3.12.10 \(\int (d x)^m (a+b x^2+c x^4)^{3/2} \, dx\) [1110]

Optimal. Leaf size=158 \[ \frac {a (d x)^{1+m} \sqrt {a+b x^2+c x^4} F_1\left (\frac {1+m}{2};-\frac {3}{2},-\frac {3}{2};\frac {3+m}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (1+m) \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}} \]

[Out]

a*(d*x)^(1+m)*AppellF1(1/2+1/2*m,-3/2,-3/2,3/2+1/2*m,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^2/(b+(-4*a*c+b^2)^
(1/2)))*(c*x^4+b*x^2+a)^(1/2)/d/(1+m)/(1+2*c*x^2/(b-(-4*a*c+b^2)^(1/2)))^(1/2)/(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2
)))^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1155, 524} \begin {gather*} \frac {a (d x)^{m+1} \sqrt {a+b x^2+c x^4} F_1\left (\frac {m+1}{2};-\frac {3}{2},-\frac {3}{2};\frac {m+3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1) \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(d*x)^(1 + m)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[(1 + m)/2, -3/2, -3/2, (3 + m)/2, (-2*c*x^2)/(b - Sqrt[b^2 -
4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(d*(1 + m)*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 +
(2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 1155

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^2 +
c*x^4)^FracPart[p]/((1 + 2*c*(x^2/(b + Rt[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^2/(b - Rt[b^2 - 4*a*c, 2
])))^FracPart[p])), Int[(d*x)^m*(1 + 2*c*(x^2/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2*c*(x^2/(b - Sqrt[b^2 - 4*a*c]
)))^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x]

Rubi steps

\begin {align*} \int (d x)^m \left (a+b x^2+c x^4\right )^{3/2} \, dx &=\frac {\left (a \sqrt {a+b x^2+c x^4}\right ) \int (d x)^m \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{3/2} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{3/2} \, dx}{\sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}}\\ &=\frac {a (d x)^{1+m} \sqrt {a+b x^2+c x^4} F_1\left (\frac {1+m}{2};-\frac {3}{2},-\frac {3}{2};\frac {3+m}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (1+m) \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(357\) vs. \(2(158)=316\).
time = 0.69, size = 357, normalized size = 2.26 \begin {gather*} \frac {x (d x)^m \sqrt {a+b x^2+c x^4} \left (a \left (15+8 m+m^2\right ) F_1\left (\frac {1+m}{2};-\frac {1}{2},-\frac {1}{2};\frac {3+m}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+(1+m) x^2 \left (b (5+m) F_1\left (\frac {3+m}{2};-\frac {1}{2},-\frac {1}{2};\frac {5+m}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+c (3+m) x^2 F_1\left (\frac {5+m}{2};-\frac {1}{2},-\frac {1}{2};\frac {7+m}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )}{(1+m) (3+m) (5+m) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(d*x)^m*Sqrt[a + b*x^2 + c*x^4]*(a*(15 + 8*m + m^2)*AppellF1[(1 + m)/2, -1/2, -1/2, (3 + m)/2, (-2*c*x^2)/(
b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + (1 + m)*x^2*(b*(5 + m)*AppellF1[(3 + m)/2, -1/2,
-1/2, (5 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + c*(3 + m)*x^2*Appe
llF1[(5 + m)/2, -1/2, -1/2, (7 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]
)))/((1 + m)*(3 + m)*(5 + m)*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^
2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((d*x)^m*(c*x^4+b*x^2+a)^(3/2),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((d*x)^m*(c*x^4+b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d x\right )^{m} \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d\,x\right )}^m\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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