\(\int (d+e x)^{5/2} (a+c x^2) \, dx\) [591]

Optimal result

Integrand size = 17, antiderivative size = 63 \[ \int (d+e x)^{5/2} \left (a+c x^2\right ) \, dx=\frac {2 \left (c d^2+a e^2\right ) (d+e x)^{7/2}}{7 e^3}-\frac {4 c d (d+e x)^{9/2}}{9 e^3}+\frac {2 c (d+e x)^{11/2}}{11 e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {711} \[ \int (d+e x)^{5/2} \left (a+c x^2\right ) \, dx=\frac {2 (d+e x)^{7/2} \left (a e^2+c d^2\right )}{7 e^3}+\frac {2 c (d+e x)^{11/2}}{11 e^3}-\frac {4 c d (d+e x)^{9/2}}{9 e^3} \]

[In]

Int[(d + e*x)^(5/2)*(a + c*x^2),x]

[Out]

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

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 ) (d+e x)^{5/2}}{e^2}-\frac {2 c d (d+e x)^{7/2}}{e^2}+\frac {c (d+e x)^{9/2}}{e^2}\right ) \, dx \\ & = \frac {2 \left (c d^2+a e^2\right ) (d+e x)^{7/2}}{7 e^3}-\frac {4 c d (d+e x)^{9/2}}{9 e^3}+\frac {2 c (d+e x)^{11/2}}{11 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int (d+e x)^{5/2} \left (a+c x^2\right ) \, dx=\frac {2 (d+e x)^{7/2} \left (99 a e^2+c \left (8 d^2-28 d e x+63 e^2 x^2\right )\right )}{693 e^3} \]

[In]

Integrate[(d + e*x)^(5/2)*(a + c*x^2),x]

[Out]

(2*(d + e*x)^(7/2)*(99*a*e^2 + c*(8*d^2 - 28*d*e*x + 63*e^2*x^2)))/(693*e^3)

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.65

method	result	size
gosper	\(\frac {2 \left (e x +d \right )^{\frac {7}{2}} \left (63 c \,x^{2} e^{2}-28 x c d e +99 e^{2} a +8 c \,d^{2}\right )}{693 e^{3}}\)	\(41\)
pseudoelliptic	\(\frac {2 \left (e x +d \right )^{\frac {7}{2}} \left (63 c \,x^{2} e^{2}-28 x c d e +99 e^{2} a +8 c \,d^{2}\right )}{693 e^{3}}\)	\(41\)
derivativedivides	\(\frac {\frac {2 c \left (e x +d \right )^{\frac {11}{2}}}{11}-\frac {4 c d \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (e^{2} a +c \,d^{2}\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}}{e^{3}}\)	\(48\)
default	\(\frac {\frac {2 c \left (e x +d \right )^{\frac {11}{2}}}{11}-\frac {4 c d \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (e^{2} a +c \,d^{2}\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}}{e^{3}}\)	\(48\)
trager	\(\frac {2 \left (63 e^{5} c \,x^{5}+161 c d \,e^{4} x^{4}+99 a \,e^{5} x^{3}+113 d^{2} e^{3} c \,x^{3}+297 a d \,e^{4} x^{2}+3 d^{3} e^{2} c \,x^{2}+297 a \,d^{2} e^{3} x -4 c \,d^{4} e x +99 a \,d^{3} e^{2}+8 c \,d^{5}\right ) \sqrt {e x +d}}{693 e^{3}}\)	\(109\)
risch	\(\frac {2 \left (63 e^{5} c \,x^{5}+161 c d \,e^{4} x^{4}+99 a \,e^{5} x^{3}+113 d^{2} e^{3} c \,x^{3}+297 a d \,e^{4} x^{2}+3 d^{3} e^{2} c \,x^{2}+297 a \,d^{2} e^{3} x -4 c \,d^{4} e x +99 a \,d^{3} e^{2}+8 c \,d^{5}\right ) \sqrt {e x +d}}{693 e^{3}}\)	\(109\)

[In]

int((e*x+d)^(5/2)*(c*x^2+a),x,method=_RETURNVERBOSE)

[Out]

2/693*(e*x+d)^(7/2)*(63*c*e^2*x^2-28*c*d*e*x+99*a*e^2+8*c*d^2)/e^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (51) = 102\).

Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.71 \[ \int (d+e x)^{5/2} \left (a+c x^2\right ) \, dx=\frac {2 \, {\left (63 \, c e^{5} x^{5} + 161 \, c d e^{4} x^{4} + 8 \, c d^{5} + 99 \, a d^{3} e^{2} + {\left (113 \, c d^{2} e^{3} + 99 \, a e^{5}\right )} x^{3} + 3 \, {\left (c d^{3} e^{2} + 99 \, a d e^{4}\right )} x^{2} - {\left (4 \, c d^{4} e - 297 \, a d^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{693 \, e^{3}} \]

[In]

integrate((e*x+d)^(5/2)*(c*x^2+a),x, algorithm="fricas")

[Out]

2/693*(63*c*e^5*x^5 + 161*c*d*e^4*x^4 + 8*c*d^5 + 99*a*d^3*e^2 + (113*c*d^2*e^3 + 99*a*e^5)*x^3 + 3*(c*d^3*e^2
+ 99*a*d*e^4)*x^2 - (4*c*d^4*e - 297*a*d^2*e^3)*x)*sqrt(e*x + d)/e^3

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (63) = 126\).

Time = 0.37 (sec) , antiderivative size = 218, normalized size of antiderivative = 3.46 \[ \int (d+e x)^{5/2} \left (a+c x^2\right ) \, dx=\begin {cases} \frac {2 a d^{3} \sqrt {d + e x}}{7 e} + \frac {6 a d^{2} x \sqrt {d + e x}}{7} + \frac {6 a d e x^{2} \sqrt {d + e x}}{7} + \frac {2 a e^{2} x^{3} \sqrt {d + e x}}{7} + \frac {16 c d^{5} \sqrt {d + e x}}{693 e^{3}} - \frac {8 c d^{4} x \sqrt {d + e x}}{693 e^{2}} + \frac {2 c d^{3} x^{2} \sqrt {d + e x}}{231 e} + \frac {226 c d^{2} x^{3} \sqrt {d + e x}}{693} + \frac {46 c d e x^{4} \sqrt {d + e x}}{99} + \frac {2 c e^{2} x^{5} \sqrt {d + e x}}{11} & \text {for}\: e \neq 0 \\d^{\frac {5}{2}} \left (a x + \frac {c x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((e*x+d)**(5/2)*(c*x**2+a),x)

[Out]

Piecewise((2*a*d**3*sqrt(d + e*x)/(7*e) + 6*a*d**2*x*sqrt(d + e*x)/7 + 6*a*d*e*x**2*sqrt(d + e*x)/7 + 2*a*e**2
*x**3*sqrt(d + e*x)/7 + 16*c*d**5*sqrt(d + e*x)/(693*e**3) - 8*c*d**4*x*sqrt(d + e*x)/(693*e**2) + 2*c*d**3*x*
*2*sqrt(d + e*x)/(231*e) + 226*c*d**2*x**3*sqrt(d + e*x)/693 + 46*c*d*e*x**4*sqrt(d + e*x)/99 + 2*c*e**2*x**5*
sqrt(d + e*x)/11, Ne(e, 0)), (d**(5/2)*(a*x + c*x**3/3), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.75 \[ \int (d+e x)^{5/2} \left (a+c x^2\right ) \, dx=\frac {2 \, {\left (63 \, {\left (e x + d\right )}^{\frac {11}{2}} c - 154 \, {\left (e x + d\right )}^{\frac {9}{2}} c d + 99 \, {\left (c d^{2} + a e^{2}\right )} {\left (e x + d\right )}^{\frac {7}{2}}\right )}}{693 \, e^{3}} \]

[In]

integrate((e*x+d)^(5/2)*(c*x^2+a),x, algorithm="maxima")

[Out]

2/693*(63*(e*x + d)^(11/2)*c - 154*(e*x + d)^(9/2)*c*d + 99*(c*d^2 + a*e^2)*(e*x + d)^(7/2))/e^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 357 vs. \(2 (51) = 102\).

Time = 0.28 (sec) , antiderivative size = 357, normalized size of antiderivative = 5.67 \[ \int (d+e x)^{5/2} \left (a+c x^2\right ) \, dx=\frac {2 \, {\left (3465 \, \sqrt {e x + d} a d^{3} + 3465 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a d^{2} + 693 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a d + \frac {231 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} c d^{3}}{e^{2}} + 99 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} a + \frac {297 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} c d^{2}}{e^{2}} + \frac {33 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} c d}{e^{2}} + \frac {5 \, {\left (63 \, {\left (e x + d\right )}^{\frac {11}{2}} - 385 \, {\left (e x + d\right )}^{\frac {9}{2}} d + 990 \, {\left (e x + d\right )}^{\frac {7}{2}} d^{2} - 1386 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{3} + 1155 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{4} - 693 \, \sqrt {e x + d} d^{5}\right )} c}{e^{2}}\right )}}{3465 \, e} \]

[In]

integrate((e*x+d)^(5/2)*(c*x^2+a),x, algorithm="giac")

[Out]

2/3465*(3465*sqrt(e*x + d)*a*d^3 + 3465*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a*d^2 + 693*(3*(e*x + d)^(5/2) -
10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a*d + 231*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*
x + d)*d^2)*c*d^3/e^2 + 99*(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 + 297*(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)*c*d
^2/e^2 + 33*(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*d/e^2 + 5*(63*(e*x + d)^(11/2) - 385*(e*x + d)^(9/2)*d + 990*(e*x + d)^(7/2)*d^2 - 13
86*(e*x + d)^(5/2)*d^3 + 1155*(e*x + d)^(3/2)*d^4 - 693*sqrt(e*x + d)*d^5)*c/e^2)/e

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int (d+e x)^{5/2} \left (a+c x^2\right ) \, dx=\frac {2\,{\left (d+e\,x\right )}^{7/2}\,\left (63\,c\,{\left (d+e\,x\right )}^2+99\,a\,e^2+99\,c\,d^2-154\,c\,d\,\left (d+e\,x\right )\right )}{693\,e^3} \]

[In]

int((a + c*x^2)*(d + e*x)^(5/2),x)

[Out]

(2*(d + e*x)^(7/2)*(63*c*(d + e*x)^2 + 99*a*e^2 + 99*c*d^2 - 154*c*d*(d + e*x)))/(693*e^3)