3.5.58 \(\int \frac {(f+g x)^2 (a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{(d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=200 \[ -\frac {8 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2} (c d f-a e g) \left (2 a e^2 g-c d (7 e f-5 d g)\right )}{315 c^3 d^3 e (d+e x)^{5/2}}+\frac {8 g \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2} (c d f-a e g)}{63 c^2 d^2 e (d+e x)^{3/2}}+\frac {2 (f+g x)^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{9 c d (d+e x)^{5/2}} \]

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Rubi [A]  time = 0.23, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {870, 794, 648} \begin {gather*} \frac {8 g \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2} (c d f-a e g)}{63 c^2 d^2 e (d+e x)^{3/2}}-\frac {8 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2} (c d f-a e g) \left (2 a e^2 g-c d (7 e f-5 d g)\right )}{315 c^3 d^3 e (d+e x)^{5/2}}+\frac {2 (f+g x)^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{9 c d (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(c*d*f - a*e*g)*(2*a*e^2*g - c*d*(7*e*f - 5*d*g))*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(315*c^3*
d^3*e*(d + e*x)^(5/2)) + (8*g*(c*d*f - a*e*g)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(63*c^2*d^2*e*(d
+ e*x)^(3/2)) + (2*(f + g*x)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(9*c*d*(d + e*x)^(5/2))

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 794

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

Rule 870

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

Rubi steps

\begin {align*} \int \frac {(f+g x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx &=\frac {2 (f+g x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{9 c d (d+e x)^{5/2}}+\frac {\left (4 \left (c d e^2 f+c d^2 e g-e \left (c d^2+a e^2\right ) g\right )\right ) \int \frac {(f+g x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx}{9 c d e^2}\\ &=\frac {8 g (c d f-a e g) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{63 c^2 d^2 e (d+e x)^{3/2}}+\frac {2 (f+g x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{9 c d (d+e x)^{5/2}}-\frac {\left (4 (c d f-a e g) \left (2 a e^2 g-c d (7 e f-5 d g)\right )\right ) \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx}{63 c^2 d^2 e}\\ &=-\frac {8 (c d f-a e g) \left (2 a e^2 g-c d (7 e f-5 d g)\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{315 c^3 d^3 e (d+e x)^{5/2}}+\frac {8 g (c d f-a e g) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{63 c^2 d^2 e (d+e x)^{3/2}}+\frac {2 (f+g x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{9 c d (d+e x)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 90, normalized size = 0.45 \begin {gather*} \frac {2 ((d+e x) (a e+c d x))^{5/2} \left (8 a^2 e^2 g^2-4 a c d e g (9 f+5 g x)+c^2 d^2 \left (63 f^2+90 f g x+35 g^2 x^2\right )\right )}{315 c^3 d^3 (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((a*e + c*d*x)*(d + e*x))^(5/2)*(8*a^2*e^2*g^2 - 4*a*c*d*e*g*(9*f + 5*g*x) + c^2*d^2*(63*f^2 + 90*f*g*x + 3
5*g^2*x^2)))/(315*c^3*d^3*(d + e*x)^(5/2))

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IntegrateAlgebraic [A]  time = 2.27, size = 120, normalized size = 0.60 \begin {gather*} \frac {2 (a e+c d x) ((d+e x) (a e+c d x))^{3/2} \left (63 a^2 e^2 g^2+90 c d f g (a e+c d x)-126 a c d e f g+35 g^2 (a e+c d x)^2-90 a e g^2 (a e+c d x)+63 c^2 d^2 f^2\right )}{315 c^3 d^3 (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*e + c*d*x)*((a*e + c*d*x)*(d + e*x))^(3/2)*(63*c^2*d^2*f^2 - 126*a*c*d*e*f*g + 63*a^2*e^2*g^2 + 90*c*d*f
*g*(a*e + c*d*x) - 90*a*e*g^2*(a*e + c*d*x) + 35*g^2*(a*e + c*d*x)^2))/(315*c^3*d^3*(d + e*x)^(3/2))

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fricas [A]  time = 0.42, size = 230, normalized size = 1.15 \begin {gather*} \frac {2 \, {\left (35 \, c^{4} d^{4} g^{2} x^{4} + 63 \, a^{2} c^{2} d^{2} e^{2} f^{2} - 36 \, a^{3} c d e^{3} f g + 8 \, a^{4} e^{4} g^{2} + 10 \, {\left (9 \, c^{4} d^{4} f g + 5 \, a c^{3} d^{3} e g^{2}\right )} x^{3} + 3 \, {\left (21 \, c^{4} d^{4} f^{2} + 48 \, a c^{3} d^{3} e f g + a^{2} c^{2} d^{2} e^{2} g^{2}\right )} x^{2} + 2 \, {\left (63 \, a c^{3} d^{3} e f^{2} + 9 \, a^{2} c^{2} d^{2} e^{2} f g - 2 \, a^{3} c d e^{3} g^{2}\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}}{315 \, {\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((g*x+f)^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(3/2),x, algorithm="fricas")

[Out]

2/315*(35*c^4*d^4*g^2*x^4 + 63*a^2*c^2*d^2*e^2*f^2 - 36*a^3*c*d*e^3*f*g + 8*a^4*e^4*g^2 + 10*(9*c^4*d^4*f*g +
5*a*c^3*d^3*e*g^2)*x^3 + 3*(21*c^4*d^4*f^2 + 48*a*c^3*d^3*e*f*g + a^2*c^2*d^2*e^2*g^2)*x^2 + 2*(63*a*c^3*d^3*e
*f^2 + 9*a^2*c^2*d^2*e^2*f*g - 2*a^3*c*d*e^3*g^2)*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 \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {3}{2}} {\left (g x + f\right )}^{2}}{{\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 116, normalized size = 0.58 \begin {gather*} \frac {2 \left (c d x +a e \right ) \left (35 g^{2} x^{2} c^{2} d^{2}-20 a c d e \,g^{2} x +90 c^{2} d^{2} f g x +8 a^{2} e^{2} g^{2}-36 a c d e f g +63 f^{2} c^{2} d^{2}\right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}}{315 \left (e x +d \right )^{\frac {3}{2}} c^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(c*d*x+a*e)*(35*c^2*d^2*g^2*x^2-20*a*c*d*e*g^2*x+90*c^2*d^2*f*g*x+8*a^2*e^2*g^2-36*a*c*d*e*f*g+63*c^2*d^
2*f^2)*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(3/2)/c^3/d^3/(e*x+d)^(3/2)

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maxima [A]  time = 0.62, size = 192, normalized size = 0.96 \begin {gather*} \frac {2 \, {\left (c^{2} d^{2} x^{2} + 2 \, a c d e x + a^{2} e^{2}\right )} \sqrt {c d x + a e} f^{2}}{5 \, c d} + \frac {4 \, {\left (5 \, c^{3} d^{3} x^{3} + 8 \, a c^{2} d^{2} e x^{2} + a^{2} c d e^{2} x - 2 \, a^{3} e^{3}\right )} \sqrt {c d x + a e} f g}{35 \, c^{2} d^{2}} + \frac {2 \, {\left (35 \, c^{4} d^{4} x^{4} + 50 \, a c^{3} d^{3} e x^{3} + 3 \, a^{2} c^{2} d^{2} e^{2} x^{2} - 4 \, a^{3} c d e^{3} x + 8 \, a^{4} e^{4}\right )} \sqrt {c d x + a e} g^{2}}{315 \, c^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(c^2*d^2*x^2 + 2*a*c*d*e*x + a^2*e^2)*sqrt(c*d*x + a*e)*f^2/(c*d) + 4/35*(5*c^3*d^3*x^3 + 8*a*c^2*d^2*e*x^
2 + a^2*c*d*e^2*x - 2*a^3*e^3)*sqrt(c*d*x + a*e)*f*g/(c^2*d^2) + 2/315*(35*c^4*d^4*x^4 + 50*a*c^3*d^3*e*x^3 +
3*a^2*c^2*d^2*e^2*x^2 - 4*a^3*c*d*e^3*x + 8*a^4*e^4)*sqrt(c*d*x + a*e)*g^2/(c^3*d^3)

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mupad [B]  time = 3.43, size = 206, normalized size = 1.03 \begin {gather*} \frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {4\,g\,x^3\,\left (5\,a\,e\,g+9\,c\,d\,f\right )}{63}+\frac {16\,a^4\,e^4\,g^2-72\,a^3\,c\,d\,e^3\,f\,g+126\,a^2\,c^2\,d^2\,e^2\,f^2}{315\,c^3\,d^3}+\frac {x^2\,\left (6\,a^2\,c^2\,d^2\,e^2\,g^2+288\,a\,c^3\,d^3\,e\,f\,g+126\,c^4\,d^4\,f^2\right )}{315\,c^3\,d^3}+\frac {2\,c\,d\,g^2\,x^4}{9}+\frac {4\,a\,e\,x\,\left (-2\,a^2\,e^2\,g^2+9\,a\,c\,d\,e\,f\,g+63\,c^2\,d^2\,f^2\right )}{315\,c^2\,d^2}\right )}{\sqrt {d+e\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*((4*g*x^3*(5*a*e*g + 9*c*d*f))/63 + (16*a^4*e^4*g^2 + 126*a^2*c
^2*d^2*e^2*f^2 - 72*a^3*c*d*e^3*f*g)/(315*c^3*d^3) + (x^2*(126*c^4*d^4*f^2 + 6*a^2*c^2*d^2*e^2*g^2 + 288*a*c^3
*d^3*e*f*g))/(315*c^3*d^3) + (2*c*d*g^2*x^4)/9 + (4*a*e*x*(63*c^2*d^2*f^2 - 2*a^2*e^2*g^2 + 9*a*c*d*e*f*g))/(3
15*c^2*d^2)))/(d + e*x)^(1/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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