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

Optimal. Leaf size=42 \[ \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} (d+e x)^{m+1}}{e (m+4)} \]

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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 {\left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} (d+e x)^{m+1}}{e (m+4)} \end {gather*}

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)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2))/(e*(4 + 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 \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx &=\frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \int (d+e x)^{3+m} \, dx}{(d+e x)^3}\\ &=\frac {(d+e x)^{1+m} \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{e (4+m)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 31, normalized size = 0.74 \begin {gather*} \frac {\left (c (d+e x)^2\right )^{3/2} (d+e x)^{m+1}}{e (m+4)} \end {gather*}

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)*(c*(d + e*x)^2)^(3/2))/(e*(4 + m))

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IntegrateAlgebraic [F]  time = 0.77, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(d + e*x)^m*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2),x]

[Out]

Defer[IntegrateAlgebraic][(d + e*x)^m*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2), x]

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fricas [A]  time = 0.41, size = 71, normalized size = 1.69 \begin {gather*} \frac {{\left (c e^{3} x^{3} + 3 \, c d e^{2} x^{2} + 3 \, c d^{2} e x + c d^{3}\right )} \sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}} {\left (e x + d\right )}^{m}}{e m + 4 \, 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)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 41, normalized size = 0.98 \begin {gather*} \frac {\left (c \,e^{2} x^{2}+2 c d e x +c \,d^{2}\right )^{\frac {3}{2}} \left (e x +d \right )^{m +1}}{\left (m +4\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)^(3/2),x)

[Out]

(e*x+d)^(m+1)*(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)/e/(m+4)

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maxima [A]  time = 1.44, size = 70, normalized size = 1.67 \begin {gather*} \frac {{\left (c^{\frac {3}{2}} e^{4} x^{4} + 4 \, c^{\frac {3}{2}} d e^{3} x^{3} + 6 \, c^{\frac {3}{2}} d^{2} e^{2} x^{2} + 4 \, c^{\frac {3}{2}} d^{3} e x + c^{\frac {3}{2}} d^{4}\right )} {\left (e x + d\right )}^{m}}{e {\left (m + 4\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)^(3/2),x, algorithm="maxima")

[Out]

(c^(3/2)*e^4*x^4 + 4*c^(3/2)*d*e^3*x^3 + 6*c^(3/2)*d^2*e^2*x^2 + 4*c^(3/2)*d^3*e*x + c^(3/2)*d^4)*(e*x + d)^m/
(e*(m + 4))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\left (d+e\,x\right )}^m\,{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^{3/2} \,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)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (c \left (d + e x\right )^{2}\right )^{\frac {3}{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)**(3/2),x)

[Out]

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

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