\(\int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx\) [225]

Optimal result

Integrand size = 22, antiderivative size = 349 \[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=-\frac {(c+d x)^{2+n} \operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c-\sqrt {-\sqrt {-a}} d}\right )}{4 b^{3/4} \left (\sqrt [4]{b} c-\sqrt {-\sqrt {-a}} d\right ) (2+n)}-\frac {(c+d x)^{2+n} \operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c+\sqrt {-\sqrt {-a}} d}\right )}{4 b^{3/4} \left (\sqrt [4]{b} c+\sqrt {-\sqrt {-a}} d\right ) (2+n)}-\frac {(c+d x)^{2+n} \operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c-\sqrt [4]{-a} d}\right )}{4 b^{3/4} \left (\sqrt [4]{b} c-\sqrt [4]{-a} d\right ) (2+n)}-\frac {(c+d x)^{2+n} \operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c+\sqrt [4]{-a} d}\right )}{4 b^{3/4} \left (\sqrt [4]{b} c+\sqrt [4]{-a} d\right ) (2+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6857, 845, 70} \[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=-\frac {(c+d x)^{n+2} \operatorname {Hypergeometric2F1}\left (1,n+2,n+3,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c-\sqrt {-\sqrt {-a}} d}\right )}{4 b^{3/4} (n+2) \left (\sqrt [4]{b} c-\sqrt {-\sqrt {-a}} d\right )}-\frac {(c+d x)^{n+2} \operatorname {Hypergeometric2F1}\left (1,n+2,n+3,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c+\sqrt {-\sqrt {-a}} d}\right )}{4 b^{3/4} (n+2) \left (\sqrt {-\sqrt {-a}} d+\sqrt [4]{b} c\right )}-\frac {(c+d x)^{n+2} \operatorname {Hypergeometric2F1}\left (1,n+2,n+3,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c-\sqrt [4]{-a} d}\right )}{4 b^{3/4} (n+2) \left (\sqrt [4]{b} c-\sqrt [4]{-a} d\right )}-\frac {(c+d x)^{n+2} \operatorname {Hypergeometric2F1}\left (1,n+2,n+3,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c+\sqrt [4]{-a} d}\right )}{4 b^{3/4} (n+2) \left (\sqrt [4]{-a} d+\sqrt [4]{b} c\right )} \]

[In]

Int[(x^3*(c + d*x)^(1 + n))/(a + b*x^4),x]

[Out]

-1/4*((c + d*x)^(2 + n)*Hypergeometric2F1[1, 2 + n, 3 + n, (b^(1/4)*(c + d*x))/(b^(1/4)*c - Sqrt[-Sqrt[-a]]*d)
])/(b^(3/4)*(b^(1/4)*c - Sqrt[-Sqrt[-a]]*d)*(2 + n)) - ((c + d*x)^(2 + n)*Hypergeometric2F1[1, 2 + n, 3 + n, (
b^(1/4)*(c + d*x))/(b^(1/4)*c + Sqrt[-Sqrt[-a]]*d)])/(4*b^(3/4)*(b^(1/4)*c + Sqrt[-Sqrt[-a]]*d)*(2 + n)) - ((c
+ d*x)^(2 + n)*Hypergeometric2F1[1, 2 + n, 3 + n, (b^(1/4)*(c + d*x))/(b^(1/4)*c - (-a)^(1/4)*d)])/(4*b^(3/4)
*(b^(1/4)*c - (-a)^(1/4)*d)*(2 + n)) - ((c + d*x)^(2 + n)*Hypergeometric2F1[1, 2 + n, 3 + n, (b^(1/4)*(c + d*x
))/(b^(1/4)*c + (-a)^(1/4)*d)])/(4*b^(3/4)*(b^(1/4)*c + (-a)^(1/4)*d)*(2 + n))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 845

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m, (f + g*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !Ration
alQ[m]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {x (c+d x)^{1+n}}{2 \left (-\sqrt {-a} \sqrt {b}+b x^2\right )}+\frac {x (c+d x)^{1+n}}{2 \left (\sqrt {-a} \sqrt {b}+b x^2\right )}\right ) \, dx \\ & = \frac {1}{2} \int \frac {x (c+d x)^{1+n}}{-\sqrt {-a} \sqrt {b}+b x^2} \, dx+\frac {1}{2} \int \frac {x (c+d x)^{1+n}}{\sqrt {-a} \sqrt {b}+b x^2} \, dx \\ & = \frac {1}{2} \int \left (-\frac {(c+d x)^{1+n}}{2 b^{3/4} \left (\sqrt {-\sqrt {-a}}-\sqrt [4]{b} x\right )}+\frac {(c+d x)^{1+n}}{2 b^{3/4} \left (\sqrt {-\sqrt {-a}}+\sqrt [4]{b} x\right )}\right ) \, dx+\frac {1}{2} \int \left (-\frac {(c+d x)^{1+n}}{2 b^{3/4} \left (\sqrt [4]{-a}-\sqrt [4]{b} x\right )}+\frac {(c+d x)^{1+n}}{2 b^{3/4} \left (\sqrt [4]{-a}+\sqrt [4]{b} x\right )}\right ) \, dx \\ & = -\frac {\int \frac {(c+d x)^{1+n}}{\sqrt {-\sqrt {-a}}-\sqrt [4]{b} x} \, dx}{4 b^{3/4}}-\frac {\int \frac {(c+d x)^{1+n}}{\sqrt [4]{-a}-\sqrt [4]{b} x} \, dx}{4 b^{3/4}}+\frac {\int \frac {(c+d x)^{1+n}}{\sqrt {-\sqrt {-a}}+\sqrt [4]{b} x} \, dx}{4 b^{3/4}}+\frac {\int \frac {(c+d x)^{1+n}}{\sqrt [4]{-a}+\sqrt [4]{b} x} \, dx}{4 b^{3/4}} \\ & = -\frac {(c+d x)^{2+n} \, _2F_1\left (1,2+n;3+n;\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c-\sqrt {-\sqrt {-a}} d}\right )}{4 b^{3/4} \left (\sqrt [4]{b} c-\sqrt {-\sqrt {-a}} d\right ) (2+n)}-\frac {(c+d x)^{2+n} \, _2F_1\left (1,2+n;3+n;\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c+\sqrt {-\sqrt {-a}} d}\right )}{4 b^{3/4} \left (\sqrt [4]{b} c+\sqrt {-\sqrt {-a}} d\right ) (2+n)}-\frac {(c+d x)^{2+n} \, _2F_1\left (1,2+n;3+n;\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c-\sqrt [4]{-a} d}\right )}{4 b^{3/4} \left (\sqrt [4]{b} c-\sqrt [4]{-a} d\right ) (2+n)}-\frac {(c+d x)^{2+n} \, _2F_1\left (1,2+n;3+n;\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c+\sqrt [4]{-a} d}\right )}{4 b^{3/4} \left (\sqrt [4]{b} c+\sqrt [4]{-a} d\right ) (2+n)} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=\frac {(c+d x)^{2+n} \left (-\frac {\operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c-\sqrt [4]{-a} d}\right )}{\sqrt [4]{b} c-\sqrt [4]{-a} d}-\frac {\operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c-i \sqrt [4]{-a} d}\right )}{\sqrt [4]{b} c-i \sqrt [4]{-a} d}-\frac {\operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c+i \sqrt [4]{-a} d}\right )}{\sqrt [4]{b} c+i \sqrt [4]{-a} d}-\frac {\operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c+\sqrt [4]{-a} d}\right )}{\sqrt [4]{b} c+\sqrt [4]{-a} d}\right )}{4 b^{3/4} (2+n)} \]

[In]

Integrate[(x^3*(c + d*x)^(1 + n))/(a + b*x^4),x]

[Out]

((c + d*x)^(2 + n)*(-(Hypergeometric2F1[1, 2 + n, 3 + n, (b^(1/4)*(c + d*x))/(b^(1/4)*c - (-a)^(1/4)*d)]/(b^(1
/4)*c - (-a)^(1/4)*d)) - Hypergeometric2F1[1, 2 + n, 3 + n, (b^(1/4)*(c + d*x))/(b^(1/4)*c - I*(-a)^(1/4)*d)]/
(b^(1/4)*c - I*(-a)^(1/4)*d) - Hypergeometric2F1[1, 2 + n, 3 + n, (b^(1/4)*(c + d*x))/(b^(1/4)*c + I*(-a)^(1/4
)*d)]/(b^(1/4)*c + I*(-a)^(1/4)*d) - Hypergeometric2F1[1, 2 + n, 3 + n, (b^(1/4)*(c + d*x))/(b^(1/4)*c + (-a)^
(1/4)*d)]/(b^(1/4)*c + (-a)^(1/4)*d)))/(4*b^(3/4)*(2 + n))

Maple [F]

\[\int \frac {x^{3} \left (d x +c \right )^{1+n}}{b \,x^{4}+a}d x\]

[In]

int(x^3*(d*x+c)^(1+n)/(b*x^4+a),x)

[Out]

int(x^3*(d*x+c)^(1+n)/(b*x^4+a),x)

Fricas [F]

\[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=\int { \frac {{\left (d x + c\right )}^{n + 1} x^{3}}{b x^{4} + a} \,d x } \]

[In]

integrate(x^3*(d*x+c)^(1+n)/(b*x^4+a),x, algorithm="fricas")

[Out]

integral((d*x + c)^(n + 1)*x^3/(b*x^4 + a), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=\text {Timed out} \]

[In]

integrate(x**3*(d*x+c)**(1+n)/(b*x**4+a),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=\int { \frac {{\left (d x + c\right )}^{n + 1} x^{3}}{b x^{4} + a} \,d x } \]

[In]

integrate(x^3*(d*x+c)^(1+n)/(b*x^4+a),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(n + 1)*x^3/(b*x^4 + a), x)

Giac [F]

\[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=\int { \frac {{\left (d x + c\right )}^{n + 1} x^{3}}{b x^{4} + a} \,d x } \]

[In]

integrate(x^3*(d*x+c)^(1+n)/(b*x^4+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^(n + 1)*x^3/(b*x^4 + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=\int \frac {x^3\,{\left (c+d\,x\right )}^{n+1}}{b\,x^4+a} \,d x \]

[In]

int((x^3*(c + d*x)^(n + 1))/(a + b*x^4),x)

[Out]

int((x^3*(c + d*x)^(n + 1))/(a + b*x^4), x)