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

Optimal. Leaf size=294 \[ \frac {a d x \sqrt {a+b x^n+c x^{2 n}} F_1\left (\frac {1}{n};-\frac {3}{2},-\frac {3}{2};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1}}+\frac {a e x^{n+1} \sqrt {a+b x^n+c x^{2 n}} F_1\left (1+\frac {1}{n};-\frac {3}{2},-\frac {3}{2};2+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{(n+1) \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1}} \]

[Out]

a*e*x^(1+n)*AppellF1(1+1/n,-3/2,-3/2,2+1/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)/(1+n)/(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*x*AppellF1(1/n,-3/2,-3/2,1+1/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)/(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 = 0.35, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1432, 1348, 429, 1385, 510} \[ \frac {a d x \sqrt {a+b x^n+c x^{2 n}} F_1\left (\frac {1}{n};-\frac {3}{2},-\frac {3}{2};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1}}+\frac {a e x^{n+1} \sqrt {a+b x^n+c x^{2 n}} F_1\left (1+\frac {1}{n};-\frac {3}{2},-\frac {3}{2};2+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{(n+1) \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*e*x^(1 + n)*Sqrt[a + b*x^n + c*x^(2*n)]*AppellF1[1 + n^(-1), -3/2, -3/2, 2 + n^(-1), (-2*c*x^n)/(b - Sqrt[b
^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((1 + n)*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])]) + (a*d*x*Sqrt[a + b*x^n + c*x^(2*n)]*AppellF1[n^(-1), -3/2, -3/2, 1 + n
^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(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 429

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

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 1348

Int[((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]))^F
racPart[p]), Int[(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, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

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]

Rule 1432

Int[((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegran
d[(d + e*x^n)*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4
*a*c, 0]

Rubi steps

\begin {align*} \int \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^{3/2} \, dx &=\int \left (d \left (a+b x^n+c x^{2 n}\right )^{3/2}+e x^n \left (a+b x^n+c x^{2 n}\right )^{3/2}\right ) \, dx\\ &=d \int \left (a+b x^n+c x^{2 n}\right )^{3/2} \, dx+e \int x^n \left (a+b x^n+c x^{2 n}\right )^{3/2} \, dx\\ &=\frac {\left (a d \sqrt {a+b x^n+c x^{2 n}}\right ) \int \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 {\left (a e \sqrt {a+b x^n+c x^{2 n}}\right ) \int x^n \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 e x^{1+n} \sqrt {a+b x^n+c x^{2 n}} F_1\left (1+\frac {1}{n};-\frac {3}{2},-\frac {3}{2};2+\frac {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 )}{(1+n) \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 \sqrt {a+b x^n+c x^{2 n}} F_1\left (\frac {1}{n};-\frac {3}{2},-\frac {3}{2};1+\frac {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 )}{\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 = 4.53, size = 690, normalized size = 2.35 \[ \frac {x \left (3 n^2 x^n \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^n}{\sqrt {b^2-4 a c}+b}} F_1\left (1+\frac {1}{n};\frac {1}{2},\frac {1}{2};2+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{\sqrt {b^2-4 a c}-b}\right ) \left (16 a^2 c^2 e \left (3 n^2+4 n+1\right )-4 a b^2 c e \left (6 n^2+14 n+5\right )+8 a b c^2 d \left (12 n^2+11 n+2\right )+b^4 e \left (3 n^2+8 n+4\right )-2 b^3 c d \left (4 n^2+9 n+2\right )\right )+2 (n+1) \left (3 a n^2 \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^n}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {1}{n};\frac {1}{2},\frac {1}{2};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{\sqrt {b^2-4 a c}-b}\right ) \left (-4 a b c e (5 n+2)+8 a c^2 d \left (8 n^2+6 n+1\right )+b^3 e (3 n+2)-2 b^2 c d (4 n+1)\right )+\left (a+x^n \left (b+c x^n\right )\right ) \left (4 a c \left (3 b e (5 n+2) n^2+2 c \left (d (2 n+1) (4 n+1)^2+e \left (15 n^3+23 n^2+9 n+1\right ) x^n\right )\right )-3 b^3 e n^2 (3 n+2)+6 b^2 c n^2 \left (4 d n+d+e (n+1) x^n\right )+4 b c^2 (n+1) x^n \left (d \left (28 n^2+15 n+2\right )+e \left (18 n^2+13 n+2\right ) x^n\right )+8 c^3 \left (2 n^2+3 n+1\right ) x^{2 n} \left (4 d n+d+e (3 n+1) x^n\right )\right )\right )\right )}{16 c^2 (n+1)^2 (2 n+1) (3 n+1) (4 n+1) \sqrt {a+x^n \left (b+c x^n\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(3*n^2*(16*a^2*c^2*e*(1 + 4*n + 3*n^2) + b^4*e*(4 + 8*n + 3*n^2) - 2*b^3*c*d*(2 + 9*n + 4*n^2) - 4*a*b^2*c*
e*(5 + 14*n + 6*n^2) + 8*a*b*c^2*d*(2 + 11*n + 12*n^2))*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 + n^(-1), 1/2, 1/2, 2
+ n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])] + 2*(1 + n)*((a + x^n*(b +
c*x^n))*(-3*b^3*e*n^2*(2 + 3*n) + 6*b^2*c*n^2*(d + 4*d*n + e*(1 + n)*x^n) + 8*c^3*(1 + 3*n + 2*n^2)*x^(2*n)*(d
+ 4*d*n + e*(1 + 3*n)*x^n) + 4*b*c^2*(1 + n)*x^n*(d*(2 + 15*n + 28*n^2) + e*(2 + 13*n + 18*n^2)*x^n) + 4*a*c*
(3*b*e*n^2*(2 + 5*n) + 2*c*(d*(1 + 2*n)*(1 + 4*n)^2 + e*(1 + 9*n + 23*n^2 + 15*n^3)*x^n))) + 3*a*n^2*(b^3*e*(2
+ 3*n) - 2*b^2*c*d*(1 + 4*n) - 4*a*b*c*e*(2 + 5*n) + 8*a*c^2*d*(1 + 6*n + 8*n^2))*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[
n^(-1), 1/2, 1/2, 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])])))/(16*c
^2*(1 + n)^2*(1 + 2*n)*(1 + 3*n)*(1 + 4*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+e*x^n)*(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 \[ \int {\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {3}{2}} {\left (e x^{n} + d\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x^n)*(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)*(e*x^n + d), x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \left (e \,x^{n}+d \right ) \left (b \,x^{n}+c \,x^{2 n}+a \right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x^n)*(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)*(e*x^n + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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