\(\int (d x)^m (a+b x^n+c x^{2 n})^{3/2} \, dx\) [602]

Optimal result

Integrand size = 24, antiderivative size = 161 \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^{3/2} \, dx=\frac {a (d x)^{1+m} \sqrt {a+b x^n+c x^{2 n}} \operatorname {AppellF1}\left (\frac {1+m}{n},-\frac {3}{2},-\frac {3}{2},\frac {1+m+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{d (1+m) \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}}} \]

[Out]

a*(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)))*(a+b*x^n+c*x^(2*n))^(1/2)/d/(1+m)/(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)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 161, 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} \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^{3/2} \, dx=\frac {a (d x)^{m+1} \sqrt {a+b x^n+c x^{2 n}} \operatorname {AppellF1}\left (\frac {m+1}{n},-\frac {3}{2},-\frac {3}{2},\frac {m+n+1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1) \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1}} \]

[In]

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

[Out]

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

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 (a \sqrt {a+b x^n+c x^{2 n}}\right ) \int (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}{\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}}}} \\ & = \frac {a (d x)^{1+m} \sqrt {a+b x^n+c x^{2 n}} 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 )}{d (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}}}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(545\) vs. \(2(161)=322\).

Time = 2.56 (sec) , antiderivative size = 545, normalized size of antiderivative = 3.39 \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^{3/2} \, dx=\frac {x (d x)^m \left (-6 a n^2 (1+m+n) \left (b^2 (1+m)-4 a c (1+m+2 n)\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1+m}{n},\frac {1}{2},\frac {1}{2},\frac {1+m+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+(1+m) \left (2 (1+m+n) \left (3 b^2 n^2+4 a c \left (1+m^2+6 n+8 n^2+m (2+6 n)\right )+2 b c \left (2+4 m+2 m^2+9 n+9 m n+7 n^2\right ) x^n+4 c^2 \left (1+2 m+m^2+3 n+3 m n+2 n^2\right ) x^{2 n}\right ) \left (a+x^n \left (b+c x^n\right )\right )-3 b n^2 \left (b^2 (2+2 m+n)-4 a c (2+2 m+3 n)\right ) x^n \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1+m+n}{n},\frac {1}{2},\frac {1}{2},\frac {1+m+2 n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )}{8 c (1+m) (1+m+n)^2 (1+m+2 n) (1+m+3 n) \sqrt {a+x^n \left (b+c x^n\right )}} \]

[In]

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

[Out]

(x*(d*x)^m*(-6*a*n^2*(1 + m + n)*(b^2*(1 + m) - 4*a*c*(1 + m + 2*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, 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])] + (1 + m)*(2*(1
+ m + n)*(3*b^2*n^2 + 4*a*c*(1 + m^2 + 6*n + 8*n^2 + m*(2 + 6*n)) + 2*b*c*(2 + 4*m + 2*m^2 + 9*n + 9*m*n + 7*n
^2)*x^n + 4*c^2*(1 + 2*m + m^2 + 3*n + 3*m*n + 2*n^2)*x^(2*n))*(a + x^n*(b + c*x^n)) - 3*b*n^2*(b^2*(2 + 2*m +
n) - 4*a*c*(2 + 2*m + 3*n))*x^n*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqr
t[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])])))/(8*c*(1 + m)*(1 + m + n)^2*(1 + m + 2*n)*(1
+ m + 3*n)*Sqrt[a + x^n*(b + c*x^n)])

Maple [F]

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

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

[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 (ha
s polynomial part)

Sympy [F]

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

[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]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)