\(\int (d+e x)^m (c d^2+2 c d e x+c e^2 x^2)^{3/2} \, dx\) [1094]

Optimal result

Integrand size = 32, antiderivative size = 42 \[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\frac {(d+e x)^{1+m} \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{e (4+m)} \]

[Out]

(e*x+d)^(1+m)*(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)/e/(4+m)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {658, 32} \[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} (d+e x)^{m+1}}{e (m+4)} \]

[In]

Int[(d + e*x)^m*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2),x]

[Out]

((d + e*x)^(1 + m)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2))/(e*(4 + m))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 658

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^p/(d
+ e*x)^(2*p), Int[(d + e*x)^(m + 2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !
IntegerQ[p] && EqQ[2*c*d - b*e, 0] &&  !IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \int (d+e x)^{3+m} \, dx}{(d+e x)^3} \\ & = \frac {(d+e x)^{1+m} \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{e (4+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.74 \[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\frac {(d+e x)^{1+m} \left (c (d+e x)^2\right )^{3/2}}{e (4+m)} \]

[In]

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

[Out]

((d + e*x)^(1 + m)*(c*(d + e*x)^2)^(3/2))/(e*(4 + m))

Maple [A] (verified)

Time = 2.54 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.98

method	result	size
gosper	\(\frac {\left (e x +d \right )^{1+m} \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {3}{2}}}{e \left (4+m \right )}\)	\(41\)
risch	\(\frac {c \sqrt {c \left (e x +d \right )^{2}}\, \left (e^{4} x^{4}+4 d \,e^{3} x^{3}+6 d^{2} e^{2} x^{2}+4 d^{3} e x +d^{4}\right ) \left (e x +d \right )^{m}}{\left (e x +d \right ) e \left (4+m \right )}\)	\(74\)

[In]

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

[Out]

(e*x+d)^(1+m)*(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)/e/(4+m)

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.69 \[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\frac {{\left (c e^{3} x^{3} + 3 \, c d e^{2} x^{2} + 3 \, c d^{2} e x + c d^{3}\right )} \sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}} {\left (e x + d\right )}^{m}}{e m + 4 \, e} \]

[In]

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

[Out]

(c*e^3*x^3 + 3*c*d*e^2*x^2 + 3*c*d^2*e*x + c*d^3)*sqrt(c*e^2*x^2 + 2*c*d*e*x + c*d^2)*(e*x + d)^m/(e*m + 4*e)

Sympy [F]

\[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\int \left (c \left (d + e x\right )^{2}\right )^{\frac {3}{2}} \left (d + e x\right )^{m}\, dx \]

[In]

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

[Out]

Integral((c*(d + e*x)**2)**(3/2)*(d + e*x)**m, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.67 \[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\frac {{\left (c^{\frac {3}{2}} e^{4} x^{4} + 4 \, c^{\frac {3}{2}} d e^{3} x^{3} + 6 \, c^{\frac {3}{2}} d^{2} e^{2} x^{2} + 4 \, c^{\frac {3}{2}} d^{3} e x + c^{\frac {3}{2}} d^{4}\right )} {\left (e x + d\right )}^{m}}{e {\left (m + 4\right )}} \]

[In]

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

[Out]

(c^(3/2)*e^4*x^4 + 4*c^(3/2)*d*e^3*x^3 + 6*c^(3/2)*d^2*e^2*x^2 + 4*c^(3/2)*d^3*e*x + c^(3/2)*d^4)*(e*x + d)^m/
(e*(m + 4))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (40) = 80\).

Time = 0.29 (sec) , antiderivative size = 133, normalized size of antiderivative = 3.17 \[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\frac {{\left (c e^{3} x^{3} e^{\left (m \log \left (e x + d\right ) + \log \left (e x + d\right )\right )} \mathrm {sgn}\left (e x + d\right ) + 3 \, c d e^{2} x^{2} e^{\left (m \log \left (e x + d\right ) + \log \left (e x + d\right )\right )} \mathrm {sgn}\left (e x + d\right ) + 3 \, c d^{2} e x e^{\left (m \log \left (e x + d\right ) + \log \left (e x + d\right )\right )} \mathrm {sgn}\left (e x + d\right ) + c d^{3} e^{\left (m \log \left (e x + d\right ) + \log \left (e x + d\right )\right )} \mathrm {sgn}\left (e x + d\right )\right )} \sqrt {c}}{e m + 4 \, e} \]

[In]

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

[Out]

(c*e^3*x^3*e^(m*log(e*x + d) + log(e*x + d))*sgn(e*x + d) + 3*c*d*e^2*x^2*e^(m*log(e*x + d) + log(e*x + d))*sg
n(e*x + d) + 3*c*d^2*e*x*e^(m*log(e*x + d) + log(e*x + d))*sgn(e*x + d) + c*d^3*e^(m*log(e*x + d) + log(e*x +
d))*sgn(e*x + d))*sqrt(c)/(e*m + 4*e)

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\int {\left (d+e\,x\right )}^m\,{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^{3/2} \,d x \]

[In]

int((d + e*x)^m*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(3/2),x)

[Out]

int((d + e*x)^m*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(3/2), x)