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

Optimal. Leaf size=125 \[ \frac {2 g \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{7 c d e (d+e x)^{3/2}}-\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2} \left (2 a e^2 g-c d (7 e f-5 d g)\right )}{35 c^2 d^2 e (d+e x)^{5/2}} \]

[Out]

-2/35*(2*a*e^2*g-c*d*(-5*d*g+7*e*f))*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/c^2/d^2/e/(e*x+d)^(5/2)+2/7*g*(a*
d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/c/d/e/(e*x+d)^(3/2)

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Rubi [A]  time = 0.10, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {794, 648} \[ \frac {2 g \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{7 c d e (d+e x)^{3/2}}-\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2} \left (2 a e^2 g-c d (7 e f-5 d g)\right )}{35 c^2 d^2 e (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x)*(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*(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))/(35*c^2*d^2*e*(d + e*x)^(
5/2)) + (2*g*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(7*c*d*e*(d + e*x)^(3/2))

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])

Rubi steps

\begin {align*} \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 &=\frac {2 g \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{7 c d e (d+e x)^{3/2}}+\frac {1}{7} \left (7 f-\frac {5 d g}{e}-\frac {2 a e g}{c d}\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\\ &=\frac {2 \left (7 f-\frac {5 d g}{e}-\frac {2 a e g}{c d}\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{35 c d (d+e x)^{5/2}}+\frac {2 g \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{7 c d e (d+e x)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 54, normalized size = 0.43 \[ \frac {2 ((d+e x) (a e+c d x))^{5/2} (c d (7 f+5 g x)-2 a e g)}{35 c^2 d^2 (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x)*(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)*(-2*a*e*g + c*d*(7*f + 5*g*x)))/(35*c^2*d^2*(d + e*x)^(5/2))

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fricas [A]  time = 0.82, size = 137, normalized size = 1.10 \[ \frac {2 \, {\left (5 \, c^{3} d^{3} g x^{3} + 7 \, a^{2} c d e^{2} f - 2 \, a^{3} e^{3} g + {\left (7 \, c^{3} d^{3} f + 8 \, a c^{2} d^{2} e g\right )} x^{2} + {\left (14 \, a c^{2} d^{2} e f + a^{2} c d e^{2} g\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{35 \, {\left (c^{2} d^{2} e x + c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(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]

2/35*(5*c^3*d^3*g*x^3 + 7*a^2*c*d*e^2*f - 2*a^3*e^3*g + (7*c^3*d^3*f + 8*a*c^2*d^2*e*g)*x^2 + (14*a*c^2*d^2*e*
f + a^2*c*d*e^2*g)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)/(c^2*d^2*e*x + c^2*d^3)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(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]

Timed out

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maple [A]  time = 0.00, size = 67, normalized size = 0.54 \[ -\frac {2 \left (c d x +a e \right ) \left (-5 c d g x +2 a e g -7 c d f \right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}}{35 \left (e x +d \right )^{\frac {3}{2}} c^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(c*d*x+a*e)*(-5*c*d*g*x+2*a*e*g-7*c*d*f)*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(3/2)/c^2/d^2/(e*x+d)^(3/2)

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maxima [A]  time = 0.57, size = 107, normalized size = 0.86 \[ \frac {2 \, {\left (c^{2} d^{2} x^{2} + 2 \, a c d e x + a^{2} e^{2}\right )} \sqrt {c d x + a e} f}{5 \, c d} + \frac {2 \, {\left (5 \, c^{3} d^{3} x^{3} + 8 \, a c^{2} d^{2} e x^{2} + a^{2} c d e^{2} x - 2 \, a^{3} e^{3}\right )} \sqrt {c d x + a e} g}{35 \, c^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(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/(c*d) + 2/35*(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)*g/(c^2*d^2)

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mupad [B]  time = 3.25, size = 109, normalized size = 0.87 \[ \frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (x^2\,\left (\frac {16\,a\,e\,g}{35}+\frac {2\,c\,d\,f}{5}\right )-\frac {4\,a^3\,e^3\,g-14\,a^2\,c\,d\,e^2\,f}{35\,c^2\,d^2}+\frac {2\,c\,d\,g\,x^3}{7}+\frac {2\,a\,e\,x\,\left (a\,e\,g+14\,c\,d\,f\right )}{35\,c\,d}\right )}{\sqrt {d+e\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)*(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)*(x^2*((16*a*e*g)/35 + (2*c*d*f)/5) - (4*a^3*e^3*g - 14*a^2*c*d*
e^2*f)/(35*c^2*d^2) + (2*c*d*g*x^3)/7 + (2*a*e*x*(a*e*g + 14*c*d*f))/(35*c*d)))/(d + e*x)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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