2} c^{2} x^{2}+456 c^{3} d x +304 c^{4}\right )}{3465 \sqrt {d x +c}\, d^{4}}\)	\(76\)
gosper	\(-\frac {2 \left (-d x +c \right ) \left (315 d^{4} x^{4}+665 c \,d^{3} x^{3}+570 d^{2} c^{2} x^{2}+456 c^{3} d x +304 c^{4}\right ) \sqrt {-b \,x^{2} d^{2}+b \,c^{2}}}{3465 d^{4} \sqrt {d x +c}}\)	\(77\)
orering	\(-\frac {2 \left (-d x +c \right ) \left (315 d^{4} x^{4}+665 c \,d^{3} x^{3}+570 d^{2} c^{2} x^{2}+456 c^{3} d x +304 c^{4}\right ) \sqrt {-b \,x^{2} d^{2}+b \,c^{2}}}{3465 d^{4} \sqrt {d x +c}}\)	\(77\)
risch	\(-\frac {2 \sqrt {-\frac {b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}\, b \left (-315 x^{5} d^{5}-350 x^{4} c \,d^{4}+95 c^{2} d^{3} x^{3}+114 c^{3} d^{2} x^{2}+152 c^{4} d x +304 c^{5}\right ) \left (-d x +c \right )}{3465 \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}\, d^{4} \sqrt {-b \left (d x -c \right )}}\)	\(127\)

Input:

int(x^3*(d*x+c)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/3465/(d*x+c)^(1/2)*(b*(-d^2*x^2+c^2))^(1/2)*(-d*x+c)*(315*d^4*x^4+665*c 
*d^3*x^3+570*c^2*d^2*x^2+456*c^3*d*x+304*c^4)/d^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.45 \[ \int x^3 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=\frac {2 \, {\left (315 \, d^{5} x^{5} + 350 \, c d^{4} x^{4} - 95 \, c^{2} d^{3} x^{3} - 114 \, c^{3} d^{2} x^{2} - 152 \, c^{4} d x - 304 \, c^{5}\right )} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c}}{3465 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

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

Output:

2/3465*(315*d^5*x^5 + 350*c*d^4*x^4 - 95*c^2*d^3*x^3 - 114*c^3*d^2*x^2 - 1 
52*c^4*d*x - 304*c^5)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)/(d^5*x + c*d^ 
4)
 

Sympy [F]

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

integrate(x**3*(d*x+c)**(1/2)*(-b*d**2*x**2+b*c**2)**(1/2),x)
 

Output:

Integral(x**3*sqrt(-b*(-c + d*x)*(c + d*x))*sqrt(c + d*x), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.49 \[ \int x^3 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=\frac {2 \, {\left (315 \, \sqrt {b} d^{5} x^{5} + 350 \, \sqrt {b} c d^{4} x^{4} - 95 \, \sqrt {b} c^{2} d^{3} x^{3} - 114 \, \sqrt {b} c^{3} d^{2} x^{2} - 152 \, \sqrt {b} c^{4} d x - 304 \, \sqrt {b} c^{5}\right )} {\left (d x + c\right )} \sqrt {-d x + c}}{3465 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

integrate(x^3*(d*x+c)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2),x, algorithm="maxima" 
)
 

Output:

2/3465*(315*sqrt(b)*d^5*x^5 + 350*sqrt(b)*c*d^4*x^4 - 95*sqrt(b)*c^2*d^3*x 
^3 - 114*sqrt(b)*c^3*d^2*x^2 - 152*sqrt(b)*c^4*d*x - 304*sqrt(b)*c^5)*(d*x 
 + c)*sqrt(-d*x + c)/(d^5*x + c*d^4)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.28 \[ \int x^3 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=-\frac {2 \, {\left (\frac {99 \, {\left (35 \, \sqrt {-b d x + b c} b^{3} c^{3} - 35 \, {\left (-b d x + b c\right )}^{\frac {3}{2}} b^{2} c^{2} + 21 \, {\left (b d x - b c\right )}^{2} \sqrt {-b d x + b c} b c + 5 \, {\left (b d x - b c\right )}^{3} \sqrt {-b d x + b c}\right )} c^{2}}{b^{3}} - \frac {5 \, {\left (693 \, \sqrt {-b d x + b c} b^{5} c^{5} - 1155 \, {\left (-b d x + b c\right )}^{\frac {3}{2}} b^{4} c^{4} + 1386 \, {\left (b d x - b c\right )}^{2} \sqrt {-b d x + b c} b^{3} c^{3} + 990 \, {\left (b d x - b c\right )}^{3} \sqrt {-b d x + b c} b^{2} c^{2} + 385 \, {\left (b d x - b c\right )}^{4} \sqrt {-b d x + b c} b c + 63 \, {\left (b d x - b c\right )}^{5} \sqrt {-b d x + b c}\right )}}{b^{5}}\right )}}{3465 \, d^{4}} \] Input:

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

Output:

-2/3465*(99*(35*sqrt(-b*d*x + b*c)*b^3*c^3 - 35*(-b*d*x + b*c)^(3/2)*b^2*c 
^2 + 21*(b*d*x - b*c)^2*sqrt(-b*d*x + b*c)*b*c + 5*(b*d*x - b*c)^3*sqrt(-b 
*d*x + b*c))*c^2/b^3 - 5*(693*sqrt(-b*d*x + b*c)*b^5*c^5 - 1155*(-b*d*x + 
b*c)^(3/2)*b^4*c^4 + 1386*(b*d*x - b*c)^2*sqrt(-b*d*x + b*c)*b^3*c^3 + 990 
*(b*d*x - b*c)^3*sqrt(-b*d*x + b*c)*b^2*c^2 + 385*(b*d*x - b*c)^4*sqrt(-b* 
d*x + b*c)*b*c + 63*(b*d*x - b*c)^5*sqrt(-b*d*x + b*c))/b^5)/d^4
 

Mupad [B] (verification not implemented)

Time = 7.24 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.62 \[ \int x^3 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=-\frac {\sqrt {b\,c^2-b\,d^2\,x^2}\,\left (\frac {608\,c^5\,\sqrt {c+d\,x}}{3465\,d^5}-\frac {2\,x^5\,\sqrt {c+d\,x}}{11}+\frac {38\,c^2\,x^3\,\sqrt {c+d\,x}}{693\,d^2}+\frac {76\,c^3\,x^2\,\sqrt {c+d\,x}}{1155\,d^3}-\frac {20\,c\,x^4\,\sqrt {c+d\,x}}{99\,d}+\frac {304\,c^4\,x\,\sqrt {c+d\,x}}{3465\,d^4}\right )}{x+\frac {c}{d}} \] Input:

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

Output:

-((b*c^2 - b*d^2*x^2)^(1/2)*((608*c^5*(c + d*x)^(1/2))/(3465*d^5) - (2*x^5 
*(c + d*x)^(1/2))/11 + (38*c^2*x^3*(c + d*x)^(1/2))/(693*d^2) + (76*c^3*x^ 
2*(c + d*x)^(1/2))/(1155*d^3) - (20*c*x^4*(c + d*x)^(1/2))/(99*d) + (304*c 
^4*x*(c + d*x)^(1/2))/(3465*d^4)))/(x + c/d)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.33 \[ \int x^3 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=\frac {2 \sqrt {b}\, \sqrt {-d x +c}\, \left (315 d^{5} x^{5}+350 c \,d^{4} x^{4}-95 c^{2} d^{3} x^{3}-114 c^{3} d^{2} x^{2}-152 c^{4} d x -304 c^{5}\right )}{3465 d^{4}} \] Input:

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

Output:

(2*sqrt(b)*sqrt(c - d*x)*( - 304*c**5 - 152*c**4*d*x - 114*c**3*d**2*x**2 
- 95*c**2*d**3*x**3 + 350*c*d**4*x**4 + 315*d**5*x**5))/(3465*d**4)