Integrand size = 19, antiderivative size = 712 \[ \int \frac {(a+b x)^{3/4}}{(c+d x)^{5/4}} \, dx=-\frac {4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}}+\frac {6 \sqrt {b} \sqrt {(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \sqrt {(a d+b (c+2 d x))^2}}{d^{3/2} (b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )}-\frac {3 \sqrt {2} \sqrt [4]{b} (b c-a d)^{3/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{d^{7/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}+\frac {3 \sqrt [4]{b} (b c-a d)^{3/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{\sqrt {2} d^{7/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \]
-4*(b*x+a)^(3/4)/d/(d*x+c)^(1/4)+6*b^(1/2)*((b*x+a)*(d*x+c))^(1/2)*((2*b*d *x+a*d+b*c)^2)^(1/2)*((a*d+b*(2*d*x+c))^2)^(1/2)/d^(3/2)/(-a*d+b*c)/(b*x+a )^(1/4)/(d*x+c)^(1/4)/(2*b*d*x+a*d+b*c)/(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x +c))^(1/2)/(-a*d+b*c))+3/2*b^(1/4)*(-a*d+b*c)^(3/2)*((b*x+a)*(d*x+c))^(1/4 )*(cos(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c) ^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^( 1/2)/(-a*d+b*c)^(1/2)))*EllipticF(sin(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d *x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2))),1/2*2^(1/2))*(1+2*b^(1/2)*d^(1/2)* ((b*x+a)*(d*x+c))^(1/2)/(-a*d+b*c))*((2*b*d*x+a*d+b*c)^2)^(1/2)*((a*d+b*(2 *d*x+c))^2/(-a*d+b*c)^2/(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2)/(-a*d +b*c))^2)^(1/2)/d^(7/4)/(b*x+a)^(1/4)/(d*x+c)^(1/4)/(2*b*d*x+a*d+b*c)*2^(1 /2)/((a*d+b*(2*d*x+c))^2)^(1/2)-3*b^(1/4)*(-a*d+b*c)^(3/2)*((b*x+a)*(d*x+c ))^(1/4)*(cos(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a *d+b*c)^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1 /4)*2^(1/2)/(-a*d+b*c)^(1/2)))*EllipticE(sin(2*arctan(b^(1/4)*d^(1/4)*((b* x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2))),1/2*2^(1/2))*2^(1/2)*(1+2*b ^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2)/(-a*d+b*c))*((2*b*d*x+a*d+b*c)^2)^( 1/2)*((a*d+b*(2*d*x+c))^2/(-a*d+b*c)^2/(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+ c))^(1/2)/(-a*d+b*c))^2)^(1/2)/d^(7/4)/(b*x+a)^(1/4)/(d*x+c)^(1/4)/(2*b*d* x+a*d+b*c)/((a*d+b*(2*d*x+c))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.10 \[ \int \frac {(a+b x)^{3/4}}{(c+d x)^{5/4}} \, dx=\frac {4 (a+b x)^{7/4} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {7}{4},\frac {11}{4},\frac {d (a+b x)}{-b c+a d}\right )}{7 b (c+d x)^{5/4}} \]
(4*(a + b*x)^(7/4)*((b*(c + d*x))/(b*c - a*d))^(5/4)*Hypergeometric2F1[5/4 , 7/4, 11/4, (d*(a + b*x))/(-(b*c) + a*d)])/(7*b*(c + d*x)^(5/4))
Time = 0.32 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.26, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {57, 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.
| \(\displaystyle \int \frac {(a+b x)^{3/4}}{(c+d x)^{5/4}} \, dx\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {3 b \int \frac {1}{\sqrt [4]{a+b x} \sqrt [4]{c+d x}}dx}{d}-\frac {4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {12 \int \frac {\sqrt {a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}d\sqrt [4]{a+b x}}{d}-\frac {4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}}\) |
| \(\Big \downarrow \) 839 |
| \(\displaystyle \frac {12 \left (\frac {(a+b x)^{3/4}}{2 \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}-\frac {1}{2} \left (c-\frac {a d}{b}\right ) \int \frac {\sqrt {a+b x}}{\left (c-\frac {a d}{b}+\frac {d (a+b x)}{b}\right )^{5/4}}d\sqrt [4]{a+b x}\right )}{d}-\frac {4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}}\) |
| \(\Big \downarrow \) 813 |
| \(\displaystyle \frac {12 \left (\frac {(a+b x)^{3/4}}{2 \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}-\frac {b \sqrt [4]{a+b x} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d (a+b x)}+1} \int \frac {1}{(a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{5/4}}d\sqrt [4]{a+b x}}{2 d \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{d}-\frac {4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}}\) |
| \(\Big \downarrow \) 858 |
| \(\displaystyle \frac {12 \left (\frac {b \sqrt [4]{a+b x} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d (a+b x)}+1} \int \frac {1}{\sqrt [4]{a+b x} \left (\frac {(b c-a d) (a+b x)}{d}+1\right )^{5/4}}d\frac {1}{\sqrt [4]{a+b x}}}{2 d \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}+\frac {(a+b x)^{3/4}}{2 \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{d}-\frac {4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}}\) |
| \(\Big \downarrow \) 807 |
| \(\displaystyle \frac {12 \left (\frac {b \sqrt [4]{a+b x} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d (a+b x)}+1} \int \frac {1}{\left (\frac {\sqrt {a+b x} (b c-a d)}{d}+1\right )^{5/4}}d\sqrt {a+b x}}{4 d \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}+\frac {(a+b x)^{3/4}}{2 \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{d}-\frac {4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}}\) |
| \(\Big \downarrow \) 212 |
| \(\displaystyle \frac {12 \left (\frac {b \sqrt [4]{a+b x} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d (a+b x)}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d} \sqrt {a+b x}}{\sqrt {d}}\right )\right |2\right )}{2 \sqrt {d} \sqrt {b c-a d} \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}+\frac {(a+b x)^{3/4}}{2 \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{d}-\frac {4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}}\) |
(-4*(a + b*x)^(3/4))/(d*(c + d*x)^(1/4)) + (12*((a + b*x)^(3/4)/(2*(c - (a *d)/b + (d*(a + b*x))/b)^(1/4)) + (b*(c - (a*d)/b)*(a + b*x)^(1/4)*(1 + (b *c - a*d)/(d*(a + b*x)))^(1/4)*EllipticE[ArcTan[(Sqrt[b*c - a*d]*Sqrt[a + b*x])/Sqrt[d]]/2, 2])/(2*Sqrt[d]*Sqrt[b*c - a*d]*(c - (a*d)/b + (d*(a + b* x))/b)^(1/4))))/d
3.18.25.3.1 Defintions of rubi rules used
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] :> 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]
\[\int \frac {\left (b x +a \right )^{\frac {3}{4}}}{\left (d x +c \right )^{\frac {5}{4}}}d x\]
\[ \int \frac {(a+b x)^{3/4}}{(c+d x)^{5/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {3}{4}}}{{\left (d x + c\right )}^{\frac {5}{4}}} \,d x } \]
\[ \int \frac {(a+b x)^{3/4}}{(c+d x)^{5/4}} \, dx=\int \frac {\left (a + b x\right )^{\frac {3}{4}}}{\left (c + d x\right )^{\frac {5}{4}}}\, dx \]
\[ \int \frac {(a+b x)^{3/4}}{(c+d x)^{5/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {3}{4}}}{{\left (d x + c\right )}^{\frac {5}{4}}} \,d x } \]
\[ \int \frac {(a+b x)^{3/4}}{(c+d x)^{5/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {3}{4}}}{{\left (d x + c\right )}^{\frac {5}{4}}} \,d x } \]
Timed out. \[ \int \frac {(a+b x)^{3/4}}{(c+d x)^{5/4}} \, dx=\int \frac {{\left (a+b\,x\right )}^{3/4}}{{\left (c+d\,x\right )}^{5/4}} \,d x \]
\[ \int \frac {(a+b x)^{3/4}}{(c+d x)^{5/4}} \, dx=\int \frac {\left (b x +a \right )^{\frac {3}{4}}}{\left (d x +c \right )^{\frac {1}{4}} c +\left (d x +c \right )^{\frac {1}{4}} d x}d x \]