Integrand size = 19, antiderivative size = 146 \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{7/4}} \, dx=-\frac {2}{(b c-a d) \sqrt {a+b x} (c+d x)^{3/4}}-\frac {10 d \sqrt {a+b x}}{3 (b c-a d)^2 (c+d x)^{3/4}}-\frac {10 b^{3/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 (b c-a d)^{7/4} \sqrt {a+b x}} \]
-2/(-a*d+b*c)/(d*x+c)^(3/4)/(b*x+a)^(1/2)-10/3*d*(b*x+a)^(1/2)/(-a*d+b*c)^ 2/(d*x+c)^(3/4)-10/3*b^(3/4)*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)/(-a*d+b*c)^(7/4)/(b*x+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.49 \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{7/4}} \, dx=-\frac {2 \left (\frac {b (c+d x)}{b c-a d}\right )^{7/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {7}{4},\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )}{b \sqrt {a+b x} (c+d x)^{7/4}} \]
(-2*((b*(c + d*x))/(b*c - a*d))^(7/4)*Hypergeometric2F1[-1/2, 7/4, 1/2, (d *(a + b*x))/(-(b*c) + a*d)])/(b*Sqrt[a + b*x]*(c + d*x)^(7/4))
Time = 0.26 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.23, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {61, 61, 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.
\(\displaystyle \int \frac {1}{(a+b x)^{3/2} (c+d x)^{7/4}} \, dx\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {5 d \int \frac {1}{\sqrt {a+b x} (c+d x)^{7/4}}dx}{2 (b c-a d)}-\frac {2}{\sqrt {a+b x} (c+d x)^{3/4} (b c-a d)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {5 d \left (\frac {b \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}}dx}{3 (b c-a d)}+\frac {4 \sqrt {a+b x}}{3 (c+d x)^{3/4} (b c-a d)}\right )}{2 (b c-a d)}-\frac {2}{\sqrt {a+b x} (c+d x)^{3/4} (b c-a d)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {5 d \left (\frac {4 b \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [4]{c+d x}}{3 d (b c-a d)}+\frac {4 \sqrt {a+b x}}{3 (c+d x)^{3/4} (b c-a d)}\right )}{2 (b c-a d)}-\frac {2}{\sqrt {a+b x} (c+d x)^{3/4} (b c-a d)}\) |
\(\Big \downarrow \) 765 |
\(\displaystyle -\frac {5 d \left (\frac {4 b \sqrt {1-\frac {b (c+d x)}{b c-a d}} \int \frac {1}{\sqrt {1-\frac {b (c+d x)}{b c-a d}}}d\sqrt [4]{c+d x}}{3 d (b c-a d) \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}+\frac {4 \sqrt {a+b x}}{3 (c+d x)^{3/4} (b c-a d)}\right )}{2 (b c-a d)}-\frac {2}{\sqrt {a+b x} (c+d x)^{3/4} (b c-a d)}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle -\frac {5 d \left (\frac {4 b^{3/4} \sqrt {1-\frac {b (c+d x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 d (b c-a d)^{3/4} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}+\frac {4 \sqrt {a+b x}}{3 (c+d x)^{3/4} (b c-a d)}\right )}{2 (b c-a d)}-\frac {2}{\sqrt {a+b x} (c+d x)^{3/4} (b c-a d)}\) |
-2/((b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/4)) - (5*d*((4*Sqrt[a + b*x])/( 3*(b*c - a*d)*(c + d*x)^(3/4)) + (4*b^(3/4)*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*d*(b*c - a*d)^(3/4)*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])))/(2*(b*c - a* d))
3.17.69.3.1 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 + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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[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]
\[\int \frac {1}{\left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {7}{4}}}d x\]
\[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{7/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {7}{4}}} \,d x } \]
integral(sqrt(b*x + a)*(d*x + c)^(1/4)/(b^2*d^2*x^4 + a^2*c^2 + 2*(b^2*c*d + a*b*d^2)*x^3 + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^2 + 2*(a*b*c^2 + a^2*c *d)*x), x)
\[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{7/4}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {7}{4}}}\, dx \]
\[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{7/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {7}{4}}} \,d x } \]
\[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{7/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {7}{4}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{7/4}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{7/4}} \,d x \]