3.6.84 \(\int \frac {(c x)^m (d+e x^n+f x^{2 n}+g x^{3 n})}{\sqrt {a+b x^n}} \, dx\) [584]

Optimal. Leaf size=305 \[ \frac {d (c x)^{1+m} \sqrt {1+\frac {b x^n}{a}} \, _2F_1\left (\frac {1}{2},\frac {1+m}{n};\frac {1+m+n}{n};-\frac {b x^n}{a}\right )}{c (1+m) \sqrt {a+b x^n}}+\frac {e x^{1+n} (c x)^m \sqrt {1+\frac {b x^n}{a}} \, _2F_1\left (\frac {1}{2},\frac {1+m+n}{n};\frac {1+m+2 n}{n};-\frac {b x^n}{a}\right )}{(1+m+n) \sqrt {a+b x^n}}+\frac {f x^{1+2 n} (c x)^m \sqrt {1+\frac {b x^n}{a}} \, _2F_1\left (\frac {1}{2},\frac {1+m+2 n}{n};\frac {1+m+3 n}{n};-\frac {b x^n}{a}\right )}{(1+m+2 n) \sqrt {a+b x^n}}+\frac {g x^{1+3 n} (c x)^m \sqrt {1+\frac {b x^n}{a}} \, _2F_1\left (\frac {1}{2},\frac {1+m+3 n}{n};\frac {1+m+4 n}{n};-\frac {b x^n}{a}\right )}{(1+m+3 n) \sqrt {a+b x^n}} \]

[Out]

d*(c*x)^(1+m)*hypergeom([1/2, (1+m)/n],[(1+m+n)/n],-b*x^n/a)*(1+b*x^n/a)^(1/2)/c/(1+m)/(a+b*x^n)^(1/2)+e*x^(1+
n)*(c*x)^m*hypergeom([1/2, (1+m+n)/n],[(1+m+2*n)/n],-b*x^n/a)*(1+b*x^n/a)^(1/2)/(1+m+n)/(a+b*x^n)^(1/2)+f*x^(1
+2*n)*(c*x)^m*hypergeom([1/2, (1+m+2*n)/n],[(1+m+3*n)/n],-b*x^n/a)*(1+b*x^n/a)^(1/2)/(1+m+2*n)/(a+b*x^n)^(1/2)
+g*x^(1+3*n)*(c*x)^m*hypergeom([1/2, (1+m+3*n)/n],[(1+m+4*n)/n],-b*x^n/a)*(1+b*x^n/a)^(1/2)/(1+m+3*n)/(a+b*x^n
)^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 305, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1858, 372, 371, 20} \begin {gather*} \frac {d (c x)^{m+1} \sqrt {\frac {b x^n}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{c (m+1) \sqrt {a+b x^n}}+\frac {e x^{n+1} (c x)^m \sqrt {\frac {b x^n}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {m+n+1}{n};\frac {m+2 n+1}{n};-\frac {b x^n}{a}\right )}{(m+n+1) \sqrt {a+b x^n}}+\frac {f x^{2 n+1} (c x)^m \sqrt {\frac {b x^n}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {m+2 n+1}{n};\frac {m+3 n+1}{n};-\frac {b x^n}{a}\right )}{(m+2 n+1) \sqrt {a+b x^n}}+\frac {g x^{3 n+1} (c x)^m \sqrt {\frac {b x^n}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {m+3 n+1}{n};\frac {m+4 n+1}{n};-\frac {b x^n}{a}\right )}{(m+3 n+1) \sqrt {a+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(c*x)^(1 + m)*Sqrt[1 + (b*x^n)/a]*Hypergeometric2F1[1/2, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(c*(1 + m
)*Sqrt[a + b*x^n]) + (e*x^(1 + n)*(c*x)^m*Sqrt[1 + (b*x^n)/a]*Hypergeometric2F1[1/2, (1 + m + n)/n, (1 + m + 2
*n)/n, -((b*x^n)/a)])/((1 + m + n)*Sqrt[a + b*x^n]) + (f*x^(1 + 2*n)*(c*x)^m*Sqrt[1 + (b*x^n)/a]*Hypergeometri
c2F1[1/2, (1 + m + 2*n)/n, (1 + m + 3*n)/n, -((b*x^n)/a)])/((1 + m + 2*n)*Sqrt[a + b*x^n]) + (g*x^(1 + 3*n)*(c
*x)^m*Sqrt[1 + (b*x^n)/a]*Hypergeometric2F1[1/2, (1 + m + 3*n)/n, (1 + m + 4*n)/n, -((b*x^n)/a)])/((1 + m + 3*
n)*Sqrt[a + b*x^n])

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[b^IntPart[n]*((b*v)^FracPart[n]/(a^IntPart[n]
*(a*v)^FracPart[n])), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 1858

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(c x)^m \left (d+e x^n+f x^{2 n}+g x^{3 n}\right )}{\sqrt {a+b x^n}} \, dx &=\int \left (\frac {d (c x)^m}{\sqrt {a+b x^n}}+\frac {e x^n (c x)^m}{\sqrt {a+b x^n}}+\frac {f x^{2 n} (c x)^m}{\sqrt {a+b x^n}}+\frac {g x^{3 n} (c x)^m}{\sqrt {a+b x^n}}\right ) \, dx\\ &=d \int \frac {(c x)^m}{\sqrt {a+b x^n}} \, dx+e \int \frac {x^n (c x)^m}{\sqrt {a+b x^n}} \, dx+f \int \frac {x^{2 n} (c x)^m}{\sqrt {a+b x^n}} \, dx+g \int \frac {x^{3 n} (c x)^m}{\sqrt {a+b x^n}} \, dx\\ &=\left (e x^{-m} (c x)^m\right ) \int \frac {x^{m+n}}{\sqrt {a+b x^n}} \, dx+\left (f x^{-m} (c x)^m\right ) \int \frac {x^{m+2 n}}{\sqrt {a+b x^n}} \, dx+\left (g x^{-m} (c x)^m\right ) \int \frac {x^{m+3 n}}{\sqrt {a+b x^n}} \, dx+\frac {\left (d \sqrt {1+\frac {b x^n}{a}}\right ) \int \frac {(c x)^m}{\sqrt {1+\frac {b x^n}{a}}} \, dx}{\sqrt {a+b x^n}}\\ &=\frac {d (c x)^{1+m} \sqrt {1+\frac {b x^n}{a}} \, _2F_1\left (\frac {1}{2},\frac {1+m}{n};\frac {1+m+n}{n};-\frac {b x^n}{a}\right )}{c (1+m) \sqrt {a+b x^n}}+\frac {\left (e x^{-m} (c x)^m \sqrt {1+\frac {b x^n}{a}}\right ) \int \frac {x^{m+n}}{\sqrt {1+\frac {b x^n}{a}}} \, dx}{\sqrt {a+b x^n}}+\frac {\left (f x^{-m} (c x)^m \sqrt {1+\frac {b x^n}{a}}\right ) \int \frac {x^{m+2 n}}{\sqrt {1+\frac {b x^n}{a}}} \, dx}{\sqrt {a+b x^n}}+\frac {\left (g x^{-m} (c x)^m \sqrt {1+\frac {b x^n}{a}}\right ) \int \frac {x^{m+3 n}}{\sqrt {1+\frac {b x^n}{a}}} \, dx}{\sqrt {a+b x^n}}\\ &=\frac {d (c x)^{1+m} \sqrt {1+\frac {b x^n}{a}} \, _2F_1\left (\frac {1}{2},\frac {1+m}{n};\frac {1+m+n}{n};-\frac {b x^n}{a}\right )}{c (1+m) \sqrt {a+b x^n}}+\frac {e x^{1+n} (c x)^m \sqrt {1+\frac {b x^n}{a}} \, _2F_1\left (\frac {1}{2},\frac {1+m+n}{n};\frac {1+m+2 n}{n};-\frac {b x^n}{a}\right )}{(1+m+n) \sqrt {a+b x^n}}+\frac {f x^{1+2 n} (c x)^m \sqrt {1+\frac {b x^n}{a}} \, _2F_1\left (\frac {1}{2},\frac {1+m+2 n}{n};\frac {1+m+3 n}{n};-\frac {b x^n}{a}\right )}{(1+m+2 n) \sqrt {a+b x^n}}+\frac {g x^{1+3 n} (c x)^m \sqrt {1+\frac {b x^n}{a}} \, _2F_1\left (\frac {1}{2},\frac {1+m+3 n}{n};\frac {1+m+4 n}{n};-\frac {b x^n}{a}\right )}{(1+m+3 n) \sqrt {a+b x^n}}\\ \end {align*}

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Mathematica [A]
time = 1.57, size = 399, normalized size = 1.31 \begin {gather*} \frac {x (c x)^m \left (2 (1+m) \left (a+b x^n\right ) \left (4 a^2 g \left (1+m^2+3 n+2 n^2+m (2+3 n)\right )-2 a b \left (f \left (2+2 m^2+7 n+5 n^2+m (4+7 n)\right )+g \left (2+2 m^2+5 n+2 n^2+m (4+5 n)\right ) x^n\right )+b^2 \left (e \left (4+4 m^2+16 n+15 n^2+8 m (1+2 n)\right )+(2+2 m+n) x^n \left (f (2+2 m+5 n)+g (2+2 m+3 n) x^n\right )\right )\right )+\left (-2 a b^2 e (1+m) \left (4+4 m^2+16 n+15 n^2+8 m (1+2 n)\right )-8 a^3 g (1+m) \left (1+m^2+3 n+2 n^2+m (2+3 n)\right )+4 a^2 b f (1+m) \left (2+2 m^2+7 n+5 n^2+m (4+7 n)\right )+b^3 d \left (8+8 m^3+36 n+46 n^2+15 n^3+12 m^2 (2+3 n)+m \left (24+72 n+46 n^2\right )\right )\right ) \sqrt {1+\frac {b x^n}{a}} \, _2F_1\left (\frac {1}{2},\frac {1+m}{n};\frac {1+m+n}{n};-\frac {b x^n}{a}\right )\right )}{b^3 (1+m) (2+2 m+n) (2+2 m+3 n) (2+2 m+5 n) \sqrt {a+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(c*x)^m*(2*(1 + m)*(a + b*x^n)*(4*a^2*g*(1 + m^2 + 3*n + 2*n^2 + m*(2 + 3*n)) - 2*a*b*(f*(2 + 2*m^2 + 7*n +
5*n^2 + m*(4 + 7*n)) + g*(2 + 2*m^2 + 5*n + 2*n^2 + m*(4 + 5*n))*x^n) + b^2*(e*(4 + 4*m^2 + 16*n + 15*n^2 + 8
*m*(1 + 2*n)) + (2 + 2*m + n)*x^n*(f*(2 + 2*m + 5*n) + g*(2 + 2*m + 3*n)*x^n))) + (-2*a*b^2*e*(1 + m)*(4 + 4*m
^2 + 16*n + 15*n^2 + 8*m*(1 + 2*n)) - 8*a^3*g*(1 + m)*(1 + m^2 + 3*n + 2*n^2 + m*(2 + 3*n)) + 4*a^2*b*f*(1 + m
)*(2 + 2*m^2 + 7*n + 5*n^2 + m*(4 + 7*n)) + b^3*d*(8 + 8*m^3 + 36*n + 46*n^2 + 15*n^3 + 12*m^2*(2 + 3*n) + m*(
24 + 72*n + 46*n^2)))*Sqrt[1 + (b*x^n)/a]*Hypergeometric2F1[1/2, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)]))/(b^
3*(1 + m)*(2 + 2*m + n)*(2 + 2*m + 3*n)*(2 + 2*m + 5*n)*Sqrt[a + b*x^n])

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (c x \right )^{m} \left (d +e \,x^{n}+f \,x^{2 n}+g \,x^{3 n}\right )}{\sqrt {a +b \,x^{n}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^m*(d+e*x^n+f*x^(2*n)+g*x^(3*n))/(a+b*x^n)^(1/2),x)

[Out]

int((c*x)^m*(d+e*x^n+f*x^(2*n)+g*x^(3*n))/(a+b*x^n)^(1/2),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((c*x)^m*(d+e*x^n+f*x^(2*n)+g*x^(3*n))/(a+b*x^n)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^m*(d+e*x^n+f*x^(2*n)+g*x^(3*n))/(a+b*x^n)^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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Sympy [C] Result contains complex when optimal does not.
time = 26.53, size = 274, normalized size = 0.90 \begin {gather*} \frac {c^{m} d x x^{m} \Gamma \left (\frac {m}{n} + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{n} + \frac {1}{n} \\ \frac {m}{n} + 1 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{\sqrt {a} n \Gamma \left (\frac {m}{n} + 1 + \frac {1}{n}\right )} + \frac {c^{m} e x x^{m} x^{n} \Gamma \left (\frac {m}{n} + 1 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{n} + 1 + \frac {1}{n} \\ \frac {m}{n} + 2 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{\sqrt {a} n \Gamma \left (\frac {m}{n} + 2 + \frac {1}{n}\right )} + \frac {c^{m} f x x^{m} x^{2 n} \Gamma \left (\frac {m}{n} + 2 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{n} + 2 + \frac {1}{n} \\ \frac {m}{n} + 3 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{\sqrt {a} n \Gamma \left (\frac {m}{n} + 3 + \frac {1}{n}\right )} + \frac {c^{m} g x x^{m} x^{3 n} \Gamma \left (\frac {m}{n} + 3 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{n} + 3 + \frac {1}{n} \\ \frac {m}{n} + 4 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{\sqrt {a} n \Gamma \left (\frac {m}{n} + 4 + \frac {1}{n}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**m*(d+e*x**n+f*x**(2*n)+g*x**(3*n))/(a+b*x**n)**(1/2),x)

[Out]

c**m*d*x*x**m*gamma(m/n + 1/n)*hyper((1/2, m/n + 1/n), (m/n + 1 + 1/n,), b*x**n*exp_polar(I*pi)/a)/(sqrt(a)*n*
gamma(m/n + 1 + 1/n)) + c**m*e*x*x**m*x**n*gamma(m/n + 1 + 1/n)*hyper((1/2, m/n + 1 + 1/n), (m/n + 2 + 1/n,),
b*x**n*exp_polar(I*pi)/a)/(sqrt(a)*n*gamma(m/n + 2 + 1/n)) + c**m*f*x*x**m*x**(2*n)*gamma(m/n + 2 + 1/n)*hyper
((1/2, m/n + 2 + 1/n), (m/n + 3 + 1/n,), b*x**n*exp_polar(I*pi)/a)/(sqrt(a)*n*gamma(m/n + 3 + 1/n)) + c**m*g*x
*x**m*x**(3*n)*gamma(m/n + 3 + 1/n)*hyper((1/2, m/n + 3 + 1/n), (m/n + 4 + 1/n,), b*x**n*exp_polar(I*pi)/a)/(s
qrt(a)*n*gamma(m/n + 4 + 1/n))

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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((c*x)^m*(d+e*x^n+f*x^(2*n)+g*x^(3*n))/(a+b*x^n)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x\right )}^m\,\left (d+e\,x^n+f\,x^{2\,n}+g\,x^{3\,n}\right )}{\sqrt {a+b\,x^n}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c*x)^m*(d + e*x^n + f*x^(2*n) + g*x^(3*n)))/(a + b*x^n)^(1/2),x)

[Out]

int(((c*x)^m*(d + e*x^n + f*x^(2*n) + g*x^(3*n)))/(a + b*x^n)^(1/2), x)

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