Integrand size = 21, antiderivative size = 56 \[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{b d \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \] Output:
-3*hypergeom([1/6, 1/2],[7/6],cos(d*x+c)^2)*sin(d*x+c)/b/d/(b*sec(d*x+c))^ (1/3)/(sin(d*x+c)^2)^(1/2)
Time = 0.00 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=\frac {3 \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sec ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sqrt {-\tan ^2(c+d x)}}{2 b^2 d} \] Input:
Integrate[Sec[c + d*x]^2/(b*Sec[c + d*x])^(4/3),x]
Output:
(3*Cot[c + d*x]*Hypergeometric2F1[1/3, 1/2, 4/3, Sec[c + d*x]^2]*(b*Sec[c + d*x])^(2/3)*Sqrt[-Tan[c + d*x]^2])/(2*b^2*d)
Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2030, 3042, 4259, 3042, 3122}
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.
\(\displaystyle \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int (b \sec (c+d x))^{2/3}dx}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx}{b^2}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {\left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{2/3}}dx}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{2/3}}dx}{b^2}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle -\frac {3 \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}\) |
Input:
Int[Sec[c + d*x]^2/(b*Sec[c + d*x])^(4/3),x]
Output:
(-3*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*(b *Sec[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2])
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^(n - 1)*((Sin[c + d*x]/b)^(n - 1) Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
\[\int \frac {\sec \left (d x +c \right )^{2}}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}d x\]
Input:
int(sec(d*x+c)^2/(b*sec(d*x+c))^(4/3),x)
Output:
int(sec(d*x+c)^2/(b*sec(d*x+c))^(4/3),x)
\[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^2/(b*sec(d*x+c))^(4/3),x, algorithm="fricas")
Output:
integral((b*sec(d*x + c))^(2/3)/b^2, x)
\[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \] Input:
integrate(sec(d*x+c)**2/(b*sec(d*x+c))**(4/3),x)
Output:
Integral(sec(c + d*x)**2/(b*sec(c + d*x))**(4/3), x)
\[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^2/(b*sec(d*x+c))^(4/3),x, algorithm="maxima")
Output:
integrate(sec(d*x + c)^2/(b*sec(d*x + c))^(4/3), x)
\[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^2/(b*sec(d*x+c))^(4/3),x, algorithm="giac")
Output:
integrate(sec(d*x + c)^2/(b*sec(d*x + c))^(4/3), x)
Timed out. \[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^2\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}} \,d x \] Input:
int(1/(cos(c + d*x)^2*(b/cos(c + d*x))^(4/3)),x)
Output:
int(1/(cos(c + d*x)^2*(b/cos(c + d*x))^(4/3)), x)
\[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=\frac {\int \sec \left (d x +c \right )^{\frac {2}{3}}d x}{b^{\frac {4}{3}}} \] Input:
int(sec(d*x+c)^2/(b*sec(d*x+c))^(4/3),x)
Output:
int(sec(c + d*x)/sec(c + d*x)**(1/3),x)/(b**(1/3)*b)