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

Optimal. Leaf size=161 \[ \frac {a (d x)^{m+1} \sqrt {a+b x^n+c x^{2 n}} F_1\left (\frac {m+1}{n};-\frac {3}{2},-\frac {3}{2};\frac {m+n+1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1) \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1}} \]

[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)

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Rubi [A]  time = 0.18, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1385, 510} \[ \frac {a (d x)^{m+1} \sqrt {a+b x^n+c x^{2 n}} F_1\left (\frac {m+1}{n};-\frac {3}{2},-\frac {3}{2};\frac {m+n+1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1) \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1}} \]

Antiderivative was successfully verified.

[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 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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 1385

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

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Mathematica [B]  time = 3.21, size = 618, normalized size = 3.84 \[ \frac {x (d x)^m \left ((m+1) \left (2 (m+n+1) \left (4 a^2 c \left (m^2+m (6 n+2)+8 n^2+6 n+1\right )+a \left (3 b^2 n^2+2 b c \left (4 m^2+m (21 n+8)+23 n^2+21 n+4\right ) x^n+4 c^2 \left (2 m^2+m (9 n+4)+10 n^2+9 n+2\right ) x^{2 n}\right )+x^n \left (b+c x^n\right ) \left (3 b^2 n^2+2 b c \left (2 m^2+m (9 n+4)+7 n^2+9 n+2\right ) x^n+4 c^2 \left (m^2+m (3 n+2)+2 n^2+3 n+1\right ) x^{2 n}\right )\right )-3 b n^2 x^n \left (b^2 (2 m+n+2)-4 a c (2 m+3 n+2)\right ) \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^n}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {m+n+1}{n};\frac {1}{2},\frac {1}{2};\frac {m+2 n+1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{\sqrt {b^2-4 a c}-b}\right )\right )-6 a n^2 (m+n+1) \left (b^2 (m+1)-4 a c (m+2 n+1)\right ) \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^n}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {m+1}{n};\frac {1}{2},\frac {1}{2};\frac {m+n+1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{\sqrt {b^2-4 a c}-b}\right )\right )}{8 c (m+1) (m+n+1)^2 (m+2 n+1) (m+3 n+1) \sqrt {a+x^n \left (b+c x^n\right )}} \]

Warning: Unable to verify antiderivative.

[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)*(4*a^2*c*(1 + m^2 + 6*n + 8*n^2 + m*(2 + 6*n)) + x^n*(b + c*x^n)*(3*b^2*n^2 + 2*b*c*(2 + 2*m^2 + 9*n
+ 7*n^2 + m*(4 + 9*n))*x^n + 4*c^2*(1 + m^2 + 3*n + 2*n^2 + m*(2 + 3*n))*x^(2*n)) + a*(3*b^2*n^2 + 2*b*c*(4 +
4*m^2 + 21*n + 23*n^2 + m*(8 + 21*n))*x^n + 4*c^2*(2 + 2*m^2 + 9*n + 10*n^2 + m*(4 + 9*n))*x^(2*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 + 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])])))/(8*c*(1 + m)*(1 + m + n)^2*
(1 + m + 2*n)*(1 + m + 3*n)*Sqrt[a + x^n*(b + c*x^n)])

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.26, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{n}+c \,x^{2 n}+a \right )^{\frac {3}{2}} \left (d x \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (d\,x\right )}^m\,{\left (a+b\,x^n+c\,x^{2\,n}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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