\(\int (d+e x)^{3/2} (a+c x^2) \, dx\) [592]

Optimal result

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

[Out]

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

[In]

Int[(d + e*x)^(3/2)*(a + c*x^2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int (d+e x)^{3/2} \left (a+c x^2\right ) \, dx=\frac {2 (d+e x)^{5/2} \left (63 a e^2+c \left (8 d^2-20 d e x+35 e^2 x^2\right )\right )}{315 e^3} \]

[In]

Integrate[(d + e*x)^(3/2)*(a + c*x^2),x]

[Out]

(2*(d + e*x)^(5/2)*(63*a*e^2 + c*(8*d^2 - 20*d*e*x + 35*e^2*x^2)))/(315*e^3)

Maple [A] (verified)

Time = 2.49 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.65

method	result	size
gosper	\(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (35 c \,x^{2} e^{2}-20 x c d e +63 e^{2} a +8 c \,d^{2}\right )}{315 e^{3}}\)	\(41\)
pseudoelliptic	\(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (35 c \,x^{2} e^{2}-20 x c d e +63 e^{2} a +8 c \,d^{2}\right )}{315 e^{3}}\)	\(41\)
derivativedivides	\(\frac {\frac {2 c \left (e x +d \right )^{\frac {9}{2}}}{9}-\frac {4 c d \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (e^{2} a +c \,d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}}{e^{3}}\)	\(48\)
default	\(\frac {\frac {2 c \left (e x +d \right )^{\frac {9}{2}}}{9}-\frac {4 c d \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (e^{2} a +c \,d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}}{e^{3}}\)	\(48\)
trager	\(\frac {2 \left (35 c \,e^{4} x^{4}+50 d \,e^{3} c \,x^{3}+63 a \,e^{4} x^{2}+3 c \,d^{2} e^{2} x^{2}+126 a d \,e^{3} x -4 c \,d^{3} e x +63 a \,d^{2} e^{2}+8 c \,d^{4}\right ) \sqrt {e x +d}}{315 e^{3}}\)	\(85\)
risch	\(\frac {2 \left (35 c \,e^{4} x^{4}+50 d \,e^{3} c \,x^{3}+63 a \,e^{4} x^{2}+3 c \,d^{2} e^{2} x^{2}+126 a d \,e^{3} x -4 c \,d^{3} e x +63 a \,d^{2} e^{2}+8 c \,d^{4}\right ) \sqrt {e x +d}}{315 e^{3}}\)	\(85\)

[In]

int((e*x+d)^(3/2)*(c*x^2+a),x,method=_RETURNVERBOSE)

[Out]

2/315*(e*x+d)^(5/2)*(35*c*e^2*x^2-20*c*d*e*x+63*a*e^2+8*c*d^2)/e^3

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.35 \[ \int (d+e x)^{3/2} \left (a+c x^2\right ) \, dx=\frac {2 \, {\left (35 \, c e^{4} x^{4} + 50 \, c d e^{3} x^{3} + 8 \, c d^{4} + 63 \, a d^{2} e^{2} + 3 \, {\left (c d^{2} e^{2} + 21 \, a e^{4}\right )} x^{2} - 2 \, {\left (2 \, c d^{3} e - 63 \, a d e^{3}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{3}} \]

[In]

integrate((e*x+d)^(3/2)*(c*x^2+a),x, algorithm="fricas")

[Out]

2/315*(35*c*e^4*x^4 + 50*c*d*e^3*x^3 + 8*c*d^4 + 63*a*d^2*e^2 + 3*(c*d^2*e^2 + 21*a*e^4)*x^2 - 2*(2*c*d^3*e -
63*a*d*e^3)*x)*sqrt(e*x + d)/e^3

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.21 \[ \int (d+e x)^{3/2} \left (a+c x^2\right ) \, dx=\begin {cases} \frac {2 \left (- \frac {2 c d \left (d + e x\right )^{\frac {7}{2}}}{7 e^{2}} + \frac {c \left (d + e x\right )^{\frac {9}{2}}}{9 e^{2}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (a e^{2} + c d^{2}\right )}{5 e^{2}}\right )}{e} & \text {for}\: e \neq 0 \\d^{\frac {3}{2}} \left (a x + \frac {c x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((e*x+d)**(3/2)*(c*x**2+a),x)

[Out]

Piecewise((2*(-2*c*d*(d + e*x)**(7/2)/(7*e**2) + c*(d + e*x)**(9/2)/(9*e**2) + (d + e*x)**(5/2)*(a*e**2 + c*d*
*2)/(5*e**2))/e, Ne(e, 0)), (d**(3/2)*(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 \[ \int (d+e x)^{3/2} \left (a+c x^2\right ) \, dx=\frac {2 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} c - 90 \, {\left (e x + d\right )}^{\frac {7}{2}} c d + 63 \, {\left (c d^{2} + a e^{2}\right )} {\left (e x + d\right )}^{\frac {5}{2}}\right )}}{315 \, e^{3}} \]

[In]

integrate((e*x+d)^(3/2)*(c*x^2+a),x, algorithm="maxima")

[Out]

2/315*(35*(e*x + d)^(9/2)*c - 90*(e*x + d)^(7/2)*c*d + 63*(c*d^2 + a*e^2)*(e*x + d)^(5/2))/e^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (51) = 102\).

Time = 0.27 (sec) , antiderivative size = 229, normalized size of antiderivative = 3.63 \[ \int (d+e x)^{3/2} \left (a+c x^2\right ) \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} a d^{2} + 210 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a d + 21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a + \frac {21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} c d^{2}}{e^{2}} + \frac {18 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} c d}{e^{2}} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} c}{e^{2}}\right )}}{315 \, e} \]

[In]

integrate((e*x+d)^(3/2)*(c*x^2+a),x, algorithm="giac")

[Out]

2/315*(315*sqrt(e*x + d)*a*d^2 + 210*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a*d + 21*(3*(e*x + d)^(5/2) - 10*(e
*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a + 21*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^
2)*c*d^2/e^2 + 18*(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/e^2 + (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/e^2)/e

Mupad [B] (verification not implemented)

Time = 9.44 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int (d+e x)^{3/2} \left (a+c x^2\right ) \, dx=\frac {2\,{\left (d+e\,x\right )}^{5/2}\,\left (35\,c\,{\left (d+e\,x\right )}^2+63\,a\,e^2+63\,c\,d^2-90\,c\,d\,\left (d+e\,x\right )\right )}{315\,e^3} \]

[In]

int((a + c*x^2)*(d + e*x)^(3/2),x)

[Out]

(2*(d + e*x)^(5/2)*(35*c*(d + e*x)^2 + 63*a*e^2 + 63*c*d^2 - 90*c*d*(d + e*x)))/(315*e^3)