Integrand size = 15, antiderivative size = 108 \[ \int \frac {x^{7/4}}{(a+b x)^{5/4}} \, dx=\frac {4 a x^{3/4}}{b^2 \sqrt [4]{a+b x}}+\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b^2}-\frac {7 a^2 \sqrt [4]{-\frac {b x}{a}-\frac {b^2 x^2}{a^2}} E\left (\left .\frac {1}{2} \arcsin \left (1+\frac {2 b x}{a}\right )\right |2\right )}{\sqrt {2} b^3 \sqrt [4]{x} \sqrt [4]{a+b x}} \] Output:
4*a*x^(3/4)/b^2/(b*x+a)^(1/4)+2/3*x^(3/4)*(b*x+a)^(3/4)/b^2-7/2*a^2*(-b*x/ a-b^2*x^2/a^2)^(1/4)*EllipticE(sin(1/2*arcsin(1+2*b*x/a)),2^(1/2))*2^(1/2) /b^3/x^(1/4)/(b*x+a)^(1/4)
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 = 0.46 \[ \int \frac {x^{7/4}}{(a+b x)^{5/4}} \, dx=\frac {4 x^{11/4} \sqrt [4]{1+\frac {b x}{a}} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {11}{4},\frac {15}{4},-\frac {b x}{a}\right )}{11 a \sqrt [4]{a+b x}} \] Input:
Integrate[x^(7/4)/(a + b*x)^(5/4),x]
Output:
(4*x^(11/4)*(1 + (b*x)/a)^(1/4)*Hypergeometric2F1[5/4, 11/4, 15/4, -((b*x) /a)])/(11*a*(a + b*x)^(1/4))
Time = 0.27 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.27, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {57, 60, 73, 839, 813, 858, 807, 212}
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.
\(\displaystyle \int \frac {x^{7/4}}{(a+b x)^{5/4}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {7 \int \frac {x^{3/4}}{\sqrt [4]{a+b x}}dx}{b}-\frac {4 x^{7/4}}{b \sqrt [4]{a+b x}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {7 \left (\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b}-\frac {a \int \frac {1}{\sqrt [4]{x} \sqrt [4]{a+b x}}dx}{2 b}\right )}{b}-\frac {4 x^{7/4}}{b \sqrt [4]{a+b x}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {7 \left (\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b}-\frac {2 a \int \frac {\sqrt {x}}{\sqrt [4]{a+b x}}d\sqrt [4]{x}}{b}\right )}{b}-\frac {4 x^{7/4}}{b \sqrt [4]{a+b x}}\) |
\(\Big \downarrow \) 839 |
\(\displaystyle \frac {7 \left (\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b}-\frac {2 a \left (\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}-\frac {1}{2} a \int \frac {\sqrt {x}}{(a+b x)^{5/4}}d\sqrt [4]{x}\right )}{b}\right )}{b}-\frac {4 x^{7/4}}{b \sqrt [4]{a+b x}}\) |
\(\Big \downarrow \) 813 |
\(\displaystyle \frac {7 \left (\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b}-\frac {2 a \left (\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}-\frac {a \sqrt [4]{x} \sqrt [4]{\frac {a}{b x}+1} \int \frac {1}{\left (\frac {a}{b x}+1\right )^{5/4} x^{3/4}}d\sqrt [4]{x}}{2 b \sqrt [4]{a+b x}}\right )}{b}\right )}{b}-\frac {4 x^{7/4}}{b \sqrt [4]{a+b x}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {7 \left (\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b}-\frac {2 a \left (\frac {a \sqrt [4]{x} \sqrt [4]{\frac {a}{b x}+1} \int \frac {1}{\sqrt [4]{x} \left (\frac {a x}{b}+1\right )^{5/4}}d\frac {1}{\sqrt [4]{x}}}{2 b \sqrt [4]{a+b x}}+\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}\right )}{b}\right )}{b}-\frac {4 x^{7/4}}{b \sqrt [4]{a+b x}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {7 \left (\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b}-\frac {2 a \left (\frac {a \sqrt [4]{x} \sqrt [4]{\frac {a}{b x}+1} \int \frac {1}{\left (\frac {\sqrt {x} a}{b}+1\right )^{5/4}}d\sqrt {x}}{4 b \sqrt [4]{a+b x}}+\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}\right )}{b}\right )}{b}-\frac {4 x^{7/4}}{b \sqrt [4]{a+b x}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {7 \left (\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b}-\frac {2 a \left (\frac {\sqrt {a} \sqrt [4]{x} \sqrt [4]{\frac {a}{b x}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )\right |2\right )}{2 \sqrt {b} \sqrt [4]{a+b x}}+\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}\right )}{b}\right )}{b}-\frac {4 x^{7/4}}{b \sqrt [4]{a+b x}}\) |
Input:
Int[x^(7/4)/(a + b*x)^(5/4),x]
Output:
(-4*x^(7/4))/(b*(a + b*x)^(1/4)) + (7*((2*x^(3/4)*(a + b*x)^(3/4))/(3*b) - (2*a*(x^(3/4)/(2*(a + b*x)^(1/4)) + (Sqrt[a]*(1 + a/(b*x))^(1/4)*x^(1/4)* EllipticE[ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]]/2, 2])/(2*Sqrt[b]*(a + b*x)^(1 /4))))/b))/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)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
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_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) )^(1/4)/(b*(a + b*x^4)^(1/4))) Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[x^3/(2*(a + b*x^4 )^(1/4)), x] - Simp[a/2 Int[x^2/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b} , x] && PosQ[b/a]
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]
\[\int \frac {x^{\frac {7}{4}}}{\left (b x +a \right )^{\frac {5}{4}}}d x\]
Input:
int(x^(7/4)/(b*x+a)^(5/4),x)
Output:
int(x^(7/4)/(b*x+a)^(5/4),x)
\[ \int \frac {x^{7/4}}{(a+b x)^{5/4}} \, dx=\int { \frac {x^{\frac {7}{4}}}{{\left (b x + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(x^(7/4)/(b*x+a)^(5/4),x, algorithm="fricas")
Output:
integral((b*x + a)^(3/4)*x^(7/4)/(b^2*x^2 + 2*a*b*x + a^2), x)
Result contains complex when optimal does not.
Time = 7.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.33 \[ \int \frac {x^{7/4}}{(a+b x)^{5/4}} \, dx=\frac {x^{\frac {11}{4}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{a^{\frac {5}{4}} \Gamma \left (\frac {15}{4}\right )} \] Input:
integrate(x**(7/4)/(b*x+a)**(5/4),x)
Output:
x**(11/4)*gamma(11/4)*hyper((5/4, 11/4), (15/4,), b*x*exp_polar(I*pi)/a)/( a**(5/4)*gamma(15/4))
\[ \int \frac {x^{7/4}}{(a+b x)^{5/4}} \, dx=\int { \frac {x^{\frac {7}{4}}}{{\left (b x + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(x^(7/4)/(b*x+a)^(5/4),x, algorithm="maxima")
Output:
integrate(x^(7/4)/(b*x + a)^(5/4), x)
\[ \int \frac {x^{7/4}}{(a+b x)^{5/4}} \, dx=\int { \frac {x^{\frac {7}{4}}}{{\left (b x + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(x^(7/4)/(b*x+a)^(5/4),x, algorithm="giac")
Output:
integrate(x^(7/4)/(b*x + a)^(5/4), x)
Timed out. \[ \int \frac {x^{7/4}}{(a+b x)^{5/4}} \, dx=\int \frac {x^{7/4}}{{\left (a+b\,x\right )}^{5/4}} \,d x \] Input:
int(x^(7/4)/(a + b*x)^(5/4),x)
Output:
int(x^(7/4)/(a + b*x)^(5/4), x)
\[ \int \frac {x^{7/4}}{(a+b x)^{5/4}} \, dx=\int \frac {x^{\frac {7}{4}}}{\left (b x +a \right )^{\frac {1}{4}} a +\left (b x +a \right )^{\frac {1}{4}} b x}d x \] Input:
int(x^(7/4)/(b*x+a)^(5/4),x)
Output:
int((x**(3/4)*x)/((a + b*x)**(1/4)*a + (a + b*x)**(1/4)*b*x),x)