\(\int (A+B x) (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \, dx\) [64]

Optimal result

Integrand size = 31, antiderivative size = 205 \[ \int (A+B x) (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \, dx=-\frac {256 c^3 (5 B c+11 A d) \left (c^2-d^2 x^2\right )^{3/2}}{3465 d^2 (c+d x)^{3/2}}-\frac {64 c^2 (5 B c+11 A d) \left (c^2-d^2 x^2\right )^{3/2}}{1155 d^2 \sqrt {c+d x}}-\frac {8 c (5 B c+11 A d) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{231 d^2}-\frac {2 (5 B c+11 A d) (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{99 d^2}-\frac {2 B (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2}}{11 d^2} \] Output:

-256/3465*c^3*(11*A*d+5*B*c)*(-d^2*x^2+c^2)^(3/2)/d^2/(d*x+c)^(3/2)-64/115 
5*c^2*(11*A*d+5*B*c)*(-d^2*x^2+c^2)^(3/2)/d^2/(d*x+c)^(1/2)-8/231*c*(11*A* 
d+5*B*c)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(3/2)/d^2-2/99*(11*A*d+5*B*c)*(d*x+c 
)^(3/2)*(-d^2*x^2+c^2)^(3/2)/d^2-2/11*B*(d*x+c)^(5/2)*(-d^2*x^2+c^2)^(3/2) 
/d^2
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.57 \[ \int (A+B x) (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \, dx=-\frac {2 (c-d x) \sqrt {c^2-d^2 x^2} \left (11 A d \left (319 c^3+321 c^2 d x+165 c d^2 x^2+35 d^3 x^3\right )+5 B \left (382 c^4+573 c^3 d x+543 c^2 d^2 x^2+287 c d^3 x^3+63 d^4 x^4\right )\right )}{3465 d^2 \sqrt {c+d x}} \] Input:

Integrate[(A + B*x)*(c + d*x)^(5/2)*Sqrt[c^2 - d^2*x^2],x]
 

Output:

(-2*(c - d*x)*Sqrt[c^2 - d^2*x^2]*(11*A*d*(319*c^3 + 321*c^2*d*x + 165*c*d 
^2*x^2 + 35*d^3*x^3) + 5*B*(382*c^4 + 573*c^3*d*x + 543*c^2*d^2*x^2 + 287* 
c*d^3*x^3 + 63*d^4*x^4)))/(3465*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.

\(\displaystyle \int (A+B x) (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \, dx\)

\(\Big \downarrow \) 672

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

\(\Big \downarrow \) 459

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

\(\Big \downarrow \) 459

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

\(\Big \downarrow \) 459

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

\(\Big \downarrow \) 458

\(\displaystyle \frac {\left (\frac {4}{3} c \left (\frac {8}{7} c \left (-\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d \sqrt {c+d x}}-\frac {8 c \left (c^2-d^2 x^2\right )^{3/2}}{15 d (c+d x)^{3/2}}\right )-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d}\right )-\frac {2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d}\right ) (11 A d+5 B c)}{11 d}-\frac {2 B (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2}}{11 d^2}\)

Input:

Int[(A + B*x)*(c + d*x)^(5/2)*Sqrt[c^2 - d^2*x^2],x]
 

Output:

(-2*B*(c + d*x)^(5/2)*(c^2 - d^2*x^2)^(3/2))/(11*d^2) + ((5*B*c + 11*A*d)* 
((-2*(c + d*x)^(3/2)*(c^2 - d^2*x^2)^(3/2))/(9*d) + (4*c*((-2*Sqrt[c + d*x 
]*(c^2 - d^2*x^2)^(3/2))/(7*d) + (8*c*((-8*c*(c^2 - d^2*x^2)^(3/2))/(15*d* 
(c + d*x)^(3/2)) - (2*(c^2 - d^2*x^2)^(3/2))/(5*d*Sqrt[c + d*x])))/7))/3)) 
/(11*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 = 0.43 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.56

method	result	size
gosper	\(-\frac {2 \left (-d x +c \right ) \left (315 B \,d^{4} x^{4}+385 A \,d^{4} x^{3}+1435 B c \,d^{3} x^{3}+1815 A c \,d^{3} x^{2}+2715 x^{2} c^{2} B \,d^{2}+3531 A \,c^{2} d^{2} x +2865 B \,c^{3} d x +3509 A \,c^{3} d +1910 B \,c^{4}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{3465 d^{2} \sqrt {d x +c}}\)	\(115\)
default	\(-\frac {2 \left (-d x +c \right ) \left (315 B \,d^{4} x^{4}+385 A \,d^{4} x^{3}+1435 B c \,d^{3} x^{3}+1815 A c \,d^{3} x^{2}+2715 x^{2} c^{2} B \,d^{2}+3531 A \,c^{2} d^{2} x +2865 B \,c^{3} d x +3509 A \,c^{3} d +1910 B \,c^{4}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{3465 d^{2} \sqrt {d x +c}}\)	\(115\)
orering	\(-\frac {2 \left (-d x +c \right ) \left (315 B \,d^{4} x^{4}+385 A \,d^{4} x^{3}+1435 B c \,d^{3} x^{3}+1815 A c \,d^{3} x^{2}+2715 x^{2} c^{2} B \,d^{2}+3531 A \,c^{2} d^{2} x +2865 B \,c^{3} d x +3509 A \,c^{3} d +1910 B \,c^{4}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{3465 d^{2} \sqrt {d x +c}}\)	\(115\)
risch	\(-\frac {2 \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}\, \left (-315 B \,d^{5} x^{5}-385 A \,d^{5} x^{4}-1120 B c \,d^{4} x^{4}-1430 A c \,d^{4} x^{3}-1280 B \,c^{2} d^{3} x^{3}-1716 A \,c^{2} d^{3} x^{2}-150 B \,c^{3} d^{2} x^{2}+22 A \,c^{3} d^{2} x +955 B \,c^{4} d x +3509 A \,c^{4} d +1910 B \,c^{5}\right ) \sqrt {-d x +c}}{3465 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}\)	\(163\)

Input:

int((B*x+A)*(d*x+c)^(5/2)*(-d^2*x^2+c^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/3465*(-d*x+c)*(315*B*d^4*x^4+385*A*d^4*x^3+1435*B*c*d^3*x^3+1815*A*c*d^ 
3*x^2+2715*B*c^2*d^2*x^2+3531*A*c^2*d^2*x+2865*B*c^3*d*x+3509*A*c^3*d+1910 
*B*c^4)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.70 \[ \int (A+B x) (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \, dx=\frac {2 \, {\left (315 \, B d^{5} x^{5} - 1910 \, B c^{5} - 3509 \, A c^{4} d + 35 \, {\left (32 \, B c d^{4} + 11 \, A d^{5}\right )} x^{4} + 10 \, {\left (128 \, B c^{2} d^{3} + 143 \, A c d^{4}\right )} x^{3} + 6 \, {\left (25 \, B c^{3} d^{2} + 286 \, A c^{2} d^{3}\right )} x^{2} - {\left (955 \, B c^{4} d + 22 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{3465 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:

integrate((B*x+A)*(d*x+c)^(5/2)*(-d^2*x^2+c^2)^(1/2),x, algorithm="fricas" 
)
 

Output:

2/3465*(315*B*d^5*x^5 - 1910*B*c^5 - 3509*A*c^4*d + 35*(32*B*c*d^4 + 11*A* 
d^5)*x^4 + 10*(128*B*c^2*d^3 + 143*A*c*d^4)*x^3 + 6*(25*B*c^3*d^2 + 286*A* 
c^2*d^3)*x^2 - (955*B*c^4*d + 22*A*c^3*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d 
*x + c)/(d^3*x + c*d^2)
 

Sympy [F]

\[ \int (A+B x) (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \, dx=\int \sqrt {- \left (- c + d x\right ) \left (c + d x\right )} \left (A + B x\right ) \left (c + d x\right )^{\frac {5}{2}}\, dx \] Input:

integrate((B*x+A)*(d*x+c)**(5/2)*(-d**2*x**2+c**2)**(1/2),x)
 

Output:

Integral(sqrt(-(-c + d*x)*(c + d*x))*(A + B*x)*(c + d*x)**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.73 \[ \int (A+B x) (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \, dx=\frac {2 \, {\left (35 \, d^{4} x^{4} + 130 \, c d^{3} x^{3} + 156 \, c^{2} d^{2} x^{2} - 2 \, c^{3} d x - 319 \, c^{4}\right )} {\left (d x + c\right )} \sqrt {-d x + c} A}{315 \, {\left (d^{2} x + c d\right )}} + \frac {2 \, {\left (63 \, d^{5} x^{5} + 224 \, c d^{4} x^{4} + 256 \, c^{2} d^{3} x^{3} + 30 \, c^{3} d^{2} x^{2} - 191 \, c^{4} d x - 382 \, c^{5}\right )} {\left (d x + c\right )} \sqrt {-d x + c} B}{693 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:

integrate((B*x+A)*(d*x+c)^(5/2)*(-d^2*x^2+c^2)^(1/2),x, algorithm="maxima" 
)
 

Output:

2/315*(35*d^4*x^4 + 130*c*d^3*x^3 + 156*c^2*d^2*x^2 - 2*c^3*d*x - 319*c^4) 
*(d*x + c)*sqrt(-d*x + c)*A/(d^2*x + c*d) + 2/693*(63*d^5*x^5 + 224*c*d^4* 
x^4 + 256*c^2*d^3*x^3 + 30*c^3*d^2*x^2 - 191*c^4*d*x - 382*c^5)*(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. 492 vs. \(2 (175) = 350\).

Time = 0.12 (sec) , antiderivative size = 492, normalized size of antiderivative = 2.40 \[ \int (A+B x) (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \, dx=-\frac {2 \, {\left (3465 \, \sqrt {-d x + c} A c^{4} d - 1155 \, {\left ({\left (-d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {-d x + c} c\right )} B c^{4} - 2310 \, {\left ({\left (-d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {-d x + c} c\right )} A c^{3} d + 462 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} - 10 \, {\left (-d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {-d x + c} c^{2}\right )} B c^{3} - 198 \, {\left (5 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} + 21 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{2} + 35 \, \sqrt {-d x + c} c^{3}\right )} A c d - 22 \, {\left (35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} + 180 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c + 378 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{2} - 420 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{3} + 315 \, \sqrt {-d x + c} c^{4}\right )} B c - 11 \, {\left (35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} + 180 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c + 378 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{2} - 420 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{3} + 315 \, \sqrt {-d x + c} c^{4}\right )} A d - 5 \, {\left (63 \, {\left (d x - c\right )}^{5} \sqrt {-d x + c} + 385 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} c + 990 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c^{2} + 1386 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{3} - 1155 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{4} + 693 \, \sqrt {-d x + c} c^{5}\right )} B\right )}}{3465 \, d^{2}} \] Input:

integrate((B*x+A)*(d*x+c)^(5/2)*(-d^2*x^2+c^2)^(1/2),x, algorithm="giac")
 

Output:

-2/3465*(3465*sqrt(-d*x + c)*A*c^4*d - 1155*((-d*x + c)^(3/2) - 3*sqrt(-d* 
x + c)*c)*B*c^4 - 2310*((-d*x + c)^(3/2) - 3*sqrt(-d*x + c)*c)*A*c^3*d + 4 
62*(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^3 - 198*(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)*A*c*d - 22*(3 
5*(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 - 11*(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 + 3 
15*sqrt(-d*x + c)*c^4)*A*d - 5*(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)*B)/d^2
 

Mupad [B] (verification not implemented)

Time = 10.36 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.83 \[ \int (A+B x) (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \, dx=\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {4\,c\,x^3\,\left (143\,A\,d+128\,B\,c\right )\,\sqrt {c+d\,x}}{693}-\frac {\left (3820\,B\,c^5+7018\,A\,d\,c^4\right )\,\sqrt {c+d\,x}}{3465\,d^3}+\frac {2\,B\,d^2\,x^5\,\sqrt {c+d\,x}}{11}+\frac {x^4\,\left (770\,A\,d^5+2240\,B\,c\,d^4\right )\,\sqrt {c+d\,x}}{3465\,d^3}-\frac {2\,c^3\,x\,\left (22\,A\,d+955\,B\,c\right )\,\sqrt {c+d\,x}}{3465\,d^2}+\frac {4\,c^2\,x^2\,\left (286\,A\,d+25\,B\,c\right )\,\sqrt {c+d\,x}}{1155\,d}\right )}{x+\frac {c}{d}} \] Input:

int((c^2 - d^2*x^2)^(1/2)*(A + B*x)*(c + d*x)^(5/2),x)
 

Output:

((c^2 - d^2*x^2)^(1/2)*((4*c*x^3*(143*A*d + 128*B*c)*(c + d*x)^(1/2))/693 
- ((3820*B*c^5 + 7018*A*c^4*d)*(c + d*x)^(1/2))/(3465*d^3) + (2*B*d^2*x^5* 
(c + d*x)^(1/2))/11 + (x^4*(770*A*d^5 + 2240*B*c*d^4)*(c + d*x)^(1/2))/(34 
65*d^3) - (2*c^3*x*(22*A*d + 955*B*c)*(c + d*x)^(1/2))/(3465*d^2) + (4*c^2 
*x^2*(286*A*d + 25*B*c)*(c + d*x)^(1/2))/(1155*d)))/(x + c/d)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.58 \[ \int (A+B x) (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \, dx=\frac {2 \sqrt {-d x +c}\, \left (315 b \,d^{5} x^{5}+385 a \,d^{5} x^{4}+1120 b c \,d^{4} x^{4}+1430 a c \,d^{4} x^{3}+1280 b \,c^{2} d^{3} x^{3}+1716 a \,c^{2} d^{3} x^{2}+150 b \,c^{3} d^{2} x^{2}-22 a \,c^{3} d^{2} x -955 b \,c^{4} d x -3509 a \,c^{4} d -1910 b \,c^{5}\right )}{3465 d^{2}} \] Input:

int((B*x+A)*(d*x+c)^(5/2)*(-d^2*x^2+c^2)^(1/2),x)
 

Output:

(2*sqrt(c - d*x)*( - 3509*a*c**4*d - 22*a*c**3*d**2*x + 1716*a*c**2*d**3*x 
**2 + 1430*a*c*d**4*x**3 + 385*a*d**5*x**4 - 1910*b*c**5 - 955*b*c**4*d*x 
+ 150*b*c**3*d**2*x**2 + 1280*b*c**2*d**3*x**3 + 1120*b*c*d**4*x**4 + 315* 
b*d**5*x**5))/(3465*d**2)