3.1095 \(\int (d+e x)^m \sqrt{c d^2+2 c d e x+c e^2 x^2} \, dx\)

Optimal. Leaf size=42 \[ \frac{\sqrt{c d^2+2 c d e x+c e^2 x^2} (d+e x)^{m+1}}{e (m+2)} \]

[Out]

((d + e*x)^(1 + m)*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2])/(e*(2 + m))

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Rubi [A]  time = 0.0181988, antiderivative size = 42, 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} \[ \frac{\sqrt{c d^2+2 c d e x+c e^2 x^2} (d+e x)^{m+1}}{e (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + m)*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2])/(e*(2 + m))

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 (d+e x)^m \sqrt{c d^2+2 c d e x+c e^2 x^2} \, dx &=\frac{\sqrt{c d^2+2 c d e x+c e^2 x^2} \int (d+e x)^{1+m} \, dx}{d+e x}\\ &=\frac{(d+e x)^{1+m} \sqrt{c d^2+2 c d e x+c e^2 x^2}}{e (2+m)}\\ \end{align*}

Mathematica [A]  time = 0.0175014, size = 31, normalized size = 0.74 \[ \frac{\sqrt{c (d+e x)^2} (d+e x)^{m+1}}{e (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + m)*Sqrt[c*(d + e*x)^2])/(e*(2 + m))

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Maple [A]  time = 0.044, size = 41, normalized size = 1. \begin{align*}{\frac{ \left ( ex+d \right ) ^{1+m}}{e \left ( 2+m \right ) }\sqrt{c{e}^{2}{x}^{2}+2\,cdex+c{d}^{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)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.20951, size = 57, normalized size = 1.36 \begin{align*} \frac{{\left (\sqrt{c} e^{2} x^{2} + 2 \, \sqrt{c} d e x + \sqrt{c} d^{2}\right )}{\left (e x + d\right )}^{m}}{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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.4213, size = 96, normalized size = 2.29 \begin{align*} \frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}}{\left (e x + d\right )}{\left (e x + d\right )}^{m}}{e m + 2 \, e} \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)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c \left (d + e x\right )^{2}} \left (d + e x\right )^{m}\, 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)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.19182, size = 84, normalized size = 2. \begin{align*} \frac{{\left (x e + d\right )}^{m} \sqrt{c} x^{2} e^{2} + 2 \,{\left (x e + d\right )}^{m} \sqrt{c} d x e +{\left (x e + d\right )}^{m} \sqrt{c} d^{2}}{m e + 2 \, e} \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)^(1/2),x, algorithm="giac")

[Out]

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