3.18.13 \(\int (d+e x)^{3/2} \sqrt {a d e+(c d^2+a e^2) x+c d e x^2} \, dx\)

Optimal. Leaf size=171 \[ \frac {16 \left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{105 c^3 d^3 (d+e x)^{3/2}}+\frac {8 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{35 c^2 d^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{7 c d} \]

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Rubi [A]  time = 0.11, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {656, 648} \begin {gather*} \frac {16 \left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{105 c^3 d^3 (d+e x)^{3/2}}+\frac {8 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{35 c^2 d^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{7 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(3/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2],x]

[Out]

(16*(c*d^2 - a*e^2)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(105*c^3*d^3*(d + e*x)^(3/2)) + (8*(c*d^2
- a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(35*c^2*d^2*Sqrt[d + e*x]) + (2*Sqrt[d + e*x]*(a*d*e
+ (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(7*c*d)

Rule 648

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rule 656

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[(Simplify[m + p]*(2*c*d - b*e))/(c*(m + 2*p + 1)), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rubi steps

\begin {align*} \int (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx &=\frac {2 \sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{7 c d}+\frac {\left (4 \left (d^2-\frac {a e^2}{c}\right )\right ) \int \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx}{7 d}\\ &=\frac {8 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{35 c^2 d^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{7 c d}+\frac {\left (8 \left (d^2-\frac {a e^2}{c}\right )^2\right ) \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x}} \, dx}{35 d^2}\\ &=\frac {16 \left (c d^2-a e^2\right )^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{105 c^3 d^3 (d+e x)^{3/2}}+\frac {8 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{35 c^2 d^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{7 c d}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 88, normalized size = 0.51 \begin {gather*} \frac {2 ((d+e x) (a e+c d x))^{3/2} \left (8 a^2 e^4-4 a c d e^2 (7 d+3 e x)+c^2 d^2 \left (35 d^2+42 d e x+15 e^2 x^2\right )\right )}{105 c^3 d^3 (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(3/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2],x]

[Out]

(2*((a*e + c*d*x)*(d + e*x))^(3/2)*(8*a^2*e^4 - 4*a*c*d*e^2*(7*d + 3*e*x) + c^2*d^2*(35*d^2 + 42*d*e*x + 15*e^
2*x^2)))/(105*c^3*d^3*(d + e*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.40, size = 197, normalized size = 1.15 \begin {gather*} \frac {2 \sqrt {a e (d+e x)-\frac {c d^2 (d+e x)}{e}+\frac {c d (d+e x)^2}{e}} \left (8 a^3 e^6-24 a^2 c d^2 e^4-4 a^2 c d e^4 (d+e x)+24 a c^2 d^4 e^2+8 a c^2 d^3 e^2 (d+e x)+3 a c^2 d^2 e^2 (d+e x)^2-8 c^3 d^6-4 c^3 d^5 (d+e x)-3 c^3 d^4 (d+e x)^2+15 c^3 d^3 (d+e x)^3\right )}{105 c^3 d^3 e \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2)/e]*(-8*c^3*d^6 + 24*a*c^2*d^4*e^2 - 24*a^2*
c*d^2*e^4 + 8*a^3*e^6 - 4*c^3*d^5*(d + e*x) + 8*a*c^2*d^3*e^2*(d + e*x) - 4*a^2*c*d*e^4*(d + e*x) - 3*c^3*d^4*
(d + e*x)^2 + 3*a*c^2*d^2*e^2*(d + e*x)^2 + 15*c^3*d^3*(d + e*x)^3))/(105*c^3*d^3*e*Sqrt[d + e*x])

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fricas [A]  time = 0.40, size = 159, normalized size = 0.93 \begin {gather*} \frac {2 \, {\left (15 \, c^{3} d^{3} e^{2} x^{3} + 35 \, a c^{2} d^{4} e - 28 \, a^{2} c d^{2} e^{3} + 8 \, a^{3} e^{5} + 3 \, {\left (14 \, c^{3} d^{4} e + a c^{2} d^{2} e^{3}\right )} x^{2} + {\left (35 \, c^{3} d^{5} + 14 \, a c^{2} d^{3} e^{2} - 4 \, a^{2} c d e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{105 \, {\left (c^{3} d^{3} e x + c^{3} d^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="fricas")

[Out]

2/105*(15*c^3*d^3*e^2*x^3 + 35*a*c^2*d^4*e - 28*a^2*c*d^2*e^3 + 8*a^3*e^5 + 3*(14*c^3*d^4*e + a*c^2*d^2*e^3)*x
^2 + (35*c^3*d^5 + 14*a*c^2*d^3*e^2 - 4*a^2*c*d*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x +
d)/(c^3*d^3*e*x + c^3*d^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 110, normalized size = 0.64 \begin {gather*} \frac {2 \left (c d x +a e \right ) \left (15 c^{2} d^{2} e^{2} x^{2}-12 a c d \,e^{3} x +42 c^{2} d^{3} e x +8 a^{2} e^{4}-28 a c \,d^{2} e^{2}+35 c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{105 \sqrt {e x +d}\, c^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2),x)

[Out]

2/105*(c*d*x+a*e)*(15*c^2*d^2*e^2*x^2-12*a*c*d*e^3*x+42*c^2*d^3*e*x+8*a^2*e^4-28*a*c*d^2*e^2+35*c^2*d^4)*(c*d*
e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)/c^3/d^3/(e*x+d)^(1/2)

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maxima [A]  time = 1.23, size = 140, normalized size = 0.82 \begin {gather*} \frac {2 \, {\left (15 \, c^{3} d^{3} e^{2} x^{3} + 35 \, a c^{2} d^{4} e - 28 \, a^{2} c d^{2} e^{3} + 8 \, a^{3} e^{5} + 3 \, {\left (14 \, c^{3} d^{4} e + a c^{2} d^{2} e^{3}\right )} x^{2} + {\left (35 \, c^{3} d^{5} + 14 \, a c^{2} d^{3} e^{2} - 4 \, a^{2} c d e^{4}\right )} x\right )} \sqrt {c d x + a e} {\left (e x + d\right )}}{105 \, {\left (c^{3} d^{3} e x + c^{3} d^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="maxima")

[Out]

2/105*(15*c^3*d^3*e^2*x^3 + 35*a*c^2*d^4*e - 28*a^2*c*d^2*e^3 + 8*a^3*e^5 + 3*(14*c^3*d^4*e + a*c^2*d^2*e^3)*x
^2 + (35*c^3*d^5 + 14*a*c^2*d^3*e^2 - 4*a^2*c*d*e^4)*x)*sqrt(c*d*x + a*e)*(e*x + d)/(c^3*d^3*e*x + c^3*d^4)

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mupad [B]  time = 0.92, size = 180, normalized size = 1.05 \begin {gather*} \frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {2\,e\,x^3\,\sqrt {d+e\,x}}{7}+\frac {\sqrt {d+e\,x}\,\left (16\,a^3\,e^5-56\,a^2\,c\,d^2\,e^3+70\,a\,c^2\,d^4\,e\right )}{105\,c^3\,d^3\,e}+\frac {2\,x^2\,\left (14\,c\,d^2+a\,e^2\right )\,\sqrt {d+e\,x}}{35\,c\,d}+\frac {x\,\sqrt {d+e\,x}\,\left (-8\,a^2\,c\,d\,e^4+28\,a\,c^2\,d^3\,e^2+70\,c^3\,d^5\right )}{105\,c^3\,d^3\,e}\right )}{x+\frac {d}{e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(3/2)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2),x)

[Out]

((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*((2*e*x^3*(d + e*x)^(1/2))/7 + ((d + e*x)^(1/2)*(16*a^3*e^5 - 5
6*a^2*c*d^2*e^3 + 70*a*c^2*d^4*e))/(105*c^3*d^3*e) + (2*x^2*(a*e^2 + 14*c*d^2)*(d + e*x)^(1/2))/(35*c*d) + (x*
(d + e*x)^(1/2)*(70*c^3*d^5 + 28*a*c^2*d^3*e^2 - 8*a^2*c*d*e^4))/(105*c^3*d^3*e)))/(x + d/e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(3/2)*(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2),x)

[Out]

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

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