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

Optimal. Leaf size=257 \[ -\frac {16 g \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right )}{5 c^4 d^4 e \sqrt {d+e x}}+\frac {16 g^2 \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)}{5 c^3 d^3 e}+\frac {12 g (f+g x)^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 c^2 d^2 \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} (f+g x)^3}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

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Rubi [A]  time = 0.33, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {866, 870, 794, 648} \begin {gather*} \frac {16 g^2 \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)}{5 c^3 d^3 e}+\frac {12 g (f+g x)^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 c^2 d^2 \sqrt {d+e x}}-\frac {16 g \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right )}{5 c^4 d^4 e \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} (f+g x)^3}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[d + e*x]*(f + g*x)^3)/(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (16*g*(c*d*f - a*e*g)*(2*a*
e^2*g - c*d*(3*e*f - d*g))*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(5*c^4*d^4*e*Sqrt[d + e*x]) + (16*g^2*
(c*d*f - a*e*g)*Sqrt[d + e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(5*c^3*d^3*e) + (12*g*(f + g*x)^2*S
qrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(5*c^2*d^2*Sqrt[d + e*x])

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 866

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*(p + 1)), x] - Dist[(e*g*n)/(c*(p + 1)), I
nt[(d + e*x)^(m - 1)*(f + g*x)^(n - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, 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] &&  !IntegerQ[p] && EqQ[m + p, 0] &
& LtQ[p, -1] && GtQ[n, 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 {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=-\frac {2 \sqrt {d+e x} (f+g x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {(6 g) \int \frac {\sqrt {d+e x} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{c d}\\ &=-\frac {2 \sqrt {d+e x} (f+g x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {12 g (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^2 d^2 \sqrt {d+e x}}+\frac {(24 g (c d f-a e g)) \int \frac {\sqrt {d+e x} (f+g x)}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{5 c^2 d^2}\\ &=-\frac {2 \sqrt {d+e x} (f+g x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {16 g^2 (c d f-a e g) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^3 d^3 e}+\frac {12 g (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^2 d^2 \sqrt {d+e x}}-\frac {\left (8 g (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right )\right ) \int \frac {\sqrt {d+e x}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{5 c^3 d^3 e}\\ &=-\frac {2 \sqrt {d+e x} (f+g x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {16 g (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^4 d^4 e \sqrt {d+e x}}+\frac {16 g^2 (c d f-a e g) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^3 d^3 e}+\frac {12 g (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^2 d^2 \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 134, normalized size = 0.52 \begin {gather*} \frac {2 \sqrt {d+e x} \left (16 a^3 e^3 g^3+8 a^2 c d e^2 g^2 (g x-5 f)-2 a c^2 d^2 e g \left (-15 f^2+10 f g x+g^2 x^2\right )+c^3 d^3 \left (-5 f^3+15 f^2 g x+5 f g^2 x^2+g^3 x^3\right )\right )}{5 c^4 d^4 \sqrt {(d+e x) (a e+c d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(16*a^3*e^3*g^3 + 8*a^2*c*d*e^2*g^2*(-5*f + g*x) - 2*a*c^2*d^2*e*g*(-15*f^2 + 10*f*g*x + g^2*
x^2) + c^3*d^3*(-5*f^3 + 15*f^2*g*x + 5*f*g^2*x^2 + g^3*x^3)))/(5*c^4*d^4*Sqrt[(a*e + c*d*x)*(d + e*x)])

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IntegrateAlgebraic [A]  time = 3.64, size = 202, normalized size = 0.79 \begin {gather*} \frac {2 (d+e x)^{3/2} (a e+c d x) \left (5 a^3 e^3 g^3-15 a^2 c d e^2 f g^2+15 a^2 e^2 g^3 (a e+c d x)+15 c^2 d^2 f^2 g (a e+c d x)+15 a c^2 d^2 e f^2 g+5 c d f g^2 (a e+c d x)^2-30 a c d e f g^2 (a e+c d x)+g^3 (a e+c d x)^3-5 a e g^3 (a e+c d x)^2-5 c^3 d^3 f^3\right )}{5 c^4 d^4 ((d+e x) (a e+c d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*e + c*d*x)*(d + e*x)^(3/2)*(-5*c^3*d^3*f^3 + 15*a*c^2*d^2*e*f^2*g - 15*a^2*c*d*e^2*f*g^2 + 5*a^3*e^3*g^3
+ 15*c^2*d^2*f^2*g*(a*e + c*d*x) - 30*a*c*d*e*f*g^2*(a*e + c*d*x) + 15*a^2*e^2*g^3*(a*e + c*d*x) + 5*c*d*f*g^
2*(a*e + c*d*x)^2 - 5*a*e*g^3*(a*e + c*d*x)^2 + g^3*(a*e + c*d*x)^3))/(5*c^4*d^4*((a*e + c*d*x)*(d + e*x))^(3/
2))

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fricas [A]  time = 0.42, size = 216, normalized size = 0.84 \begin {gather*} \frac {2 \, {\left (c^{3} d^{3} g^{3} x^{3} - 5 \, c^{3} d^{3} f^{3} + 30 \, a c^{2} d^{2} e f^{2} g - 40 \, a^{2} c d e^{2} f g^{2} + 16 \, a^{3} e^{3} g^{3} + {\left (5 \, c^{3} d^{3} f g^{2} - 2 \, a c^{2} d^{2} e g^{3}\right )} x^{2} + {\left (15 \, c^{3} d^{3} f^{2} g - 20 \, a c^{2} d^{2} e f g^{2} + 8 \, a^{2} c d e^{2} g^{3}\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}}{5 \, {\left (c^{5} d^{5} e x^{2} + a c^{4} d^{5} e + {\left (c^{5} d^{6} + a c^{4} d^{4} e^{2}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(c^3*d^3*g^3*x^3 - 5*c^3*d^3*f^3 + 30*a*c^2*d^2*e*f^2*g - 40*a^2*c*d*e^2*f*g^2 + 16*a^3*e^3*g^3 + (5*c^3*d
^3*f*g^2 - 2*a*c^2*d^2*e*g^3)*x^2 + (15*c^3*d^3*f^2*g - 20*a*c^2*d^2*e*f*g^2 + 8*a^2*c*d*e^2*g^3)*x)*sqrt(c*d*
e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)/(c^5*d^5*e*x^2 + a*c^4*d^5*e + (c^5*d^6 + a*c^4*d^4*e^2)*x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 1.71Unable to transpose Error: Bad Argument Value

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maple [A]  time = 0.01, size = 187, normalized size = 0.73 \begin {gather*} \frac {2 \left (c d x +a e \right ) \left (g^{3} x^{3} c^{3} d^{3}-2 a \,c^{2} d^{2} e \,g^{3} x^{2}+5 c^{3} d^{3} f \,g^{2} x^{2}+8 a^{2} c d \,e^{2} g^{3} x -20 a \,c^{2} d^{2} e f \,g^{2} x +15 c^{3} d^{3} f^{2} g x +16 a^{3} e^{3} g^{3}-40 a^{2} c d \,e^{2} f \,g^{2}+30 a \,c^{2} d^{2} e \,f^{2} g -5 f^{3} c^{3} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}}{5 \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}} c^{4} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(c*d*x+a*e)*(c^3*d^3*g^3*x^3-2*a*c^2*d^2*e*g^3*x^2+5*c^3*d^3*f*g^2*x^2+8*a^2*c*d*e^2*g^3*x-20*a*c^2*d^2*e*
f*g^2*x+15*c^3*d^3*f^2*g*x+16*a^3*e^3*g^3-40*a^2*c*d*e^2*f*g^2+30*a*c^2*d^2*e*f^2*g-5*c^3*d^3*f^3)*(e*x+d)^(3/
2)/c^4/d^4/(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(3/2)

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maxima [A]  time = 0.70, size = 165, normalized size = 0.64 \begin {gather*} -\frac {2 \, f^{3}}{\sqrt {c d x + a e} c d} + \frac {6 \, {\left (c d x + 2 \, a e\right )} f^{2} g}{\sqrt {c d x + a e} c^{2} d^{2}} + \frac {2 \, {\left (c^{2} d^{2} x^{2} - 4 \, a c d e x - 8 \, a^{2} e^{2}\right )} f g^{2}}{\sqrt {c d x + a e} c^{3} d^{3}} + \frac {2 \, {\left (c^{3} d^{3} x^{3} - 2 \, a c^{2} d^{2} e x^{2} + 8 \, a^{2} c d e^{2} x + 16 \, a^{3} e^{3}\right )} g^{3}}{5 \, \sqrt {c d x + a e} c^{4} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*f^3/(sqrt(c*d*x + a*e)*c*d) + 6*(c*d*x + 2*a*e)*f^2*g/(sqrt(c*d*x + a*e)*c^2*d^2) + 2*(c^2*d^2*x^2 - 4*a*c*
d*e*x - 8*a^2*e^2)*f*g^2/(sqrt(c*d*x + a*e)*c^3*d^3) + 2/5*(c^3*d^3*x^3 - 2*a*c^2*d^2*e*x^2 + 8*a^2*c*d*e^2*x
+ 16*a^3*e^3)*g^3/(sqrt(c*d*x + a*e)*c^4*d^4)

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mupad [B]  time = 3.61, size = 252, normalized size = 0.98 \begin {gather*} \frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {\sqrt {d+e\,x}\,\left (32\,a^3\,e^3\,g^3-80\,a^2\,c\,d\,e^2\,f\,g^2+60\,a\,c^2\,d^2\,e\,f^2\,g-10\,c^3\,d^3\,f^3\right )}{5\,c^5\,d^5\,e}+\frac {2\,g^3\,x^3\,\sqrt {d+e\,x}}{5\,c^2\,d^2\,e}-\frac {2\,g^2\,x^2\,\left (2\,a\,e\,g-5\,c\,d\,f\right )\,\sqrt {d+e\,x}}{5\,c^3\,d^3\,e}+\frac {2\,g\,x\,\sqrt {d+e\,x}\,\left (8\,a^2\,e^2\,g^2-20\,a\,c\,d\,e\,f\,g+15\,c^2\,d^2\,f^2\right )}{5\,c^4\,d^4\,e}\right )}{\frac {a}{c}+x^2+\frac {x\,\left (5\,c^5\,d^6+5\,a\,c^4\,d^4\,e^2\right )}{5\,c^5\,d^5\,e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*(((d + e*x)^(1/2)*(32*a^3*e^3*g^3 - 10*c^3*d^3*f^3 + 60*a*c^2*d
^2*e*f^2*g - 80*a^2*c*d*e^2*f*g^2))/(5*c^5*d^5*e) + (2*g^3*x^3*(d + e*x)^(1/2))/(5*c^2*d^2*e) - (2*g^2*x^2*(2*
a*e*g - 5*c*d*f)*(d + e*x)^(1/2))/(5*c^3*d^3*e) + (2*g*x*(d + e*x)^(1/2)*(8*a^2*e^2*g^2 + 15*c^2*d^2*f^2 - 20*
a*c*d*e*f*g))/(5*c^4*d^4*e)))/(a/c + x^2 + (x*(5*c^5*d^6 + 5*a*c^4*d^4*e^2))/(5*c^5*d^5*e))

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

[Out]

Timed out

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