Integrand size = 19, antiderivative size = 207 \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{7/4}} \, dx=-\frac {4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}+\frac {160 b (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{33 d^4}-\frac {80 b (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{33 d^3}+\frac {56 b (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 d^2}-\frac {320 b^{3/4} (b c-a d)^{13/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{33 d^5 \sqrt {a+b x}} \]
-4/3*(b*x+a)^(7/2)/d/(d*x+c)^(3/4)-80/33*b*(-a*d+b*c)*(b*x+a)^(3/2)*(d*x+c )^(1/4)/d^3+56/33*b*(b*x+a)^(5/2)*(d*x+c)^(1/4)/d^2+160/33*b*(-a*d+b*c)^2* (d*x+c)^(1/4)*(b*x+a)^(1/2)/d^4-320/33*b^(3/4)*(-a*d+b*c)^(13/4)*EllipticF (b^(1/4)*(d*x+c)^(1/4)/(-a*d+b*c)^(1/4),I)*(-d*(b*x+a)/(-a*d+b*c))^(1/2)/d ^5/(b*x+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.35 \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{7/4}} \, dx=\frac {2 (a+b x)^{9/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{7/4} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {9}{2},\frac {11}{2},\frac {d (a+b x)}{-b c+a d}\right )}{9 b (c+d x)^{7/4}} \]
(2*(a + b*x)^(9/2)*((b*(c + d*x))/(b*c - a*d))^(7/4)*Hypergeometric2F1[7/4 , 9/2, 11/2, (d*(a + b*x))/(-(b*c) + a*d)])/(9*b*(c + d*x)^(7/4))
Time = 0.29 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {57, 60, 60, 60, 73, 765, 762}
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.
\(\displaystyle \int \frac {(a+b x)^{7/2}}{(c+d x)^{7/4}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {14 b \int \frac {(a+b x)^{5/2}}{(c+d x)^{3/4}}dx}{3 d}-\frac {4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {14 b \left (\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 d}-\frac {10 (b c-a d) \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}}dx}{11 d}\right )}{3 d}-\frac {4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {14 b \left (\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 d}-\frac {10 (b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}}dx}{7 d}\right )}{11 d}\right )}{3 d}-\frac {4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {14 b \left (\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 d}-\frac {10 (b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {2 (b c-a d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}}dx}{3 d}\right )}{7 d}\right )}{11 d}\right )}{3 d}-\frac {4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {14 b \left (\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 d}-\frac {10 (b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d) \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [4]{c+d x}}{3 d^2}\right )}{7 d}\right )}{11 d}\right )}{3 d}-\frac {4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}\) |
\(\Big \downarrow \) 765 |
\(\displaystyle \frac {14 b \left (\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 d}-\frac {10 (b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d) \sqrt {1-\frac {b (c+d x)}{b c-a d}} \int \frac {1}{\sqrt {1-\frac {b (c+d x)}{b c-a d}}}d\sqrt [4]{c+d x}}{3 d^2 \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{7 d}\right )}{11 d}\right )}{3 d}-\frac {4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {14 b \left (\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 d}-\frac {10 (b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d)^{5/4} \sqrt {1-\frac {b (c+d x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 \sqrt [4]{b} d^2 \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{7 d}\right )}{11 d}\right )}{3 d}-\frac {4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}\) |
(-4*(a + b*x)^(7/2))/(3*d*(c + d*x)^(3/4)) + (14*b*((4*(a + b*x)^(5/2)*(c + d*x)^(1/4))/(11*d) - (10*(b*c - a*d)*((4*(a + b*x)^(3/2)*(c + d*x)^(1/4) )/(7*d) - (6*(b*c - a*d)*((4*Sqrt[a + b*x]*(c + d*x)^(1/4))/(3*d) - (8*(b* c - a*d)^(5/4)*Sqrt[1 - (b*(c + d*x))/(b*c - a*d)]*EllipticF[ArcSin[(b^(1/ 4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], -1])/(3*b^(1/4)*d^2*Sqrt[a - (b*c) /d + (b*(c + d*x))/d])))/(7*d)))/(11*d)))/(3*d)
3.17.65.3.1 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] :> 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[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) )*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt [a + b*x^4] Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ [b/a] && !GtQ[a, 0]
\[\int \frac {\left (b x +a \right )^{\frac {7}{2}}}{\left (d x +c \right )^{\frac {7}{4}}}d x\]
\[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{7/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {7}{2}}}{{\left (d x + c\right )}^{\frac {7}{4}}} \,d x } \]
integral((b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(b*x + a)*(d*x + c) ^(1/4)/(d^2*x^2 + 2*c*d*x + c^2), x)
\[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{7/4}} \, dx=\int \frac {\left (a + b x\right )^{\frac {7}{2}}}{\left (c + d x\right )^{\frac {7}{4}}}\, dx \]
\[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{7/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {7}{2}}}{{\left (d x + c\right )}^{\frac {7}{4}}} \,d x } \]
\[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{7/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {7}{2}}}{{\left (d x + c\right )}^{\frac {7}{4}}} \,d x } \]
Timed out. \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{7/4}} \, dx=\int \frac {{\left (a+b\,x\right )}^{7/2}}{{\left (c+d\,x\right )}^{7/4}} \,d x \]