\(\int (A+B x) (c+d x)^{3/2} (c^2-d^2 x^2)^{3/2} \, dx\) [73]

Optimal result

Integrand size = 31, antiderivative size = 205 \[ \int (A+B x) (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {256 c^3 (3 B c+13 A d) \left (c^2-d^2 x^2\right )^{5/2}}{15015 d^2 (c+d x)^{5/2}}-\frac {64 c^2 (3 B c+13 A d) \left (c^2-d^2 x^2\right )^{5/2}}{3003 d^2 (c+d x)^{3/2}}-\frac {8 c (3 B c+13 A d) \left (c^2-d^2 x^2\right )^{5/2}}{429 d^2 \sqrt {c+d x}}-\frac {2 (3 B c+13 A d) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{143 d^2}-\frac {2 B (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d^2} \] Output:

-256/15015*c^3*(13*A*d+3*B*c)*(-d^2*x^2+c^2)^(5/2)/d^2/(d*x+c)^(5/2)-64/30 
03*c^2*(13*A*d+3*B*c)*(-d^2*x^2+c^2)^(5/2)/d^2/(d*x+c)^(3/2)-8/429*c*(13*A 
*d+3*B*c)*(-d^2*x^2+c^2)^(5/2)/d^2/(d*x+c)^(1/2)-2/143*(13*A*d+3*B*c)*(d*x 
+c)^(1/2)*(-d^2*x^2+c^2)^(5/2)/d^2-2/13*B*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(5/ 
2)/d^2
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.58 \[ \int (A+B x) (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {2 (c-d x)^2 \sqrt {c^2-d^2 x^2} \left (13 A d \left (533 c^3+755 c^2 d x+455 c d^2 x^2+105 d^3 x^3\right )+3 B \left (918 c^4+2295 c^3 d x+2765 c^2 d^2 x^2+1645 c d^3 x^3+385 d^4 x^4\right )\right )}{15015 d^2 \sqrt {c+d x}} \] Input:

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

Output:

(-2*(c - d*x)^2*Sqrt[c^2 - d^2*x^2]*(13*A*d*(533*c^3 + 755*c^2*d*x + 455*c 
*d^2*x^2 + 105*d^3*x^3) + 3*B*(918*c^4 + 2295*c^3*d*x + 2765*c^2*d^2*x^2 + 
 1645*c*d^3*x^3 + 385*d^4*x^4)))/(15015*d^2*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {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.

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

\(\Big \downarrow \) 672

\(\displaystyle \frac {(13 A d+3 B c) \int (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}dx}{13 d}-\frac {2 B (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d^2}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {(13 A d+3 B c) \left (\frac {12}{11} c \int \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}dx-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d}\right )}{13 d}-\frac {2 B (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d^2}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {(13 A d+3 B c) \left (\frac {12}{11} c \left (\frac {8}{9} c \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{\sqrt {c+d x}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{9 d \sqrt {c+d x}}\right )-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d}\right )}{13 d}-\frac {2 B (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d^2}\)

\(\Big \downarrow \) 459

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

\(\Big \downarrow \) 458

\(\displaystyle \frac {\left (\frac {12}{11} c \left (\frac {8}{9} c \left (-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{7 d (c+d x)^{3/2}}-\frac {8 c \left (c^2-d^2 x^2\right )^{5/2}}{35 d (c+d x)^{5/2}}\right )-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{9 d \sqrt {c+d x}}\right )-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d}\right ) (13 A d+3 B c)}{13 d}-\frac {2 B (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d^2}\)

Input:

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

Output:

(-2*B*(c + d*x)^(3/2)*(c^2 - d^2*x^2)^(5/2))/(13*d^2) + ((3*B*c + 13*A*d)* 
((-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))/1 
1))/(13*d)
 

Defintions of rubi rules used

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 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]
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.56

method	result	size
gosper	\(-\frac {2 \left (-d x +c \right ) \left (1155 B \,d^{4} x^{4}+1365 A \,d^{4} x^{3}+4935 B c \,d^{3} x^{3}+5915 A c \,d^{3} x^{2}+8295 x^{2} c^{2} B \,d^{2}+9815 A \,c^{2} d^{2} x +6885 B \,c^{3} d x +6929 A \,c^{3} d +2754 B \,c^{4}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{15015 d^{2} \left (d x +c \right )^{\frac {3}{2}}}\)	\(115\)
orering	\(-\frac {2 \left (-d x +c \right ) \left (1155 B \,d^{4} x^{4}+1365 A \,d^{4} x^{3}+4935 B c \,d^{3} x^{3}+5915 A c \,d^{3} x^{2}+8295 x^{2} c^{2} B \,d^{2}+9815 A \,c^{2} d^{2} x +6885 B \,c^{3} d x +6929 A \,c^{3} d +2754 B \,c^{4}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{15015 d^{2} \left (d x +c \right )^{\frac {3}{2}}}\)	\(115\)
default	\(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (-d x +c \right )^{2} \left (1155 B \,d^{4} x^{4}+1365 A \,d^{4} x^{3}+4935 B c \,d^{3} x^{3}+5915 A c \,d^{3} x^{2}+8295 x^{2} c^{2} B \,d^{2}+9815 A \,c^{2} d^{2} x +6885 B \,c^{3} d x +6929 A \,c^{3} d +2754 B \,c^{4}\right )}{15015 \sqrt {d x +c}\, d^{2}}\)	\(117\)
risch	\(-\frac {2 \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}\, \left (1155 B \,d^{6} x^{6}+1365 A \,d^{6} x^{5}+2625 B c \,d^{5} x^{5}+3185 A c \,d^{5} x^{4}-420 B \,c^{2} d^{4} x^{4}-650 A \,c^{2} d^{4} x^{3}-4770 B \,c^{3} d^{3} x^{3}-6786 A \,c^{3} d^{3} x^{2}-2721 B \,c^{4} d^{2} x^{2}-4043 A \,c^{4} d^{2} x +1377 B \,c^{5} d x +6929 A \,c^{5} d +2754 B \,c^{6}\right ) \sqrt {-d x +c}}{15015 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}\)	\(187\)

Input:

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

Output:

-2/15015*(-d*x+c)*(1155*B*d^4*x^4+1365*A*d^4*x^3+4935*B*c*d^3*x^3+5915*A*c 
*d^3*x^2+8295*B*c^2*d^2*x^2+9815*A*c^2*d^2*x+6885*B*c^3*d*x+6929*A*c^3*d+2 
754*B*c^4)*(-d^2*x^2+c^2)^(3/2)/d^2/(d*x+c)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.81 \[ \int (A+B x) (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {2 \, {\left (1155 \, B d^{6} x^{6} + 2754 \, B c^{6} + 6929 \, A c^{5} d + 105 \, {\left (25 \, B c d^{5} + 13 \, A d^{6}\right )} x^{5} - 35 \, {\left (12 \, B c^{2} d^{4} - 91 \, A c d^{5}\right )} x^{4} - 10 \, {\left (477 \, B c^{3} d^{3} + 65 \, A c^{2} d^{4}\right )} x^{3} - 3 \, {\left (907 \, B c^{4} d^{2} + 2262 \, A c^{3} d^{3}\right )} x^{2} + {\left (1377 \, B c^{5} d - 4043 \, A c^{4} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{15015 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:

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

Output:

-2/15015*(1155*B*d^6*x^6 + 2754*B*c^6 + 6929*A*c^5*d + 105*(25*B*c*d^5 + 1 
3*A*d^6)*x^5 - 35*(12*B*c^2*d^4 - 91*A*c*d^5)*x^4 - 10*(477*B*c^3*d^3 + 65 
*A*c^2*d^4)*x^3 - 3*(907*B*c^4*d^2 + 2262*A*c^3*d^3)*x^2 + (1377*B*c^5*d - 
 4043*A*c^4*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)/(d^3*x + c*d^2)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.84 \[ \int (A+B x) (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {2 \, {\left (105 \, d^{5} x^{5} + 245 \, c d^{4} x^{4} - 50 \, c^{2} d^{3} x^{3} - 522 \, c^{3} d^{2} x^{2} - 311 \, c^{4} d x + 533 \, c^{5}\right )} {\left (d x + c\right )} \sqrt {-d x + c} A}{1155 \, {\left (d^{2} x + c d\right )}} - \frac {2 \, {\left (385 \, d^{6} x^{6} + 875 \, c d^{5} x^{5} - 140 \, c^{2} d^{4} x^{4} - 1590 \, c^{3} d^{3} x^{3} - 907 \, c^{4} d^{2} x^{2} + 459 \, c^{5} d x + 918 \, c^{6}\right )} {\left (d x + c\right )} \sqrt {-d x + c} B}{5005 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:

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

Output:

-2/1155*(105*d^5*x^5 + 245*c*d^4*x^4 - 50*c^2*d^3*x^3 - 522*c^3*d^2*x^2 - 
311*c^4*d*x + 533*c^5)*(d*x + c)*sqrt(-d*x + c)*A/(d^2*x + c*d) - 2/5005*( 
385*d^6*x^6 + 875*c*d^5*x^5 - 140*c^2*d^4*x^4 - 1590*c^3*d^3*x^3 - 907*c^4 
*d^2*x^2 + 459*c^5*d*x + 918*c^6)*(d*x + c)*sqrt(-d*x + c)*B/(d^3*x + c*d^ 
2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 870 vs. \(2 (175) = 350\).

Time = 0.12 (sec) , antiderivative size = 870, normalized size of antiderivative = 4.24 \[ \int (A+B x) (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

-2/45045*(45045*sqrt(-d*x + c)*A*c^5*d - 15015*((-d*x + c)^(3/2) - 3*sqrt( 
-d*x + c)*c)*B*c^5 - 15015*((-d*x + c)^(3/2) - 3*sqrt(-d*x + c)*c)*A*c^4*d 
 + 3003*(3*(d*x - c)^2*sqrt(-d*x + c) - 10*(-d*x + c)^(3/2)*c + 15*sqrt(-d 
*x + c)*c^2)*B*c^4 - 6006*(3*(d*x - c)^2*sqrt(-d*x + c) - 10*(-d*x + c)^(3 
/2)*c + 15*sqrt(-d*x + c)*c^2)*A*c^3*d - 2574*(5*(d*x - c)^3*sqrt(-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)*B*c^3 - 2574*(5*(d*x - c)^3*sqrt(-d*x + c) + 21*(d*x - c)^2*s 
qrt(-d*x + c)*c - 35*(-d*x + c)^(3/2)*c^2 + 35*sqrt(-d*x + c)*c^3)*A*c^2*d 
 - 286*(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)*B*c^2 + 143*(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)*A*c*d + 65*(63*(d*x - c)^5*sqrt(-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*s 
qrt(-d*x + c)*c^5)*B*c + 65*(63*(d*x - c)^5*sqrt(-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) 
*A*d + 15*(231*(d*x - c)^6*sqrt(-d*x + c) + 1638*(d*x - c)^5*sqrt(-d*x + c 
)*c + 5005*(d*x - c)^4*sqrt(-d*x + c)*c^2 + 8580*(d*x - c)^3*sqrt(-d*x ...
 

Mupad [B] (verification not implemented)

Time = 9.64 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.93 \[ \int (A+B x) (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {\left (5508\,B\,c^6+13858\,A\,d\,c^5\right )\,\sqrt {c+d\,x}}{15015\,d^3}+\frac {2\,B\,d^3\,x^6\,\sqrt {c+d\,x}}{13}-\frac {4\,c^2\,x^3\,\left (65\,A\,d+477\,B\,c\right )\,\sqrt {c+d\,x}}{3003}+\frac {2\,d^2\,x^5\,\left (13\,A\,d+25\,B\,c\right )\,\sqrt {c+d\,x}}{143}-\frac {2\,c^4\,x\,\left (4043\,A\,d-1377\,B\,c\right )\,\sqrt {c+d\,x}}{15015\,d^2}-\frac {2\,c^3\,x^2\,\left (2262\,A\,d+907\,B\,c\right )\,\sqrt {c+d\,x}}{5005\,d}+\frac {2\,c\,d\,x^4\,\left (91\,A\,d-12\,B\,c\right )\,\sqrt {c+d\,x}}{429}\right )}{x+\frac {c}{d}} \] Input:

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

Output:

-((c^2 - d^2*x^2)^(1/2)*(((5508*B*c^6 + 13858*A*c^5*d)*(c + d*x)^(1/2))/(1 
5015*d^3) + (2*B*d^3*x^6*(c + d*x)^(1/2))/13 - (4*c^2*x^3*(65*A*d + 477*B* 
c)*(c + d*x)^(1/2))/3003 + (2*d^2*x^5*(13*A*d + 25*B*c)*(c + d*x)^(1/2))/1 
43 - (2*c^4*x*(4043*A*d - 1377*B*c)*(c + d*x)^(1/2))/(15015*d^2) - (2*c^3* 
x^2*(2262*A*d + 907*B*c)*(c + d*x)^(1/2))/(5005*d) + (2*c*d*x^4*(91*A*d - 
12*B*c)*(c + d*x)^(1/2))/429))/(x + c/d)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.69 \[ \int (A+B x) (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \, dx=\frac {2 \sqrt {-d x +c}\, \left (-1155 b \,d^{6} x^{6}-1365 a \,d^{6} x^{5}-2625 b c \,d^{5} x^{5}-3185 a c \,d^{5} x^{4}+420 b \,c^{2} d^{4} x^{4}+650 a \,c^{2} d^{4} x^{3}+4770 b \,c^{3} d^{3} x^{3}+6786 a \,c^{3} d^{3} x^{2}+2721 b \,c^{4} d^{2} x^{2}+4043 a \,c^{4} d^{2} x -1377 b \,c^{5} d x -6929 a \,c^{5} d -2754 b \,c^{6}\right )}{15015 d^{2}} \] Input:

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

Output:

(2*sqrt(c - d*x)*( - 6929*a*c**5*d + 4043*a*c**4*d**2*x + 6786*a*c**3*d**3 
*x**2 + 650*a*c**2*d**4*x**3 - 3185*a*c*d**5*x**4 - 1365*a*d**6*x**5 - 275 
4*b*c**6 - 1377*b*c**5*d*x + 2721*b*c**4*d**2*x**2 + 4770*b*c**3*d**3*x**3 
 + 420*b*c**2*d**4*x**4 - 2625*b*c*d**5*x**5 - 1155*b*d**6*x**6))/(15015*d 
**2)