3.8.86 \(\int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [786]

Optimal. Leaf size=209 \[ -\frac {4 \left (c d^2-a e^2\right ) \left (4 a e^2 g-c d (5 e f-d g)\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^3 d^3 e \sqrt {d+e x}}-\frac {2 \left (4 a e^2 g-c d (5 e f-d g)\right ) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^2 d^2 e}+\frac {2 g (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d e} \]

[Out]

2/5*g*(e*x+d)^(3/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c/d/e-4/15*(-a*e^2+c*d^2)*(4*a*e^2*g-c*d*(-d*g+5*e
*f))*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c^3/d^3/e/(e*x+d)^(1/2)-2/15*(4*a*e^2*g-c*d*(-d*g+5*e*f))*(e*x+d)
^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c^2/d^2/e

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Rubi [A]
time = 0.13, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {808, 670, 662} \begin {gather*} -\frac {4 \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (4 a e^2 g-c d (5 e f-d g)\right )}{15 c^3 d^3 e \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (4 a e^2 g-c d (5 e f-d g)\right )}{15 c^2 d^2 e}+\frac {2 g (d+e x)^{3/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 c d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 662

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 670

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*(m + 2*p + 1))), x] + Dist[Simplify[m + p]*((2*c*d - b*e)/(c*(m + 2*p + 1))), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rule 808

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 {(d+e x)^{3/2} (f+g x)}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx &=\frac {2 g (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d e}+\frac {1}{5} \left (5 f-\frac {d g}{e}-\frac {4 a e g}{c d}\right ) \int \frac {(d+e x)^{3/2}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx\\ &=-\frac {2 \left (4 a e^2 g-c d (5 e f-d g)\right ) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^2 d^2 e}+\frac {2 g (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d e}+\frac {\left (2 \left (d^2-\frac {a e^2}{c}\right ) \left (5 f-\frac {d g}{e}-\frac {4 a e g}{c d}\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}{15 d}\\ &=-\frac {4 \left (c d^2-a e^2\right ) \left (4 a e^2 g-c d (5 e f-d g)\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^3 d^3 e \sqrt {d+e x}}-\frac {2 \left (4 a e^2 g-c d (5 e f-d g)\right ) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^2 d^2 e}+\frac {2 g (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d e}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 96, normalized size = 0.46 \begin {gather*} \frac {2 \sqrt {(a e+c d x) (d+e x)} \left (8 a^2 e^3 g-2 a c d e (5 e f+5 d g+2 e g x)+c^2 d^2 (5 d (3 f+g x)+e x (5 f+3 g x))\right )}{15 c^3 d^3 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.13, size = 113, normalized size = 0.54

method	result	size
default	\(\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (3 e g \,x^{2} c^{2} d^{2}-4 a c d \,e^{2} g x +5 c^{2} d^{3} g x +5 c^{2} d^{2} e f x +8 a^{2} e^{3} g -10 a c \,d^{2} e g -10 a c d \,e^{2} f +15 d^{3} f \,c^{2}\right )}{15 \sqrt {e x +d}\, c^{3} d^{3}}\)	\(113\)
gosper	\(\frac {2 \left (c d x +a e \right ) \left (3 e g \,x^{2} c^{2} d^{2}-4 a c d \,e^{2} g x +5 c^{2} d^{3} g x +5 c^{2} d^{2} e f x +8 a^{2} e^{3} g -10 a c \,d^{2} e g -10 a c d \,e^{2} f +15 d^{3} f \,c^{2}\right ) \sqrt {e x +d}}{15 c^{3} d^{3} \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)*(g*x+f)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.31, size = 168, normalized size = 0.80 \begin {gather*} \frac {2 \, {\left (c^{2} d^{2} x^{2} e + 3 \, a c d^{2} e - 2 \, a^{2} e^{3} + {\left (3 \, c^{2} d^{3} - a c d e^{2}\right )} x\right )} f}{3 \, \sqrt {c d x + a e} c^{2} d^{2}} + \frac {2 \, {\left (3 \, c^{3} d^{3} x^{3} e - 10 \, a^{2} c d^{2} e^{2} + 8 \, a^{3} e^{4} + {\left (5 \, c^{3} d^{4} - a c^{2} d^{2} e^{2}\right )} x^{2} - {\left (5 \, a c^{2} d^{3} e - 4 \, a^{2} c d e^{3}\right )} x\right )} g}{15 \, \sqrt {c d x + a e} c^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(g*x+f)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.93, size = 137, normalized size = 0.66 \begin {gather*} \frac {2 \, {\left (5 \, c^{2} d^{3} g x + 15 \, c^{2} d^{3} f + 8 \, a^{2} g e^{3} - 2 \, {\left (2 \, a c d g x + 5 \, a c d f\right )} e^{2} + {\left (3 \, c^{2} d^{2} g x^{2} + 5 \, c^{2} d^{2} f x - 10 \, a c d^{2} g\right )} e\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d}}{15 \, {\left (c^{3} d^{3} x e + c^{3} d^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(g*x+f)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(3/2)*(g*x+f)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2),x)

[Out]

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

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Giac [A]
time = 6.25, size = 342, normalized size = 1.64 \begin {gather*} \frac {2 \, {\left (c^{2} d^{3} f - a c d^{2} g e - a c d f e^{2} + a^{2} g e^{3}\right )} \sqrt {{\left (x e + d\right )} c d e - c d^{2} e + a e^{3}} e^{\left (-1\right )}}{c^{3} d^{3}} + \frac {4 \, {\left (\sqrt {-c d^{2} e + a e^{3}} c^{2} d^{4} g - 5 \, \sqrt {-c d^{2} e + a e^{3}} c^{2} d^{3} f e + 3 \, \sqrt {-c d^{2} e + a e^{3}} a c d^{2} g e^{2} + 5 \, \sqrt {-c d^{2} e + a e^{3}} a c d f e^{3} - 4 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} g e^{4}\right )} e^{\left (-2\right )}}{15 \, c^{3} d^{3}} + \frac {2 \, {\left (5 \, {\left ({\left (x e + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c d^{2} g e + 5 \, {\left ({\left (x e + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c d f e^{2} - 10 \, {\left ({\left (x e + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a g e^{3} + 3 \, {\left ({\left (x e + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}} g\right )} e^{\left (-4\right )}}{15 \, c^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(g*x+f)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="giac")

[Out]

2*(c^2*d^3*f - a*c*d^2*g*e - a*c*d*f*e^2 + a^2*g*e^3)*sqrt((x*e + d)*c*d*e - c*d^2*e + a*e^3)*e^(-1)/(c^3*d^3)
+ 4/15*(sqrt(-c*d^2*e + a*e^3)*c^2*d^4*g - 5*sqrt(-c*d^2*e + a*e^3)*c^2*d^3*f*e + 3*sqrt(-c*d^2*e + a*e^3)*a*
c*d^2*g*e^2 + 5*sqrt(-c*d^2*e + a*e^3)*a*c*d*f*e^3 - 4*sqrt(-c*d^2*e + a*e^3)*a^2*g*e^4)*e^(-2)/(c^3*d^3) + 2/
15*(5*((x*e + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*c*d^2*g*e + 5*((x*e + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*c*d*f*
e^2 - 10*((x*e + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a*g*e^3 + 3*((x*e + d)*c*d*e - c*d^2*e + a*e^3)^(5/2)*g)*e^
(-4)/(c^3*d^3)

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Mupad [B]
time = 3.48, size = 152, normalized size = 0.73 \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 (16\,g\,a^2\,e^3-20\,g\,a\,c\,d^2\,e-20\,f\,a\,c\,d\,e^2+30\,f\,c^2\,d^3\right )}{15\,c^3\,d^3\,e}+\frac {2\,g\,x^2\,\sqrt {d+e\,x}}{5\,c\,d}+\frac {2\,x\,\sqrt {d+e\,x}\,\left (5\,c\,g\,d^2+5\,c\,f\,d\,e-4\,a\,g\,e^2\right )}{15\,c^2\,d^2\,e}\right )}{x+\frac {d}{e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)*(d + e*x)^(3/2))/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2),x)

[Out]

((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*(((d + e*x)^(1/2)*(16*a^2*e^3*g + 30*c^2*d^3*f - 20*a*c*d*e^2*f
- 20*a*c*d^2*e*g))/(15*c^3*d^3*e) + (2*g*x^2*(d + e*x)^(1/2))/(5*c*d) + (2*x*(d + e*x)^(1/2)*(5*c*d^2*g - 4*a
*e^2*g + 5*c*d*e*f))/(15*c^2*d^2*e)))/(x + d/e)

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