3.5.98 \(\int \frac {(d+e x)^{5/2}}{(f+g x)^{3/2} (a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=194 \[ \frac {16 g^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 \sqrt {d+e x} \sqrt {f+g x} (c d f-a e g)^3}+\frac {8 g \sqrt {d+e x}}{3 \sqrt {f+g x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)^2}-\frac {2 (d+e x)^{3/2}}{3 \sqrt {f+g x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} (c d f-a e g)} \]

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Rubi [A]  time = 0.22, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {868, 860} \begin {gather*} \frac {16 g^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 \sqrt {d+e x} \sqrt {f+g x} (c d f-a e g)^3}+\frac {8 g \sqrt {d+e x}}{3 \sqrt {f+g 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)^2}-\frac {2 (d+e x)^{3/2}}{3 \sqrt {f+g x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} (c d f-a e g)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^(3/2))/(3*(c*d*f - a*e*g)*Sqrt[f + g*x]*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) + (8*g*Sq
rt[d + e*x])/(3*(c*d*f - a*e*g)^2*Sqrt[f + g*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (16*g^2*Sqrt[a*
d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(3*(c*d*f - a*e*g)^3*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]

Rubi steps

\begin {align*} \int \frac {(d+e x)^{5/2}}{(f+g x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx &=-\frac {2 (d+e x)^{3/2}}{3 (c d f-a e g) \sqrt {f+g x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {(4 g) \int \frac {(d+e x)^{3/2}}{(f+g x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{3 (c d f-a e g)}\\ &=-\frac {2 (d+e x)^{3/2}}{3 (c d f-a e g) \sqrt {f+g x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac {8 g \sqrt {d+e x}}{3 (c d f-a e g)^2 \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {\left (8 g^2\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}{3 (c d f-a e g)^2}\\ &=-\frac {2 (d+e x)^{3/2}}{3 (c d f-a e g) \sqrt {f+g x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac {8 g \sqrt {d+e x}}{3 (c d f-a e g)^2 \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {16 g^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 (c d f-a e g)^3 \sqrt {d+e x} \sqrt {f+g x}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 103, normalized size = 0.53 \begin {gather*} \frac {2 (d+e x)^{3/2} \left (3 a^2 e^2 g^2+6 a c d e g (f+2 g x)+c^2 d^2 \left (-f^2+4 f g x+8 g^2 x^2\right )\right )}{3 \sqrt {f+g x} ((d+e x) (a e+c d x))^{3/2} (c d f-a e g)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(5/2)/((f + g*x)^(3/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)),x]

[Out]

(2*(d + e*x)^(3/2)*(3*a^2*e^2*g^2 + 6*a*c*d*e*g*(f + 2*g*x) + c^2*d^2*(-f^2 + 4*f*g*x + 8*g^2*x^2)))/(3*(c*d*f
- a*e*g)^3*((a*e + c*d*x)*(d + e*x))^(3/2)*Sqrt[f + g*x])

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IntegrateAlgebraic [A]  time = 1.20, size = 198, normalized size = 1.02 \begin {gather*} \frac {2 (d+e x)^{5/2} (a e g+c d g x)^{5/2} \left (3 a^2 e^2 g^{7/2}+12 a c d e g^{5/2} (f+g x)-6 a c d e f g^{5/2}+3 c^2 d^2 f^2 g^{3/2}+8 c^2 d^2 g^{3/2} (f+g x)^2-12 c^2 d^2 f g^{3/2} (f+g x)\right )}{3 g^{5/2} \sqrt {f+g x} (c d f-a e g)^3 \left (\frac {(d g+e g x) (a e g+c d g x)}{g^2}\right )^{5/2} (a e g+c d (f+g x)-c d f)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^(5/2)/((f + g*x)^(3/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)),x]

[Out]

(2*(d + e*x)^(5/2)*(a*e*g + c*d*g*x)^(5/2)*(3*c^2*d^2*f^2*g^(3/2) - 6*a*c*d*e*f*g^(5/2) + 3*a^2*e^2*g^(7/2) -
12*c^2*d^2*f*g^(3/2)*(f + g*x) + 12*a*c*d*e*g^(5/2)*(f + g*x) + 8*c^2*d^2*g^(3/2)*(f + g*x)^2))/(3*g^(5/2)*(c*
d*f - a*e*g)^3*Sqrt[f + g*x]*(((a*e*g + c*d*g*x)*(d*g + e*g*x))/g^2)^(5/2)*(-(c*d*f) + a*e*g + c*d*(f + g*x))^
(3/2))

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fricas [B]  time = 0.45, size = 667, normalized size = 3.44 \begin {gather*} \frac {2 \, {\left (8 \, c^{2} d^{2} g^{2} x^{2} - c^{2} d^{2} f^{2} + 6 \, a c d e f g + 3 \, a^{2} e^{2} g^{2} + 4 \, {\left (c^{2} d^{2} f g + 3 \, a c d e 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} \sqrt {g x + f}}{3 \, {\left (a^{2} c^{3} d^{4} e^{2} f^{4} - 3 \, a^{3} c^{2} d^{3} e^{3} f^{3} g + 3 \, a^{4} c d^{2} e^{4} f^{2} g^{2} - a^{5} d e^{5} f g^{3} + {\left (c^{5} d^{5} e f^{3} g - 3 \, a c^{4} d^{4} e^{2} f^{2} g^{2} + 3 \, a^{2} c^{3} d^{3} e^{3} f g^{3} - a^{3} c^{2} d^{2} e^{4} g^{4}\right )} x^{4} + {\left (c^{5} d^{5} e f^{4} + {\left (c^{5} d^{6} - a c^{4} d^{4} e^{2}\right )} f^{3} g - 3 \, {\left (a c^{4} d^{5} e + a^{2} c^{3} d^{3} e^{3}\right )} f^{2} g^{2} + {\left (3 \, a^{2} c^{3} d^{4} e^{2} + 5 \, a^{3} c^{2} d^{2} e^{4}\right )} f g^{3} - {\left (a^{3} c^{2} d^{3} e^{3} + 2 \, a^{4} c d e^{5}\right )} g^{4}\right )} x^{3} + {\left ({\left (c^{5} d^{6} + 2 \, a c^{4} d^{4} e^{2}\right )} f^{4} - {\left (a c^{4} d^{5} e + 5 \, a^{2} c^{3} d^{3} e^{3}\right )} f^{3} g - 3 \, {\left (a^{2} c^{3} d^{4} e^{2} - a^{3} c^{2} d^{2} e^{4}\right )} f^{2} g^{2} + {\left (5 \, a^{3} c^{2} d^{3} e^{3} + a^{4} c d e^{5}\right )} f g^{3} - {\left (2 \, a^{4} c d^{2} e^{4} + a^{5} e^{6}\right )} g^{4}\right )} x^{2} - {\left (a^{5} d e^{5} g^{4} - {\left (2 \, a c^{4} d^{5} e + a^{2} c^{3} d^{3} e^{3}\right )} f^{4} + {\left (5 \, a^{2} c^{3} d^{4} e^{2} + 3 \, a^{3} c^{2} d^{2} e^{4}\right )} f^{3} g - 3 \, {\left (a^{3} c^{2} d^{3} e^{3} + a^{4} c d e^{5}\right )} f^{2} g^{2} - {\left (a^{4} c d^{2} e^{4} - a^{5} e^{6}\right )} f g^{3}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(g*x+f)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="fricas")

[Out]

2/3*(8*c^2*d^2*g^2*x^2 - c^2*d^2*f^2 + 6*a*c*d*e*f*g + 3*a^2*e^2*g^2 + 4*(c^2*d^2*f*g + 3*a*c*d*e*g^2)*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^2*c^3*d^4*e^2*f^4 - 3*a^3*c^2*d^3*e^3*f
^3*g + 3*a^4*c*d^2*e^4*f^2*g^2 - a^5*d*e^5*f*g^3 + (c^5*d^5*e*f^3*g - 3*a*c^4*d^4*e^2*f^2*g^2 + 3*a^2*c^3*d^3*
e^3*f*g^3 - a^3*c^2*d^2*e^4*g^4)*x^4 + (c^5*d^5*e*f^4 + (c^5*d^6 - a*c^4*d^4*e^2)*f^3*g - 3*(a*c^4*d^5*e + a^2
*c^3*d^3*e^3)*f^2*g^2 + (3*a^2*c^3*d^4*e^2 + 5*a^3*c^2*d^2*e^4)*f*g^3 - (a^3*c^2*d^3*e^3 + 2*a^4*c*d*e^5)*g^4)
*x^3 + ((c^5*d^6 + 2*a*c^4*d^4*e^2)*f^4 - (a*c^4*d^5*e + 5*a^2*c^3*d^3*e^3)*f^3*g - 3*(a^2*c^3*d^4*e^2 - a^3*c
^2*d^2*e^4)*f^2*g^2 + (5*a^3*c^2*d^3*e^3 + a^4*c*d*e^5)*f*g^3 - (2*a^4*c*d^2*e^4 + a^5*e^6)*g^4)*x^2 - (a^5*d*
e^5*g^4 - (2*a*c^4*d^5*e + a^2*c^3*d^3*e^3)*f^4 + (5*a^2*c^3*d^4*e^2 + 3*a^3*c^2*d^2*e^4)*f^3*g - 3*(a^3*c^2*d
^3*e^3 + a^4*c*d*e^5)*f^2*g^2 - (a^4*c*d^2*e^4 - a^5*e^6)*f*g^3)*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)^(5/2)/(g*x+f)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 169, normalized size = 0.87 \begin {gather*} -\frac {2 \left (c d x +a e \right ) \left (8 g^{2} x^{2} c^{2} d^{2}+12 a c d e \,g^{2} x +4 c^{2} d^{2} f g x +3 a^{2} e^{2} g^{2}+6 a c d e f g -f^{2} c^{2} d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{3 \sqrt {g x +f}\, \left (a^{3} e^{3} g^{3}-3 a^{2} c d \,e^{2} f \,g^{2}+3 a \,c^{2} d^{2} e \,f^{2} g -f^{3} c^{3} d^{3}\right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)/(g*x+f)^(3/2)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(5/2),x)

[Out]

-2/3*(c*d*x+a*e)*(8*c^2*d^2*g^2*x^2+12*a*c*d*e*g^2*x+4*c^2*d^2*f*g*x+3*a^2*e^2*g^2+6*a*c*d*e*f*g-c^2*d^2*f^2)*
(e*x+d)^(5/2)/(g*x+f)^(1/2)/(a^3*e^3*g^3-3*a^2*c*d*e^2*f*g^2+3*a*c^2*d^2*e*f^2*g-c^3*d^3*f^3)/(c*d*e*x^2+a*e^2
*x+c*d^2*x+a*d*e)^(5/2)

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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 {5}{2}}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{2}} {\left (g x + f\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(g*x+f)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(5/2)/((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)*(g*x + f)^(3/2)), x)

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mupad [B]  time = 5.28, size = 255, normalized size = 1.31 \begin {gather*} -\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {16\,g^2\,x^2\,\sqrt {d+e\,x}}{3\,e\,{\left (a\,e\,g-c\,d\,f\right )}^3}+\frac {\sqrt {d+e\,x}\,\left (6\,a^2\,e^2\,g^2+12\,a\,c\,d\,e\,f\,g-2\,c^2\,d^2\,f^2\right )}{3\,c^2\,d^2\,e\,{\left (a\,e\,g-c\,d\,f\right )}^3}+\frac {8\,g\,x\,\left (3\,a\,e\,g+c\,d\,f\right )\,\sqrt {d+e\,x}}{3\,c\,d\,e\,{\left (a\,e\,g-c\,d\,f\right )}^3}\right )}{x^3\,\sqrt {f+g\,x}+\frac {a^2\,e\,\sqrt {f+g\,x}}{c^2\,d}+\frac {x^2\,\sqrt {f+g\,x}\,\left (c\,d^2+2\,a\,e^2\right )}{c\,d\,e}+\frac {a\,x\,\sqrt {f+g\,x}\,\left (2\,c\,d^2+a\,e^2\right )}{c^2\,d^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(5/2)/((f + g*x)^(3/2)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)),x)

[Out]

-((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*((16*g^2*x^2*(d + e*x)^(1/2))/(3*e*(a*e*g - c*d*f)^3) + ((d +
e*x)^(1/2)*(6*a^2*e^2*g^2 - 2*c^2*d^2*f^2 + 12*a*c*d*e*f*g))/(3*c^2*d^2*e*(a*e*g - c*d*f)^3) + (8*g*x*(3*a*e*g
+ c*d*f)*(d + e*x)^(1/2))/(3*c*d*e*(a*e*g - c*d*f)^3)))/(x^3*(f + g*x)^(1/2) + (a^2*e*(f + g*x)^(1/2))/(c^2*d
) + (x^2*(f + g*x)^(1/2)*(2*a*e^2 + c*d^2))/(c*d*e) + (a*x*(f + g*x)^(1/2)*(a*e^2 + 2*c*d^2))/(c^2*d^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)**(5/2)/(g*x+f)**(3/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2),x)

[Out]

Timed out

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