3.2548 \(\int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=239 \[ \frac {2 \left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \sqrt [4]{\frac {\left (\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}{\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}} \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {1}{2};-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt {b^2-4 a c}\right )}\right )}{\sqrt {d+e x} \sqrt [4]{a+b x+c x^2} \left (e \sqrt {b^2-4 a c}-b e+2 c d\right )} \]

[Out]

2*hypergeom([-1/2, 1/4],[1/2],-4*c*(e*x+d)*(-4*a*c+b^2)^(1/2)/(b+2*c*x-(-4*a*c+b^2)^(1/2))/(2*c*d-e*(b+(-4*a*c
+b^2)^(1/2))))*(b+2*c*x-(-4*a*c+b^2)^(1/2))*((2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))*(b+2*c*x+(-4*a*c+b^2)^(1/2))/(b+
2*c*x-(-4*a*c+b^2)^(1/2))/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(1/4)/(c*x^2+b*x+a)^(1/4)/(2*c*d-b*e+e*(-4*a*c+b^2
)^(1/2))/(e*x+d)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {726} \[ \frac {2 \left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \sqrt [4]{\frac {\left (\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}{\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}} \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {1}{2};-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt {b^2-4 a c}\right )}\right )}{\sqrt {d+e x} \sqrt [4]{a+b x+c x^2} \left (e \sqrt {b^2-4 a c}-b e+2 c d\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^(3/2)*(a + b*x + c*x^2)^(1/4)),x]

[Out]

(2*(b - Sqrt[b^2 - 4*a*c] + 2*c*x)*(((2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((2*
c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(b - Sqrt[b^2 - 4*a*c] + 2*c*x)))^(1/4)*Hypergeometric2F1[-1/2, 1/4, 1/2, (-4
*c*Sqrt[b^2 - 4*a*c]*(d + e*x))/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))])/((2*c*
d - b*e + Sqrt[b^2 - 4*a*c]*e)*Sqrt[d + e*x]*(a + b*x + c*x^2)^(1/4))

Rule 726

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b - Rt[b^2 - 4*a*
c, 2] + 2*c*x)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Hypergeometric2F1[m + 1, -p, m + 2, (-4*c*Rt[b^2 - 4*a*c,
2]*(d + e*x))/((2*c*d - b*e - e*Rt[b^2 - 4*a*c, 2])*(b - Rt[b^2 - 4*a*c, 2] + 2*c*x))])/((m + 1)*(2*c*d - b*e
+ e*Rt[b^2 - 4*a*c, 2])*(((2*c*d - b*e + e*Rt[b^2 - 4*a*c, 2])*(b + Rt[b^2 - 4*a*c, 2] + 2*c*x))/((2*c*d - b*
e - e*Rt[b^2 - 4*a*c, 2])*(b - Rt[b^2 - 4*a*c, 2] + 2*c*x)))^p), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[
b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0
]

Rubi steps

\begin {align*} \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx &=\frac {2 \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \sqrt [4]{\frac {\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}} \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {1}{2};-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )}{\left (2 c d-b e+\sqrt {b^2-4 a c} e\right ) \sqrt {d+e x} \sqrt [4]{a+b x+c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 231, normalized size = 0.97 \[ -\frac {2 \left (\sqrt {b^2-4 a c}+b+2 c x\right ) \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {1}{2};-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (\left (b+\sqrt {b^2-4 a c}\right ) e-2 c d\right ) \left (-b-2 c x+\sqrt {b^2-4 a c}\right )}\right )}{\sqrt {d+e x} \sqrt [4]{a+x (b+c x)} \left (e \left (\sqrt {b^2-4 a c}+b\right )-2 c d\right ) \left (\frac {\left (\sqrt {b^2-4 a c}+b+2 c x\right ) \left (e \left (\sqrt {b^2-4 a c}-b\right )+2 c d\right )}{\left (\sqrt {b^2-4 a c}-b-2 c x\right ) \left (e \left (\sqrt {b^2-4 a c}+b\right )-2 c d\right )}\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^(3/2)*(a + b*x + c*x^2)^(1/4)),x]

[Out]

(-2*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)*Hypergeometric2F1[-1/2, 1/4, 1/2, (-4*c*Sqrt[b^2 - 4*a*c]*(d + e*x))/((-2*
c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))])/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(((
2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(-b
+ Sqrt[b^2 - 4*a*c] - 2*c*x)))^(3/4)*Sqrt[d + e*x]*(a + x*(b + c*x))^(1/4))

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fricas [F]  time = 1.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{2} + b x + a\right )}^{\frac {3}{4}} \sqrt {e x + d}}{c e^{2} x^{4} + {\left (2 \, c d e + b e^{2}\right )} x^{3} + a d^{2} + {\left (c d^{2} + 2 \, b d e + a e^{2}\right )} x^{2} + {\left (b d^{2} + 2 \, a d e\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(3/2)/(c*x^2+b*x+a)^(1/4),x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^(3/4)*sqrt(e*x + d)/(c*e^2*x^4 + (2*c*d*e + b*e^2)*x^3 + a*d^2 + (c*d^2 + 2*b*d*e +
a*e^2)*x^2 + (b*d^2 + 2*a*d*e)*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} {\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(3/2)/(c*x^2+b*x+a)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((c*x^2 + b*x + a)^(1/4)*(e*x + d)^(3/2)), x)

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maple [F]  time = 2.24, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e x +d \right )^{\frac {3}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(3/2)/(c*x^2+b*x+a)^(1/4),x)

[Out]

int(1/(e*x+d)^(3/2)/(c*x^2+b*x+a)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} {\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(3/2)/(c*x^2+b*x+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((c*x^2 + b*x + a)^(1/4)*(e*x + d)^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (d+e\,x\right )}^{3/2}\,{\left (c\,x^2+b\,x+a\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d + e*x)^(3/2)*(a + b*x + c*x^2)^(1/4)),x)

[Out]

int(1/((d + e*x)^(3/2)*(a + b*x + c*x^2)^(1/4)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d + e x\right )^{\frac {3}{2}} \sqrt [4]{a + b x + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**(3/2)/(c*x**2+b*x+a)**(1/4),x)

[Out]

Integral(1/((d + e*x)**(3/2)*(a + b*x + c*x**2)**(1/4)), x)

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