3.1096 \(\int \frac {(d+e x)^m}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx\)

Optimal. Leaf size=40 \[ \frac {(d+e x)^{m+1}}{e m \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 40, 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} \[ \frac {(d+e x)^{m+1}}{e m \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

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

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

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Mathematica [A]  time = 0.01, size = 29, normalized size = 0.72 \[ \frac {(d+e x)^{m+1}}{e m \sqrt {c (d+e x)^2}} \]

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)/(e*m*Sqrt[c*(d + e*x)^2])

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fricas [A]  time = 1.21, size = 45, normalized size = 1.12 \[ \frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}} {\left (e x + d\right )}^{m}}{c e^{2} m x + c d e m} \]

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)^m/(c*e^2*m*x + c*d*e*m)

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

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

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maple [A]  time = 0.05, size = 39, normalized size = 0.98 \[ \frac {\left (e x +d \right )^{m +1}}{\sqrt {c \,e^{2} x^{2}+2 c d e x +c \,d^{2}}\, e m} \]

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

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maxima [A]  time = 1.43, size = 17, normalized size = 0.42 \[ \frac {{\left (e x + d\right )}^{m}}{\sqrt {c} e m} \]

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]

(e*x + d)^m/(sqrt(c)*e*m)

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mupad [B]  time = 0.49, size = 48, normalized size = 1.20 \[ \frac {{\left (d+e\,x\right )}^m\,\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}}{c\,e^2\,m\,\left (x+\frac {d}{e}\right )} \]

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]

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

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

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

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