3.8.22 \(\int \frac {(d+e x)^{3/2} \sqrt {f+g x}}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\) [722]

Optimal. Leaf size=161 \[ -\frac {2 \sqrt {d+e x} \sqrt {f+g x}}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {2 \sqrt {g} \sqrt {a e+c d x} \sqrt {d+e x} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{c^{3/2} d^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {880, 905, 65, 223, 212} \begin {gather*} \frac {2 \sqrt {g} \sqrt {d+e x} \sqrt {a e+c d x} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{c^{3/2} d^{3/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 \sqrt {d+e x} \sqrt {f+g x}}{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)*Sqrt[f + g*x])/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2),x]

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 880

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 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} \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}} \, dx &=-\frac {2 \sqrt {d+e x} \sqrt {f+g x}}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {g \int \frac {\sqrt {d+e x}}{\sqrt {f+g x} \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} \sqrt {f+g x}}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {\left (g \sqrt {a e+c d x} \sqrt {d+e x}\right ) \int \frac {1}{\sqrt {a e+c d x} \sqrt {f+g x}} \, dx}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=-\frac {2 \sqrt {d+e x} \sqrt {f+g x}}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {\left (2 g \sqrt {a e+c d x} \sqrt {d+e x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {f-\frac {a e g}{c d}+\frac {g x^2}{c d}}} \, dx,x,\sqrt {a e+c d x}\right )}{c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=-\frac {2 \sqrt {d+e x} \sqrt {f+g x}}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {\left (2 g \sqrt {a e+c d x} \sqrt {d+e x}\right ) \text {Subst}\left (\int \frac {1}{1-\frac {g x^2}{c d}} \, dx,x,\frac {\sqrt {a e+c d x}}{\sqrt {f+g x}}\right )}{c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=-\frac {2 \sqrt {d+e x} \sqrt {f+g x}}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {2 \sqrt {g} \sqrt {a e+c d x} \sqrt {d+e x} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{c^{3/2} d^{3/2} \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 = 0.12, size = 117, normalized size = 0.73 \begin {gather*} -\frac {2 \sqrt {d+e x} \left (\sqrt {c} \sqrt {d} \sqrt {f+g x}-\sqrt {g} \sqrt {a e+c d x} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {f+g x}}{\sqrt {g} \sqrt {a e+c d x}}\right )\right )}{c^{3/2} d^{3/2} \sqrt {(a e+c d x) (d+e x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.14, size = 200, normalized size = 1.24

method	result	size
default	\(\frac {\sqrt {g x +f}\, \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (\ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}}{2 \sqrt {d g c}}\right ) c d g x +\ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}}{2 \sqrt {d g c}}\right ) a e g -2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}\right )}{\sqrt {d g c}\, \left (c d x +a e \right ) \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, d c \sqrt {e x +d}}\)	\(200\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(g*x+f)^(1/2)*((c*d*x+a*e)*(e*x+d))^(1/2)*(ln(1/2*(2*c*d*g*x+a*e*g+c*d*f+2*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)
^(1/2))/(d*g*c)^(1/2))*c*d*g*x+ln(1/2*(2*c*d*g*x+a*e*g+c*d*f+2*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2))/(d*g
*c)^(1/2))*a*e*g-2*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2))/(d*g*c)^(1/2)/(c*d*x+a*e)/((g*x+f)*(c*d*x+a*e))^
(1/2)/d/c/(e*x+d)^(1/2)

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

[Out]

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

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Fricas [A]
time = 4.83, size = 577, normalized size = 3.58 \begin {gather*} \left [\frac {{\left (c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e\right )} \sqrt {\frac {g}{c d}} \log \left (-\frac {8 \, c^{2} d^{3} g^{2} x^{2} + 8 \, c^{2} d^{3} f g x + c^{2} d^{3} f^{2} + a^{2} g^{2} x e^{3} + 4 \, {\left (2 \, c^{2} d^{2} g x + c^{2} d^{2} f + a c d g e\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {g x + f} \sqrt {x e + d} \sqrt {\frac {g}{c d}} + {\left (8 \, a c d g^{2} x^{2} + 6 \, a c d f g x + a^{2} d g^{2}\right )} e^{2} + {\left (8 \, c^{2} d^{2} g^{2} x^{3} + 8 \, c^{2} d^{2} f g x^{2} + 6 \, a c d^{2} f g + {\left (c^{2} d^{2} f^{2} + 8 \, a c d^{2} g^{2}\right )} x\right )} e}{x e + d}\right ) - 4 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {g x + f} \sqrt {x e + d}}{2 \, {\left (c^{2} d^{3} x + a c d x e^{2} + {\left (c^{2} d^{2} x^{2} + a c d^{2}\right )} e\right )}}, -\frac {{\left (c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e\right )} \sqrt {-\frac {g}{c d}} \arctan \left (\frac {2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {g x + f} \sqrt {x e + d} c d \sqrt {-\frac {g}{c d}}}{2 \, c d^{2} g x + c d^{2} f + a g x e^{2} + {\left (2 \, c d g x^{2} + c d f x + a d g\right )} e}\right ) + 2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {g x + f} \sqrt {x e + d}}{c^{2} d^{3} x + a c d x e^{2} + {\left (c^{2} d^{2} x^{2} + a c d^{2}\right )} e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(g/(c*d))*log(-(8*c^2*d^3*g^2*x^2 + 8*c^2*d^3*f*g*x + c^2*d^
3*f^2 + a^2*g^2*x*e^3 + 4*(2*c^2*d^2*g*x + c^2*d^2*f + a*c*d*g*e)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*
sqrt(g*x + f)*sqrt(x*e + d)*sqrt(g/(c*d)) + (8*a*c*d*g^2*x^2 + 6*a*c*d*f*g*x + a^2*d*g^2)*e^2 + (8*c^2*d^2*g^2
*x^3 + 8*c^2*d^2*f*g*x^2 + 6*a*c*d^2*f*g + (c^2*d^2*f^2 + 8*a*c*d^2*g^2)*x)*e)/(x*e + d)) - 4*sqrt(c*d^2*x + a
*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(g*x + f)*sqrt(x*e + d))/(c^2*d^3*x + a*c*d*x*e^2 + (c^2*d^2*x^2 + a*c*d^2)*e)
, -((c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(-g/(c*d))*arctan(2*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e
)*sqrt(g*x + f)*sqrt(x*e + d)*c*d*sqrt(-g/(c*d))/(2*c*d^2*g*x + c*d^2*f + a*g*x*e^2 + (2*c*d*g*x^2 + c*d*f*x +
a*d*g)*e)) + 2*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(g*x + f)*sqrt(x*e + d))/(c^2*d^3*x + a*c*d*x*
e^2 + (c^2*d^2*x^2 + a*c*d^2)*e)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [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)^(1/2)/(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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {f+g\,x}\,{\left (d+e\,x\right )}^{3/2}}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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