3.8.82 \(\int \frac {(d+e x)^{3/2} (f+g x)^n}{\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [782]

Optimal. Leaf size=213 \[ \frac {2 e (f+g x)^{1+n} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c d g (3+2 n) \sqrt {d+e x}}+\frac {\left (2 a e^2 g (1+n)+c d (e f-d g (3+2 n))\right ) (a e+c d x) \sqrt {d+e x} (f+g x)^{1+n} \, _2F_1\left (1,\frac {3}{2}+n;2+n;\frac {c d (f+g x)}{c d f-a e g}\right )}{c d g (c d f-a e g) (1+n) (3+2 n) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \]

[Out]

(2*a*e^2*g*(1+n)+c*d*(e*f-d*g*(3+2*n)))*(c*d*x+a*e)*(g*x+f)^(1+n)*hypergeom([1, 3/2+n],[2+n],c*d*(g*x+f)/(-a*e
*g+c*d*f))*(e*x+d)^(1/2)/c/d/g/(-a*e*g+c*d*f)/(1+n)/(3+2*n)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+2*e*(g*x+f
)^(1+n)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c/d/g/(3+2*n)/(e*x+d)^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 222, normalized size of antiderivative = 1.04, 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 = {894, 905, 72, 71} \begin {gather*} \frac {2 e (f+g x)^{n+1} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d g (2 n+3) \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} (f+g x)^n (a e+c d x) \left (2 a e^2 g (n+1)+c d (e f-d g (2 n+3))\right ) \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};-\frac {g (a e+c d x)}{c d f-a e g}\right )}{c^2 d^2 g (2 n+3) \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)^n)/Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2],x]

[Out]

(2*e*(f + g*x)^(1 + n)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(c*d*g*(3 + 2*n)*Sqrt[d + e*x]) - (2*(2*a*
e^2*g*(1 + n) + c*d*(e*f - d*g*(3 + 2*n)))*(a*e + c*d*x)*Sqrt[d + e*x]*(f + g*x)^n*Hypergeometric2F1[1/2, -n,
3/2, -((g*(a*e + c*d*x))/(c*d*f - a*e*g))])/(c^2*d^2*g*(3 + 2*n)*((c*d*(f + g*x))/(c*d*f - a*e*g))^n*Sqrt[a*d*
e + (c*d^2 + a*e^2)*x + c*d*e*x^2])

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 894

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 - 2)*(f + g*x)^(n + 1)*((a + b*x + c*x^2)^(p + 1)/(c*g*(n + p + 2))), x] - Dist[(b*e*g*(
n + 1) + c*e*f*(p + 1) - c*d*g*(2*n + p + 3))/(c*g*(n + p + 2)), Int[(d + e*x)^(m - 1)*(f + g*x)^n*(a + b*x +
c*x^2)^p, x], 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] && Eq
Q[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p - 1, 0] &&  !LtQ[n, -1] && IntegerQ[2*p]

Rule 905

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^{3/2} (f+g x)^n}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx &=\frac {2 e (f+g x)^{1+n} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c d g (3+2 n) \sqrt {d+e x}}-\frac {\left (2 a e^2 g (1+n)+c d (e f-d g (3+2 n))\right ) \int \frac {\sqrt {d+e x} (f+g x)^n}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{c d g (3+2 n)}\\ &=\frac {2 e (f+g x)^{1+n} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c d g (3+2 n) \sqrt {d+e x}}-\frac {\left (\left (2 a e^2 g (1+n)+c d (e f-d g (3+2 n))\right ) \sqrt {a e+c d x} \sqrt {d+e x}\right ) \int \frac {(f+g x)^n}{\sqrt {a e+c d x}} \, dx}{c d g (3+2 n) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac {2 e (f+g x)^{1+n} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c d g (3+2 n) \sqrt {d+e x}}-\frac {\left (\left (2 a e^2 g (1+n)+c d (e f-d g (3+2 n))\right ) \sqrt {a e+c d x} \sqrt {d+e x} (f+g x)^n \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n}\right ) \int \frac {\left (\frac {c d f}{c d f-a e g}+\frac {c d g x}{c d f-a e g}\right )^n}{\sqrt {a e+c d x}} \, dx}{c d g (3+2 n) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac {2 e (f+g x)^{1+n} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c d g (3+2 n) \sqrt {d+e x}}-\frac {2 \left (2 a e^2 g (1+n)+c d (e f-d g (3+2 n))\right ) (a e+c d x) \sqrt {d+e x} (f+g x)^n \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};-\frac {g (a e+c d x)}{c d f-a e g}\right )}{c^2 d^2 g (3+2 n) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 145, normalized size = 0.68 \begin {gather*} \frac {\sqrt {(a e+c d x) (d+e x)} (f+g x)^n \left (c d e (f+g x)+\left (-2 a e^2 g (1+n)+c d (-e f+d g (3+2 n))\right ) \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};\frac {g (a e+c d x)}{-c d f+a e g}\right )\right )}{c^2 d^2 g \left (\frac {3}{2}+n\right ) \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a*e + c*d*x)*(d + e*x)]*(f + g*x)^n*(c*d*e*(f + g*x) + ((-2*a*e^2*g*(1 + n) + c*d*(-(e*f) + d*g*(3 + 2*
n)))*Hypergeometric2F1[1/2, -n, 3/2, (g*(a*e + c*d*x))/(-(c*d*f) + a*e*g)])/((c*d*(f + g*x))/(c*d*f - a*e*g))^
n))/(c^2*d^2*g*(3/2 + n)*Sqrt[d + e*x])

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{\frac {3}{2}} \left (g x +f \right )^{n}}{\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
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)**n/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (f+g\,x\right )}^n\,{\left (d+e\,x\right )}^{3/2}}{\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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