3.1097 \(\int \frac{(d+e x)^m}{(c d^2+2 c d e x+c e^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=45 \[ -\frac{(d+e x)^{m+1}}{e (2-m) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0202965, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {644, 32} \[ -\frac{(d+e x)^{m+1}}{e (2-m) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[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)/(e*(2 - m)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2)))

Rule 644

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]

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]

Rubi steps

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

Mathematica [A]  time = 0.019931, size = 31, normalized size = 0.69 \[ \frac{(d+e x)^{m+1}}{e (m-2) \left (c (d+e x)^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[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)/(e*(-2 + m)*(c*(d + e*x)^2)^(3/2))

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Maple [A]  time = 0.039, size = 41, normalized size = 0.9 \begin{align*}{\frac{ \left ( ex+d \right ) ^{1+m}}{e \left ( -2+m \right ) } \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.07959, size = 69, normalized size = 1.53 \begin{align*} \frac{{\left (e x + d\right )}^{m} \sqrt{c}}{c^{2} e^{3}{\left (m - 2\right )} x^{2} + 2 \, c^{2} d e^{2}{\left (m - 2\right )} x + c^{2} d^{2} e{\left (m - 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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Fricas [B]  time = 2.47161, size = 244, normalized size = 5.42 \begin{align*} \frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}}{\left (e x + d\right )}^{m}}{c^{2} d^{3} e m - 2 \, c^{2} d^{3} e +{\left (c^{2} e^{4} m - 2 \, c^{2} e^{4}\right )} x^{3} + 3 \,{\left (c^{2} d e^{3} m - 2 \, c^{2} d e^{3}\right )} x^{2} + 3 \,{\left (c^{2} d^{2} e^{2} m - 2 \, c^{2} d^{2} e^{2}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{m}}{\left (c \left (d + e x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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