3.20.27 \(\int \frac {(a+b x) (d+e x)^m}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=43 \[ \frac {(a+b x) (d+e x)^{m+1}}{e (m+1) \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {770, 21, 32} \begin {gather*} \frac {(a+b x) (d+e x)^{m+1}}{e (m+1) \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x)^m)/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {(a+b x) (d+e x)^m}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {(a+b x) (d+e x)^m}{a b+b^2 x} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int (d+e x)^m \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(a+b x) (d+e x)^{1+m}}{e (1+m) \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 34, normalized size = 0.79 \begin {gather*} \frac {(a+b x) (d+e x)^{m+1}}{e (m+1) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x)^m)/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.59, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x) (d+e x)^m}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((a + b*x)*(d + e*x)^m)/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

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fricas [A]  time = 0.44, size = 20, normalized size = 0.47 \begin {gather*} \frac {{\left (e x + d\right )} {\left (e x + d\right )}^{m}}{e m + e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(1/2),x, algorithm="fricas")

[Out]

(e*x + d)*(e*x + d)^m/(e*m + e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x + a)*(e*x + d)^m/sqrt(b^2*x^2 + 2*a*b*x + a^2), x)

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maple [A]  time = 0.04, size = 33, normalized size = 0.77 \begin {gather*} \frac {\left (b x +a \right ) \left (e x +d \right )^{m +1}}{\left (m +1\right ) \sqrt {\left (b x +a \right )^{2}}\, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.75, size = 21, normalized size = 0.49 \begin {gather*} \frac {{\left (e x + d\right )} {\left (e x + d\right )}^{m}}{e {\left (m + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(1/2),x, algorithm="maxima")

[Out]

(e*x + d)*(e*x + d)^m/(e*(m + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)**m/(b**2*x**2+2*a*b*x+a**2)**(1/2),x)

[Out]

Integral((a + b*x)*(d + e*x)**m/sqrt((a + b*x)**2), x)

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