3.5.93 \(\int \frac {(d+e x)^{3/2}}{(f+g x)^{7/2} (a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=262 \[ -\frac {32 c^2 d^2 g \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 \sqrt {d+e x} \sqrt {f+g x} (c d f-a e g)^4}-\frac {16 c d g \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 \sqrt {d+e x} (f+g x)^{3/2} (c d f-a e g)^3}-\frac {12 g \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 \sqrt {d+e x} (f+g x)^{5/2} (c d f-a e g)^2}-\frac {2 \sqrt {d+e x}}{(f+g x)^{5/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)} \]

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Rubi [A]  time = 0.33, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {868, 872, 860} \begin {gather*} -\frac {32 c^2 d^2 g \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 \sqrt {d+e x} \sqrt {f+g x} (c d f-a e g)^4}-\frac {16 c d g \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 \sqrt {d+e x} (f+g x)^{3/2} (c d f-a e g)^3}-\frac {12 g \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 \sqrt {d+e x} (f+g x)^{5/2} (c d f-a e g)^2}-\frac {2 \sqrt {d+e x}}{(f+g x)^{5/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[d + e*x])/((c*d*f - a*e*g)*(f + g*x)^(5/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (12*g*Sqrt[
a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(5*(c*d*f - a*e*g)^2*Sqrt[d + e*x]*(f + g*x)^(5/2)) - (16*c*d*g*Sqrt[a
*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(5*(c*d*f - a*e*g)^3*Sqrt[d + e*x]*(f + g*x)^(3/2)) - (32*c^2*d^2*g*Sqr
t[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(5*(c*d*f - a*e*g)^4*Sqrt[d + e*x]*Sqrt[f + g*x])

Rule 860

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

Rule 868

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

Rule 872

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^{3/2}}{(f+g x)^{7/2} \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}}{(c d f-a e g) (f+g x)^{5/2} \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)^{7/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{c d f-a e g}\\ &=-\frac {2 \sqrt {d+e x}}{(c d f-a e g) (f+g x)^{5/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {12 g \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 (c d f-a e g)^2 \sqrt {d+e x} (f+g x)^{5/2}}-\frac {(24 c d g) \int \frac {\sqrt {d+e x}}{(f+g x)^{5/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{5 (c d f-a e g)^2}\\ &=-\frac {2 \sqrt {d+e x}}{(c d f-a e g) (f+g x)^{5/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {12 g \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 (c d f-a e g)^2 \sqrt {d+e x} (f+g x)^{5/2}}-\frac {16 c d g \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 (c d f-a e g)^3 \sqrt {d+e x} (f+g x)^{3/2}}-\frac {\left (16 c^2 d^2 g\right ) \int \frac {\sqrt {d+e x}}{(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{5 (c d f-a e g)^3}\\ &=-\frac {2 \sqrt {d+e x}}{(c d f-a e g) (f+g x)^{5/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {12 g \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 (c d f-a e g)^2 \sqrt {d+e x} (f+g x)^{5/2}}-\frac {16 c d g \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 (c d f-a e g)^3 \sqrt {d+e x} (f+g x)^{3/2}}-\frac {32 c^2 d^2 g \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 (c d f-a e g)^4 \sqrt {d+e x} \sqrt {f+g x}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 150, normalized size = 0.57 \begin {gather*} -\frac {2 \sqrt {d+e x} \left (a^3 e^3 g^3-a^2 c d e^2 g^2 (5 f+2 g x)+a c^2 d^2 e g \left (15 f^2+20 f g x+8 g^2 x^2\right )+c^3 d^3 \left (5 f^3+30 f^2 g x+40 f g^2 x^2+16 g^3 x^3\right )\right )}{5 (f+g x)^{5/2} \sqrt {(d+e x) (a e+c d x)} (c d f-a e g)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(3/2)/((f + g*x)^(7/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 1.31, size = 289, normalized size = 1.10 \begin {gather*} -\frac {2 (d+e x)^{3/2} (a e g+c d g x)^{3/2} \left (a^3 e^3 g^{7/2}-2 a^2 c d e^2 g^{5/2} (f+g x)-3 a^2 c d e^2 f g^{5/2}+3 a c^2 d^2 e f^2 g^{3/2}+8 a c^2 d^2 e g^{3/2} (f+g x)^2+4 a c^2 d^2 e f g^{3/2} (f+g x)-c^3 d^3 f^3 \sqrt {g}-2 c^3 d^3 f^2 \sqrt {g} (f+g x)+16 c^3 d^3 \sqrt {g} (f+g x)^3-8 c^3 d^3 f \sqrt {g} (f+g x)^2\right )}{5 g^{3/2} (f+g x)^{5/2} (c d f-a e g)^4 \left (\frac {(d g+e g x) (a e g+c d g x)}{g^2}\right )^{3/2} \sqrt {a e g+c d (f+g x)-c d f}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^(3/2)/((f + g*x)^(7/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]

[Out]

(-2*(d + e*x)^(3/2)*(a*e*g + c*d*g*x)^(3/2)*(-(c^3*d^3*f^3*Sqrt[g]) + 3*a*c^2*d^2*e*f^2*g^(3/2) - 3*a^2*c*d*e^
2*f*g^(5/2) + a^3*e^3*g^(7/2) - 2*c^3*d^3*f^2*Sqrt[g]*(f + g*x) + 4*a*c^2*d^2*e*f*g^(3/2)*(f + g*x) - 2*a^2*c*
d*e^2*g^(5/2)*(f + g*x) - 8*c^3*d^3*f*Sqrt[g]*(f + g*x)^2 + 8*a*c^2*d^2*e*g^(3/2)*(f + g*x)^2 + 16*c^3*d^3*Sqr
t[g]*(f + g*x)^3))/(5*g^(3/2)*(c*d*f - a*e*g)^4*(f + g*x)^(5/2)*(((a*e*g + c*d*g*x)*(d*g + e*g*x))/g^2)^(3/2)*
Sqrt[-(c*d*f) + a*e*g + c*d*(f + g*x)])

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fricas [B]  time = 0.51, size = 1062, normalized size = 4.05 \begin {gather*} -\frac {2 \, {\left (16 \, c^{3} d^{3} g^{3} x^{3} + 5 \, c^{3} d^{3} f^{3} + 15 \, a c^{2} d^{2} e f^{2} g - 5 \, a^{2} c d e^{2} f g^{2} + a^{3} e^{3} g^{3} + 8 \, {\left (5 \, c^{3} d^{3} f g^{2} + a c^{2} d^{2} e g^{3}\right )} x^{2} + 2 \, {\left (15 \, c^{3} d^{3} f^{2} g + 10 \, a c^{2} d^{2} e f g^{2} - 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} \sqrt {g x + f}}{5 \, {\left (a c^{4} d^{5} e f^{7} - 4 \, a^{2} c^{3} d^{4} e^{2} f^{6} g + 6 \, a^{3} c^{2} d^{3} e^{3} f^{5} g^{2} - 4 \, a^{4} c d^{2} e^{4} f^{4} g^{3} + a^{5} d e^{5} f^{3} g^{4} + {\left (c^{5} d^{5} e f^{4} g^{3} - 4 \, a c^{4} d^{4} e^{2} f^{3} g^{4} + 6 \, a^{2} c^{3} d^{3} e^{3} f^{2} g^{5} - 4 \, a^{3} c^{2} d^{2} e^{4} f g^{6} + a^{4} c d e^{5} g^{7}\right )} x^{5} + {\left (3 \, c^{5} d^{5} e f^{5} g^{2} + {\left (c^{5} d^{6} - 11 \, a c^{4} d^{4} e^{2}\right )} f^{4} g^{3} - 2 \, {\left (2 \, a c^{4} d^{5} e - 7 \, a^{2} c^{3} d^{3} e^{3}\right )} f^{3} g^{4} + 6 \, {\left (a^{2} c^{3} d^{4} e^{2} - a^{3} c^{2} d^{2} e^{4}\right )} f^{2} g^{5} - {\left (4 \, a^{3} c^{2} d^{3} e^{3} + a^{4} c d e^{5}\right )} f g^{6} + {\left (a^{4} c d^{2} e^{4} + a^{5} e^{6}\right )} g^{7}\right )} x^{4} + {\left (3 \, c^{5} d^{5} e f^{6} g + a^{5} d e^{5} g^{7} + 3 \, {\left (c^{5} d^{6} - 3 \, a c^{4} d^{4} e^{2}\right )} f^{5} g^{2} - {\left (11 \, a c^{4} d^{5} e - 6 \, a^{2} c^{3} d^{3} e^{3}\right )} f^{4} g^{3} + 2 \, {\left (7 \, a^{2} c^{3} d^{4} e^{2} + 3 \, a^{3} c^{2} d^{2} e^{4}\right )} f^{3} g^{4} - 3 \, {\left (2 \, a^{3} c^{2} d^{3} e^{3} + 3 \, a^{4} c d e^{5}\right )} f^{2} g^{5} - {\left (a^{4} c d^{2} e^{4} - 3 \, a^{5} e^{6}\right )} f g^{6}\right )} x^{3} + {\left (c^{5} d^{5} e f^{7} + 3 \, a^{5} d e^{5} f g^{6} + {\left (3 \, c^{5} d^{6} - a c^{4} d^{4} e^{2}\right )} f^{6} g - 3 \, {\left (3 \, a c^{4} d^{5} e + 2 \, a^{2} c^{3} d^{3} e^{3}\right )} f^{5} g^{2} + 2 \, {\left (3 \, a^{2} c^{3} d^{4} e^{2} + 7 \, a^{3} c^{2} d^{2} e^{4}\right )} f^{4} g^{3} + {\left (6 \, a^{3} c^{2} d^{3} e^{3} - 11 \, a^{4} c d e^{5}\right )} f^{3} g^{4} - 3 \, {\left (3 \, a^{4} c d^{2} e^{4} - a^{5} e^{6}\right )} f^{2} g^{5}\right )} x^{2} + {\left (3 \, a^{5} d e^{5} f^{2} g^{5} + {\left (c^{5} d^{6} + a c^{4} d^{4} e^{2}\right )} f^{7} - {\left (a c^{4} d^{5} e + 4 \, a^{2} c^{3} d^{3} e^{3}\right )} f^{6} g - 6 \, {\left (a^{2} c^{3} d^{4} e^{2} - a^{3} c^{2} d^{2} e^{4}\right )} f^{5} g^{2} + 2 \, {\left (7 \, a^{3} c^{2} d^{3} e^{3} - 2 \, a^{4} c d e^{5}\right )} f^{4} g^{3} - {\left (11 \, a^{4} c d^{2} e^{4} - a^{5} e^{6}\right )} f^{3} g^{4}\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)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

-2/5*(16*c^3*d^3*g^3*x^3 + 5*c^3*d^3*f^3 + 15*a*c^2*d^2*e*f^2*g - 5*a^2*c*d*e^2*f*g^2 + a^3*e^3*g^3 + 8*(5*c^3
*d^3*f*g^2 + a*c^2*d^2*e*g^3)*x^2 + 2*(15*c^3*d^3*f^2*g + 10*a*c^2*d^2*e*f*g^2 - 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)*sqrt(g*x + f)/(a*c^4*d^5*e*f^7 - 4*a^2*c^3*d^4*e^2*f^6*g + 6*
a^3*c^2*d^3*e^3*f^5*g^2 - 4*a^4*c*d^2*e^4*f^4*g^3 + a^5*d*e^5*f^3*g^4 + (c^5*d^5*e*f^4*g^3 - 4*a*c^4*d^4*e^2*f
^3*g^4 + 6*a^2*c^3*d^3*e^3*f^2*g^5 - 4*a^3*c^2*d^2*e^4*f*g^6 + a^4*c*d*e^5*g^7)*x^5 + (3*c^5*d^5*e*f^5*g^2 + (
c^5*d^6 - 11*a*c^4*d^4*e^2)*f^4*g^3 - 2*(2*a*c^4*d^5*e - 7*a^2*c^3*d^3*e^3)*f^3*g^4 + 6*(a^2*c^3*d^4*e^2 - a^3
*c^2*d^2*e^4)*f^2*g^5 - (4*a^3*c^2*d^3*e^3 + a^4*c*d*e^5)*f*g^6 + (a^4*c*d^2*e^4 + a^5*e^6)*g^7)*x^4 + (3*c^5*
d^5*e*f^6*g + a^5*d*e^5*g^7 + 3*(c^5*d^6 - 3*a*c^4*d^4*e^2)*f^5*g^2 - (11*a*c^4*d^5*e - 6*a^2*c^3*d^3*e^3)*f^4
*g^3 + 2*(7*a^2*c^3*d^4*e^2 + 3*a^3*c^2*d^2*e^4)*f^3*g^4 - 3*(2*a^3*c^2*d^3*e^3 + 3*a^4*c*d*e^5)*f^2*g^5 - (a^
4*c*d^2*e^4 - 3*a^5*e^6)*f*g^6)*x^3 + (c^5*d^5*e*f^7 + 3*a^5*d*e^5*f*g^6 + (3*c^5*d^6 - a*c^4*d^4*e^2)*f^6*g -
3*(3*a*c^4*d^5*e + 2*a^2*c^3*d^3*e^3)*f^5*g^2 + 2*(3*a^2*c^3*d^4*e^2 + 7*a^3*c^2*d^2*e^4)*f^4*g^3 + (6*a^3*c^
2*d^3*e^3 - 11*a^4*c*d*e^5)*f^3*g^4 - 3*(3*a^4*c*d^2*e^4 - a^5*e^6)*f^2*g^5)*x^2 + (3*a^5*d*e^5*f^2*g^5 + (c^5
*d^6 + a*c^4*d^4*e^2)*f^7 - (a*c^4*d^5*e + 4*a^2*c^3*d^3*e^3)*f^6*g - 6*(a^2*c^3*d^4*e^2 - a^3*c^2*d^2*e^4)*f^
5*g^2 + 2*(7*a^3*c^2*d^3*e^3 - 2*a^4*c*d*e^5)*f^4*g^3 - (11*a^4*c*d^2*e^4 - a^5*e^6)*f^3*g^4)*x)

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giac [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)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 259, normalized size = 0.99 \begin {gather*} -\frac {2 \left (c d x +a e \right ) \left (16 g^{3} x^{3} c^{3} d^{3}+8 a \,c^{2} d^{2} e \,g^{3} x^{2}+40 c^{3} d^{3} f \,g^{2} x^{2}-2 a^{2} c d \,e^{2} g^{3} x +20 a \,c^{2} d^{2} e f \,g^{2} x +30 c^{3} d^{3} f^{2} g x +a^{3} e^{3} g^{3}-5 a^{2} c d \,e^{2} f \,g^{2}+15 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 (g x +f \right )^{\frac {5}{2}} \left (g^{4} e^{4} a^{4}-4 a^{3} c d \,e^{3} f \,g^{3}+6 a^{2} c^{2} d^{2} e^{2} f^{2} g^{2}-4 a \,c^{3} d^{3} e \,f^{3} g +f^{4} c^{4} d^{4}\right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)/(g*x+f)^(7/2)/(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)*(16*c^3*d^3*g^3*x^3+8*a*c^2*d^2*e*g^3*x^2+40*c^3*d^3*f*g^2*x^2-2*a^2*c*d*e^2*g^3*x+20*a*c^2*d
^2*e*f*g^2*x+30*c^3*d^3*f^2*g*x+a^3*e^3*g^3-5*a^2*c*d*e^2*f*g^2+15*a*c^2*d^2*e*f^2*g+5*c^3*d^3*f^3)*(e*x+d)^(3
/2)/(g*x+f)^(5/2)/(a^4*e^4*g^4-4*a^3*c*d*e^3*f*g^3+6*a^2*c^2*d^2*e^2*f^2*g^2-4*a*c^3*d^3*e*f^3*g+c^4*d^4*f^4)/
(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (e x + d\right )}^{\frac {3}{2}}}{{\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 )}^{\frac {7}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)/(g*x+f)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(3/2)/((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(3/2)*(g*x + f)^(7/2)), x)

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mupad [B]  time = 5.70, size = 414, normalized size = 1.58 \begin {gather*} -\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {4\,x\,\sqrt {d+e\,x}\,\left (-a^2\,e^2\,g^2+10\,a\,c\,d\,e\,f\,g+15\,c^2\,d^2\,f^2\right )}{5\,e\,g\,{\left (a\,e\,g-c\,d\,f\right )}^4}+\frac {\sqrt {d+e\,x}\,\left (\frac {2\,a^3\,e^3\,g^3}{5}-2\,a^2\,c\,d\,e^2\,f\,g^2+6\,a\,c^2\,d^2\,e\,f^2\,g+2\,c^3\,d^3\,f^3\right )}{c\,d\,e\,g^2\,{\left (a\,e\,g-c\,d\,f\right )}^4}+\frac {32\,c^2\,d^2\,g\,x^3\,\sqrt {d+e\,x}}{5\,e\,{\left (a\,e\,g-c\,d\,f\right )}^4}+\frac {16\,c\,d\,x^2\,\left (a\,e\,g+5\,c\,d\,f\right )\,\sqrt {d+e\,x}}{5\,e\,{\left (a\,e\,g-c\,d\,f\right )}^4}\right )}{x^4\,\sqrt {f+g\,x}+\frac {a\,f^2\,\sqrt {f+g\,x}}{c\,g^2}+\frac {x^2\,\sqrt {f+g\,x}\,\left (2\,c\,d^2\,f\,g+c\,d\,e\,f^2+a\,d\,e\,g^2+2\,a\,e^2\,f\,g\right )}{c\,d\,e\,g^2}+\frac {x^3\,\sqrt {f+g\,x}\,\left (c\,g\,d^2+2\,c\,f\,d\,e+a\,g\,e^2\right )}{c\,d\,e\,g}+\frac {f\,x\,\sqrt {f+g\,x}\,\left (c\,f\,d^2+2\,a\,g\,d\,e+a\,f\,e^2\right )}{c\,d\,e\,g^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(3/2)/((f + g*x)^(7/2)*(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)*((4*x*(d + e*x)^(1/2)*(15*c^2*d^2*f^2 - a^2*e^2*g^2 + 10*a*c*d
*e*f*g))/(5*e*g*(a*e*g - c*d*f)^4) + ((d + e*x)^(1/2)*((2*a^3*e^3*g^3)/5 + 2*c^3*d^3*f^3 + 6*a*c^2*d^2*e*f^2*g
- 2*a^2*c*d*e^2*f*g^2))/(c*d*e*g^2*(a*e*g - c*d*f)^4) + (32*c^2*d^2*g*x^3*(d + e*x)^(1/2))/(5*e*(a*e*g - c*d*
f)^4) + (16*c*d*x^2*(a*e*g + 5*c*d*f)*(d + e*x)^(1/2))/(5*e*(a*e*g - c*d*f)^4)))/(x^4*(f + g*x)^(1/2) + (a*f^2
*(f + g*x)^(1/2))/(c*g^2) + (x^2*(f + g*x)^(1/2)*(a*d*e*g^2 + c*d*e*f^2 + 2*a*e^2*f*g + 2*c*d^2*f*g))/(c*d*e*g
^2) + (x^3*(f + g*x)^(1/2)*(a*e^2*g + c*d^2*g + 2*c*d*e*f))/(c*d*e*g) + (f*x*(f + g*x)^(1/2)*(a*e^2*f + c*d^2*
f + 2*a*d*e*g))/(c*d*e*g^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((e*x+d)**(3/2)/(g*x+f)**(7/2)/(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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