\(\int \frac {(d x)^m}{(a+b x^n+c x^{2 n})^{3/2}} \, dx\) [605]

Optimal result

Integrand size = 24, antiderivative size = 163 \[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\frac {(d x)^{1+m} \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1+m}{n},\frac {3}{2},\frac {3}{2},\frac {1+m+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a d (1+m) \sqrt {a+b x^n+c x^{2 n}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1399, 524} \[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\frac {(d x)^{m+1} \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {m+1}{n},\frac {3}{2},\frac {3}{2},\frac {m+n+1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a d (m+1) \sqrt {a+b x^n+c x^{2 n}}} \]

[In]

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

[Out]

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

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 1399

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a +
b*x^n + c*x^(2*n))^FracPart[p]/((1 + 2*c*(x^n/(b + Rt[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^n/(b - Rt[b^
2 - 4*a*c, 2])))^FracPart[p])), Int[(d*x)^m*(1 + 2*c*(x^n/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2*c*(x^n/(b - Sqrt[
b^2 - 4*a*c])))^p, x], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(428\) vs. \(2(163)=326\).

Time = 1.45 (sec) , antiderivative size = 428, normalized size of antiderivative = 2.63 \[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\frac {x (d x)^m \left (\left (-4 a c (1+m-n)+b^2 (2+2 m-n)\right ) (1+m+n) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1+m}{n},\frac {1}{2},\frac {1}{2},\frac {1+m+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )-2 (1+m) \left ((1+m+n) \left (b^2-2 a c+b c x^n\right )-b c (1+m) x^n \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1+m+n}{n},\frac {1}{2},\frac {1}{2},\frac {1+m+2 n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )}{a \left (-b^2+4 a c\right ) (1+m) n (1+m+n) \sqrt {a+x^n \left (b+c x^n\right )}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (d x \right )^{m}}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{\frac {3}{2}}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int \frac {\left (d x\right )^{m}}{\left (a + b x^{n} + c x^{2 n}\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int \frac {{\left (d\,x\right )}^m}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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