Integrand size = 15, antiderivative size = 154 \[ \int \frac {x^{13/4}}{(a+b x)^{7/4}} \, dx=-\frac {4 a^2 x^{5/4}}{3 b^3 (a+b x)^{3/4}}+\frac {13 a^2 \sqrt [4]{x} \sqrt [4]{a+b x}}{2 b^4}-\frac {19 a x^{5/4} \sqrt [4]{a+b x}}{15 b^3}+\frac {2 x^{9/4} \sqrt [4]{a+b x}}{5 b^2}+\frac {13 a^{5/2} \left (\frac {b x}{a+b x}\right )^{3/4} (a+b x)^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {a}}{\sqrt {a+b x}}\right ),2\right )}{2 b^5 x^{3/4}} \] Output:
-4/3*a^2*x^(5/4)/b^3/(b*x+a)^(3/4)+13/2*a^2*x^(1/4)*(b*x+a)^(1/4)/b^4-19/1 5*a*x^(5/4)*(b*x+a)^(1/4)/b^3+2/5*x^(9/4)*(b*x+a)^(1/4)/b^2+13/2*a^(5/2)*( b*x/(b*x+a))^(3/4)*(b*x+a)^(3/4)*InverseJacobiAM(1/2*arcsin(a^(1/2)/(b*x+a )^(1/2)),2^(1/2))/b^5/x^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.32 \[ \int \frac {x^{13/4}}{(a+b x)^{7/4}} \, dx=\frac {4 x^{17/4} \left (1+\frac {b x}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {17}{4},\frac {21}{4},-\frac {b x}{a}\right )}{17 a (a+b x)^{3/4}} \] Input:
Integrate[x^(13/4)/(a + b*x)^(7/4),x]
Output:
(4*x^(17/4)*(1 + (b*x)/a)^(3/4)*Hypergeometric2F1[7/4, 17/4, 21/4, -((b*x) /a)])/(17*a*(a + b*x)^(3/4))
Time = 0.27 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {57, 60, 60, 60, 73, 768, 858, 807, 229}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^{13/4}}{(a+b x)^{7/4}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {13 \int \frac {x^{9/4}}{(a+b x)^{3/4}}dx}{3 b}-\frac {4 x^{13/4}}{3 b (a+b x)^{3/4}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {13 \left (\frac {2 x^{9/4} \sqrt [4]{a+b x}}{5 b}-\frac {9 a \int \frac {x^{5/4}}{(a+b x)^{3/4}}dx}{10 b}\right )}{3 b}-\frac {4 x^{13/4}}{3 b (a+b x)^{3/4}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {13 \left (\frac {2 x^{9/4} \sqrt [4]{a+b x}}{5 b}-\frac {9 a \left (\frac {2 x^{5/4} \sqrt [4]{a+b x}}{3 b}-\frac {5 a \int \frac {\sqrt [4]{x}}{(a+b x)^{3/4}}dx}{6 b}\right )}{10 b}\right )}{3 b}-\frac {4 x^{13/4}}{3 b (a+b x)^{3/4}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {13 \left (\frac {2 x^{9/4} \sqrt [4]{a+b x}}{5 b}-\frac {9 a \left (\frac {2 x^{5/4} \sqrt [4]{a+b x}}{3 b}-\frac {5 a \left (\frac {2 \sqrt [4]{x} \sqrt [4]{a+b x}}{b}-\frac {a \int \frac {1}{x^{3/4} (a+b x)^{3/4}}dx}{2 b}\right )}{6 b}\right )}{10 b}\right )}{3 b}-\frac {4 x^{13/4}}{3 b (a+b x)^{3/4}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {13 \left (\frac {2 x^{9/4} \sqrt [4]{a+b x}}{5 b}-\frac {9 a \left (\frac {2 x^{5/4} \sqrt [4]{a+b x}}{3 b}-\frac {5 a \left (\frac {2 \sqrt [4]{x} \sqrt [4]{a+b x}}{b}-\frac {2 a \int \frac {1}{(a+b x)^{3/4}}d\sqrt [4]{x}}{b}\right )}{6 b}\right )}{10 b}\right )}{3 b}-\frac {4 x^{13/4}}{3 b (a+b x)^{3/4}}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle \frac {13 \left (\frac {2 x^{9/4} \sqrt [4]{a+b x}}{5 b}-\frac {9 a \left (\frac {2 x^{5/4} \sqrt [4]{a+b x}}{3 b}-\frac {5 a \left (\frac {2 \sqrt [4]{x} \sqrt [4]{a+b x}}{b}-\frac {2 a x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x}+1\right )^{3/4} x^{3/4}}d\sqrt [4]{x}}{b (a+b x)^{3/4}}\right )}{6 b}\right )}{10 b}\right )}{3 b}-\frac {4 x^{13/4}}{3 b (a+b x)^{3/4}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {13 \left (\frac {2 x^{9/4} \sqrt [4]{a+b x}}{5 b}-\frac {9 a \left (\frac {2 x^{5/4} \sqrt [4]{a+b x}}{3 b}-\frac {5 a \left (\frac {2 a x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\sqrt [4]{x} \left (\frac {a x}{b}+1\right )^{3/4}}d\frac {1}{\sqrt [4]{x}}}{b (a+b x)^{3/4}}+\frac {2 \sqrt [4]{x} \sqrt [4]{a+b x}}{b}\right )}{6 b}\right )}{10 b}\right )}{3 b}-\frac {4 x^{13/4}}{3 b (a+b x)^{3/4}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {13 \left (\frac {2 x^{9/4} \sqrt [4]{a+b x}}{5 b}-\frac {9 a \left (\frac {2 x^{5/4} \sqrt [4]{a+b x}}{3 b}-\frac {5 a \left (\frac {a x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\left (\frac {\sqrt {x} a}{b}+1\right )^{3/4}}d\sqrt {x}}{b (a+b x)^{3/4}}+\frac {2 \sqrt [4]{x} \sqrt [4]{a+b x}}{b}\right )}{6 b}\right )}{10 b}\right )}{3 b}-\frac {4 x^{13/4}}{3 b (a+b x)^{3/4}}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \frac {13 \left (\frac {2 x^{9/4} \sqrt [4]{a+b x}}{5 b}-\frac {9 a \left (\frac {2 x^{5/4} \sqrt [4]{a+b x}}{3 b}-\frac {5 a \left (\frac {2 \sqrt {a} x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right ),2\right )}{\sqrt {b} (a+b x)^{3/4}}+\frac {2 \sqrt [4]{x} \sqrt [4]{a+b x}}{b}\right )}{6 b}\right )}{10 b}\right )}{3 b}-\frac {4 x^{13/4}}{3 b (a+b x)^{3/4}}\) |
Input:
Int[x^(13/4)/(a + b*x)^(7/4),x]
Output:
(-4*x^(13/4))/(3*b*(a + b*x)^(3/4)) + (13*((2*x^(9/4)*(a + b*x)^(1/4))/(5* b) - (9*a*((2*x^(5/4)*(a + b*x)^(1/4))/(3*b) - (5*a*((2*x^(1/4)*(a + b*x)^ (1/4))/b + (2*Sqrt[a]*(1 + a/(b*x))^(3/4)*x^(3/4)*EllipticF[ArcTan[(Sqrt[a ]*Sqrt[x])/Sqrt[b]]/2, 2])/(Sqrt[b]*(a + b*x)^(3/4))))/(6*b)))/(10*b)))/(3 *b)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) )*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 /4)/(a + b*x^4)^(3/4)) Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ [{a, b}, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
\[\int \frac {x^{\frac {13}{4}}}{\left (b x +a \right )^{\frac {7}{4}}}d x\]
Input:
int(x^(13/4)/(b*x+a)^(7/4),x)
Output:
int(x^(13/4)/(b*x+a)^(7/4),x)
\[ \int \frac {x^{13/4}}{(a+b x)^{7/4}} \, dx=\int { \frac {x^{\frac {13}{4}}}{{\left (b x + a\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(x^(13/4)/(b*x+a)^(7/4),x, algorithm="fricas")
Output:
integral((b*x + a)^(1/4)*x^(13/4)/(b^2*x^2 + 2*a*b*x + a^2), x)
Result contains complex when optimal does not.
Time = 159.90 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.23 \[ \int \frac {x^{13/4}}{(a+b x)^{7/4}} \, dx=\frac {x^{\frac {17}{4}} \Gamma \left (\frac {17}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {17}{4} \\ \frac {21}{4} \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{a^{\frac {7}{4}} \Gamma \left (\frac {21}{4}\right )} \] Input:
integrate(x**(13/4)/(b*x+a)**(7/4),x)
Output:
x**(17/4)*gamma(17/4)*hyper((7/4, 17/4), (21/4,), b*x*exp_polar(I*pi)/a)/( a**(7/4)*gamma(21/4))
\[ \int \frac {x^{13/4}}{(a+b x)^{7/4}} \, dx=\int { \frac {x^{\frac {13}{4}}}{{\left (b x + a\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(x^(13/4)/(b*x+a)^(7/4),x, algorithm="maxima")
Output:
integrate(x^(13/4)/(b*x + a)^(7/4), x)
Exception generated. \[ \int \frac {x^{13/4}}{(a+b x)^{7/4}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x^(13/4)/(b*x+a)^(7/4),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{-1,[0,1,3,2,0]%%%} / %%%{1,[0,0,0,0,4]%%%} Error: Bad Argu ment Valu
Timed out. \[ \int \frac {x^{13/4}}{(a+b x)^{7/4}} \, dx=\int \frac {x^{13/4}}{{\left (a+b\,x\right )}^{7/4}} \,d x \] Input:
int(x^(13/4)/(a + b*x)^(7/4),x)
Output:
int(x^(13/4)/(a + b*x)^(7/4), x)
\[ \int \frac {x^{13/4}}{(a+b x)^{7/4}} \, dx=\int \frac {x^{\frac {13}{4}}}{\left (b x +a \right )^{\frac {3}{4}} a +\left (b x +a \right )^{\frac {3}{4}} b x}d x \] Input:
int(x^(13/4)/(b*x+a)^(7/4),x)
Output:
int((x**(1/4)*x**3)/((a + b*x)**(3/4)*a + (a + b*x)**(3/4)*b*x),x)