3.8.88 \(\int \frac {(d+e x)^{3/2}}{(f+g x) \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [788]

Optimal. Leaf size=139 \[ \frac {2 e \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c d g \sqrt {d+e x}}-\frac {2 (e f-d g) \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d f-a e g} \sqrt {d+e x}}\right )}{g^{3/2} \sqrt {c d f-a e g}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {894, 888, 211} \begin {gather*} \frac {2 e \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d g \sqrt {d+e x}}-\frac {2 (e f-d g) \text {ArcTan}\left (\frac {\sqrt {g} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d f-a e g}}\right )}{g^{3/2} \sqrt {c d f-a e g}} \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]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 888

Int[Sqrt[(d_) + (e_.)*(x_)]/(((f_.) + (g_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[
2*e^2, Subst[Int[1/(c*(e*f + d*g) - b*e*g + e^2*g*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; Fre
eQ[{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]

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]

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

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Mathematica [A]
time = 0.16, size = 140, normalized size = 1.01 \begin {gather*} \frac {2 \sqrt {d+e x} \left (e \sqrt {g} \sqrt {c d f-a e g} (a e+c d x)+c d (-e f+d g) \sqrt {a e+c d x} \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c d f-a e g}}\right )\right )}{c d g^{3/2} \sqrt {c d f-a e g} \sqrt {(a e+c d x) (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[d + e*x]*(e*Sqrt[g]*Sqrt[c*d*f - a*e*g]*(a*e + c*d*x) + c*d*(-(e*f) + d*g)*Sqrt[a*e + c*d*x]*ArcTan[(S
qrt[g]*Sqrt[a*e + c*d*x])/Sqrt[c*d*f - a*e*g]]))/(c*d*g^(3/2)*Sqrt[c*d*f - a*e*g]*Sqrt[(a*e + c*d*x)*(d + e*x)
])

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Maple [A]
time = 0.14, size = 153, normalized size = 1.10

method	result	size
default	\(-\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (\arctanh \left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -c d f \right ) g}}\right ) c \,d^{2} g -\arctanh \left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -c d f \right ) g}}\right ) c d e f -e \sqrt {c d x +a e}\, \sqrt {\left (a e g -c d f \right ) g}\right )}{\sqrt {e x +d}\, \sqrt {c d x +a e}\, c d g \sqrt {\left (a e g -c d f \right ) g}}\)	\(153\)

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

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Fricas [A]
time = 2.89, size = 520, normalized size = 3.74 \begin {gather*} \left [\frac {{\left (c d^{3} g - c d f x e^{2} + {\left (c d^{2} g x - c d^{2} f\right )} e\right )} \sqrt {-c d f g + a g^{2} e} \log \left (-\frac {c d^{2} g x - c d^{2} f + 2 \, a g x e^{2} + {\left (c d g x^{2} - c d f x + 2 \, a d g\right )} e + 2 \, \sqrt {-c d f g + a g^{2} e} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d}}{d g x + d f + {\left (g x^{2} + f x\right )} e}\right ) + 2 \, {\left (c d f g e - a g^{2} e^{2}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d}}{c^{2} d^{3} f g^{2} - a c d g^{3} x e^{2} + {\left (c^{2} d^{2} f g^{2} x - a c d^{2} g^{3}\right )} e}, -\frac {2 \, {\left ({\left (c d^{3} g - c d f x e^{2} + {\left (c d^{2} g x - c d^{2} f\right )} e\right )} \sqrt {c d f g - a g^{2} e} \arctan \left (\frac {\sqrt {c d f g - a g^{2} e} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d}}{c d^{2} g x + a g x e^{2} + {\left (c d g x^{2} + a d g\right )} e}\right ) - {\left (c d f g e - a g^{2} e^{2}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d}\right )}}{c^{2} d^{3} f g^{2} - a c d g^{3} x e^{2} + {\left (c^{2} d^{2} f g^{2} x - a c d^{2} g^{3}\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)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="fricas")

[Out]

[((c*d^3*g - c*d*f*x*e^2 + (c*d^2*g*x - c*d^2*f)*e)*sqrt(-c*d*f*g + a*g^2*e)*log(-(c*d^2*g*x - c*d^2*f + 2*a*g
*x*e^2 + (c*d*g*x^2 - c*d*f*x + 2*a*d*g)*e + 2*sqrt(-c*d*f*g + a*g^2*e)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*
d)*e)*sqrt(x*e + d))/(d*g*x + d*f + (g*x^2 + f*x)*e)) + 2*(c*d*f*g*e - a*g^2*e^2)*sqrt(c*d^2*x + a*x*e^2 + (c*
d*x^2 + a*d)*e)*sqrt(x*e + d))/(c^2*d^3*f*g^2 - a*c*d*g^3*x*e^2 + (c^2*d^2*f*g^2*x - a*c*d^2*g^3)*e), -2*((c*d
^3*g - c*d*f*x*e^2 + (c*d^2*g*x - c*d^2*f)*e)*sqrt(c*d*f*g - a*g^2*e)*arctan(sqrt(c*d*f*g - a*g^2*e)*sqrt(c*d^
2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(x*e + d)/(c*d^2*g*x + a*g*x*e^2 + (c*d*g*x^2 + a*d*g)*e)) - (c*d*f*g*e
- a*g^2*e^2)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(x*e + d))/(c^2*d^3*f*g^2 - a*c*d*g^3*x*e^2 + (c
^2*d^2*f*g^2*x - a*c*d^2*g^3)*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 {\left (d + e x\right ) \left (a e + c d x\right )} \left (f + g 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)/(sqrt((d + e*x)*(a*e + c*d*x))*(f + g*x)), 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)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),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 {{\left (d+e\,x\right )}^{3/2}}{\left (f+g\,x\right )\,\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((d + e*x)^(3/2)/((f + g*x)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)),x)

[Out]

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

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