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

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

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Rubi [A]  time = 0.02, antiderivative size = 42, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {\sqrt {c d^2+2 c d e x+c e^2 x^2} (d+e x)^{m+1}}{e (m+2)} \end {gather*}

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 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 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]

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*}

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Mathematica [A]  time = 0.02, size = 31, normalized size = 0.74 \begin {gather*} \frac {\sqrt {c (d+e x)^2} (d+e x)^{m+1}}{e (m+2)} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.56, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x)^m \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 44, normalized size = 1.05 \begin {gather*} \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 {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}} {\left (e x + d\right )}^{m}\,{d x} \end {gather*}

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]

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

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maple [A]  time = 0.05, size = 41, normalized size = 0.98 \begin {gather*} \frac {\sqrt {c \,e^{2} x^{2}+2 c d e x +c \,d^{2}}\, \left (e x +d \right )^{m +1}}{\left (m +2\right ) e} \end {gather*}

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

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maxima [A]  time = 1.35, size = 42, normalized size = 1.00 \begin {gather*} \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 {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\left (d+e\,x\right )}^m\,\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {c \left (d + e x\right )^{2}} \left (d + e x\right )^{m}\, dx \end {gather*}

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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