\(\int \sqrt {d+e x} (a+c x^2)^2 \, dx\) [600]

Optimal result

Integrand size = 19, antiderivative size = 127 \[ \int \sqrt {d+e x} \left (a+c x^2\right )^2 \, dx=\frac {2 \left (c d^2+a e^2\right )^2 (d+e x)^{3/2}}{3 e^5}-\frac {8 c d \left (c d^2+a e^2\right ) (d+e x)^{5/2}}{5 e^5}+\frac {4 c \left (3 c d^2+a e^2\right ) (d+e x)^{7/2}}{7 e^5}-\frac {8 c^2 d (d+e x)^{9/2}}{9 e^5}+\frac {2 c^2 (d+e x)^{11/2}}{11 e^5} \]

[Out]

2/3*(a*e^2+c*d^2)^2*(e*x+d)^(3/2)/e^5-8/5*c*d*(a*e^2+c*d^2)*(e*x+d)^(5/2)/e^5+4/7*c*(a*e^2+3*c*d^2)*(e*x+d)^(7
/2)/e^5-8/9*c^2*d*(e*x+d)^(9/2)/e^5+2/11*c^2*(e*x+d)^(11/2)/e^5

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {711} \[ \int \sqrt {d+e x} \left (a+c x^2\right )^2 \, dx=\frac {4 c (d+e x)^{7/2} \left (a e^2+3 c d^2\right )}{7 e^5}-\frac {8 c d (d+e x)^{5/2} \left (a e^2+c d^2\right )}{5 e^5}+\frac {2 (d+e x)^{3/2} \left (a e^2+c d^2\right )^2}{3 e^5}+\frac {2 c^2 (d+e x)^{11/2}}{11 e^5}-\frac {8 c^2 d (d+e x)^{9/2}}{9 e^5} \]

[In]

Int[Sqrt[d + e*x]*(a + c*x^2)^2,x]

[Out]

(2*(c*d^2 + a*e^2)^2*(d + e*x)^(3/2))/(3*e^5) - (8*c*d*(c*d^2 + a*e^2)*(d + e*x)^(5/2))/(5*e^5) + (4*c*(3*c*d^
2 + a*e^2)*(d + e*x)^(7/2))/(7*e^5) - (8*c^2*d*(d + e*x)^(9/2))/(9*e^5) + (2*c^2*(d + e*x)^(11/2))/(11*e^5)

Rule 711

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (c d^2+a e^2\right )^2 \sqrt {d+e x}}{e^4}-\frac {4 c d \left (c d^2+a e^2\right ) (d+e x)^{3/2}}{e^4}+\frac {2 c \left (3 c d^2+a e^2\right ) (d+e x)^{5/2}}{e^4}-\frac {4 c^2 d (d+e x)^{7/2}}{e^4}+\frac {c^2 (d+e x)^{9/2}}{e^4}\right ) \, dx \\ & = \frac {2 \left (c d^2+a e^2\right )^2 (d+e x)^{3/2}}{3 e^5}-\frac {8 c d \left (c d^2+a e^2\right ) (d+e x)^{5/2}}{5 e^5}+\frac {4 c \left (3 c d^2+a e^2\right ) (d+e x)^{7/2}}{7 e^5}-\frac {8 c^2 d (d+e x)^{9/2}}{9 e^5}+\frac {2 c^2 (d+e x)^{11/2}}{11 e^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76 \[ \int \sqrt {d+e x} \left (a+c x^2\right )^2 \, dx=\frac {2 (d+e x)^{3/2} \left (1155 a^2 e^4+66 a c e^2 \left (8 d^2-12 d e x+15 e^2 x^2\right )+c^2 \left (128 d^4-192 d^3 e x+240 d^2 e^2 x^2-280 d e^3 x^3+315 e^4 x^4\right )\right )}{3465 e^5} \]

[In]

Integrate[Sqrt[d + e*x]*(a + c*x^2)^2,x]

[Out]

(2*(d + e*x)^(3/2)*(1155*a^2*e^4 + 66*a*c*e^2*(8*d^2 - 12*d*e*x + 15*e^2*x^2) + c^2*(128*d^4 - 192*d^3*e*x + 2
40*d^2*e^2*x^2 - 280*d*e^3*x^3 + 315*e^4*x^4)))/(3465*e^5)

Maple [A] (verified)

Time = 2.07 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.69

method	result	size
pseudoelliptic	\(\frac {2 \left (\left (\frac {3}{11} x^{4} c^{2}+\frac {6}{7} a c \,x^{2}+a^{2}\right ) e^{4}-\frac {24 x c d \left (\frac {35 c \,x^{2}}{99}+a \right ) e^{3}}{35}+\frac {16 c \left (\frac {5 c \,x^{2}}{11}+a \right ) d^{2} e^{2}}{35}-\frac {64 x \,c^{2} d^{3} e}{385}+\frac {128 c^{2} d^{4}}{1155}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3 e^{5}}\)	\(88\)
gosper	\(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (315 c^{2} x^{4} e^{4}-280 x^{3} c^{2} d \,e^{3}+990 x^{2} a c \,e^{4}+240 x^{2} c^{2} d^{2} e^{2}-792 x a c d \,e^{3}-192 x \,c^{2} d^{3} e +1155 a^{2} e^{4}+528 a c \,d^{2} e^{2}+128 c^{2} d^{4}\right )}{3465 e^{5}}\)	\(106\)
derivativedivides	\(\frac {\frac {2 c^{2} \left (e x +d \right )^{\frac {11}{2}}}{11}-\frac {8 c^{2} d \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (2 \left (e^{2} a +c \,d^{2}\right ) c +4 c^{2} d^{2}\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}-\frac {8 \left (e^{2} a +c \,d^{2}\right ) c d \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (e^{2} a +c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{5}}\)	\(108\)
default	\(\frac {\frac {2 c^{2} \left (e x +d \right )^{\frac {11}{2}}}{11}-\frac {8 c^{2} d \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (2 \left (e^{2} a +c \,d^{2}\right ) c +4 c^{2} d^{2}\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}-\frac {8 \left (e^{2} a +c \,d^{2}\right ) c d \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (e^{2} a +c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{5}}\)	\(108\)
trager	\(\frac {2 \left (315 c^{2} e^{5} x^{5}+35 c^{2} d \,e^{4} x^{4}+990 a c \,e^{5} x^{3}-40 d^{2} e^{3} x^{3} c^{2}+198 a c d \,e^{4} x^{2}+48 c^{2} d^{3} e^{2} x^{2}+1155 a^{2} e^{5} x -264 a c \,d^{2} e^{3} x -64 c^{2} d^{4} e x +1155 a^{2} d \,e^{4}+528 a c \,d^{3} e^{2}+128 c^{2} d^{5}\right ) \sqrt {e x +d}}{3465 e^{5}}\)	\(143\)
risch	\(\frac {2 \left (315 c^{2} e^{5} x^{5}+35 c^{2} d \,e^{4} x^{4}+990 a c \,e^{5} x^{3}-40 d^{2} e^{3} x^{3} c^{2}+198 a c d \,e^{4} x^{2}+48 c^{2} d^{3} e^{2} x^{2}+1155 a^{2} e^{5} x -264 a c \,d^{2} e^{3} x -64 c^{2} d^{4} e x +1155 a^{2} d \,e^{4}+528 a c \,d^{3} e^{2}+128 c^{2} d^{5}\right ) \sqrt {e x +d}}{3465 e^{5}}\)	\(143\)

[In]

int((c*x^2+a)^2*(e*x+d)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/3*((3/11*x^4*c^2+6/7*a*c*x^2+a^2)*e^4-24/35*x*c*d*(35/99*c*x^2+a)*e^3+16/35*c*(5/11*c*x^2+a)*d^2*e^2-64/385*
x*c^2*d^3*e+128/1155*c^2*d^4)*(e*x+d)^(3/2)/e^5

Fricas [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.13 \[ \int \sqrt {d+e x} \left (a+c x^2\right )^2 \, dx=\frac {2 \, {\left (315 \, c^{2} e^{5} x^{5} + 35 \, c^{2} d e^{4} x^{4} + 128 \, c^{2} d^{5} + 528 \, a c d^{3} e^{2} + 1155 \, a^{2} d e^{4} - 10 \, {\left (4 \, c^{2} d^{2} e^{3} - 99 \, a c e^{5}\right )} x^{3} + 6 \, {\left (8 \, c^{2} d^{3} e^{2} + 33 \, a c d e^{4}\right )} x^{2} - {\left (64 \, c^{2} d^{4} e + 264 \, a c d^{2} e^{3} - 1155 \, a^{2} e^{5}\right )} x\right )} \sqrt {e x + d}}{3465 \, e^{5}} \]

[In]

integrate((c*x^2+a)^2*(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

2/3465*(315*c^2*e^5*x^5 + 35*c^2*d*e^4*x^4 + 128*c^2*d^5 + 528*a*c*d^3*e^2 + 1155*a^2*d*e^4 - 10*(4*c^2*d^2*e^
3 - 99*a*c*e^5)*x^3 + 6*(8*c^2*d^3*e^2 + 33*a*c*d*e^4)*x^2 - (64*c^2*d^4*e + 264*a*c*d^2*e^3 - 1155*a^2*e^5)*x
)*sqrt(e*x + d)/e^5

Sympy [A] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.39 \[ \int \sqrt {d+e x} \left (a+c x^2\right )^2 \, dx=\begin {cases} \frac {2 \left (- \frac {4 c^{2} d \left (d + e x\right )^{\frac {9}{2}}}{9 e^{4}} + \frac {c^{2} \left (d + e x\right )^{\frac {11}{2}}}{11 e^{4}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \cdot \left (2 a c e^{2} + 6 c^{2} d^{2}\right )}{7 e^{4}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (- 4 a c d e^{2} - 4 c^{2} d^{3}\right )}{5 e^{4}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (a^{2} e^{4} + 2 a c d^{2} e^{2} + c^{2} d^{4}\right )}{3 e^{4}}\right )}{e} & \text {for}\: e \neq 0 \\\sqrt {d} \left (a^{2} x + \frac {2 a c x^{3}}{3} + \frac {c^{2} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((c*x**2+a)**2*(e*x+d)**(1/2),x)

[Out]

Piecewise((2*(-4*c**2*d*(d + e*x)**(9/2)/(9*e**4) + c**2*(d + e*x)**(11/2)/(11*e**4) + (d + e*x)**(7/2)*(2*a*c
*e**2 + 6*c**2*d**2)/(7*e**4) + (d + e*x)**(5/2)*(-4*a*c*d*e**2 - 4*c**2*d**3)/(5*e**4) + (d + e*x)**(3/2)*(a*
*2*e**4 + 2*a*c*d**2*e**2 + c**2*d**4)/(3*e**4))/e, Ne(e, 0)), (sqrt(d)*(a**2*x + 2*a*c*x**3/3 + c**2*x**5/5),
True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.89 \[ \int \sqrt {d+e x} \left (a+c x^2\right )^2 \, dx=\frac {2 \, {\left (315 \, {\left (e x + d\right )}^{\frac {11}{2}} c^{2} - 1540 \, {\left (e x + d\right )}^{\frac {9}{2}} c^{2} d + 990 \, {\left (3 \, c^{2} d^{2} + a c e^{2}\right )} {\left (e x + d\right )}^{\frac {7}{2}} - 2772 \, {\left (c^{2} d^{3} + a c d e^{2}\right )} {\left (e x + d\right )}^{\frac {5}{2}} + 1155 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} {\left (e x + d\right )}^{\frac {3}{2}}\right )}}{3465 \, e^{5}} \]

[In]

integrate((c*x^2+a)^2*(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

2/3465*(315*(e*x + d)^(11/2)*c^2 - 1540*(e*x + d)^(9/2)*c^2*d + 990*(3*c^2*d^2 + a*c*e^2)*(e*x + d)^(7/2) - 27
72*(c^2*d^3 + a*c*d*e^2)*(e*x + d)^(5/2) + 1155*(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*(e*x + d)^(3/2))/e^5

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (107) = 214\).

Time = 0.26 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.16 \[ \int \sqrt {d+e x} \left (a+c x^2\right )^2 \, dx=\frac {2 \, {\left (3465 \, \sqrt {e x + d} a^{2} d + 1155 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{2} + \frac {462 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a c d}{e^{2}} + \frac {198 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} a c}{e^{2}} + \frac {11 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} c^{2} d}{e^{4}} + \frac {5 \, {\left (63 \, {\left (e x + d\right )}^{\frac {11}{2}} - 385 \, {\left (e x + d\right )}^{\frac {9}{2}} d + 990 \, {\left (e x + d\right )}^{\frac {7}{2}} d^{2} - 1386 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{3} + 1155 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{4} - 693 \, \sqrt {e x + d} d^{5}\right )} c^{2}}{e^{4}}\right )}}{3465 \, e} \]

[In]

integrate((c*x^2+a)^2*(e*x+d)^(1/2),x, algorithm="giac")

[Out]

2/3465*(3465*sqrt(e*x + d)*a^2*d + 1155*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a^2 + 462*(3*(e*x + d)^(5/2) - 1
0*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a*c*d/e^2 + 198*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*
x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*a*c/e^2 + 11*(35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x +
d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*c^2*d/e^4 + 5*(63*(e*x + d)^(11/2) - 385*(e*x
+ d)^(9/2)*d + 990*(e*x + d)^(7/2)*d^2 - 1386*(e*x + d)^(5/2)*d^3 + 1155*(e*x + d)^(3/2)*d^4 - 693*sqrt(e*x +
d)*d^5)*c^2/e^4)/e

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.90 \[ \int \sqrt {d+e x} \left (a+c x^2\right )^2 \, dx=\frac {2\,c^2\,{\left (d+e\,x\right )}^{11/2}}{11\,e^5}-\frac {\left (8\,c^2\,d^3+8\,a\,c\,d\,e^2\right )\,{\left (d+e\,x\right )}^{5/2}}{5\,e^5}+\frac {2\,{\left (c\,d^2+a\,e^2\right )}^2\,{\left (d+e\,x\right )}^{3/2}}{3\,e^5}+\frac {\left (12\,c^2\,d^2+4\,a\,c\,e^2\right )\,{\left (d+e\,x\right )}^{7/2}}{7\,e^5}-\frac {8\,c^2\,d\,{\left (d+e\,x\right )}^{9/2}}{9\,e^5} \]

[In]

int((a + c*x^2)^2*(d + e*x)^(1/2),x)

[Out]

(2*c^2*(d + e*x)^(11/2))/(11*e^5) - ((8*c^2*d^3 + 8*a*c*d*e^2)*(d + e*x)^(5/2))/(5*e^5) + (2*(a*e^2 + c*d^2)^2
*(d + e*x)^(3/2))/(3*e^5) + ((12*c^2*d^2 + 4*a*c*e^2)*(d + e*x)^(7/2))/(7*e^5) - (8*c^2*d*(d + e*x)^(9/2))/(9*
e^5)