Integrand size = 33, antiderivative size = 146 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b C \sec ^{2+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (7+3 m)}+\frac {3 b (C (4+3 m)+A (7+3 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1-3 m),\frac {1}{6} (5-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (1+3 m) (7+3 m) \sqrt {\sin ^2(c+d x)}} \]
3*b*C*sec(d*x+c)^(2+m)*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(7+3*m)+3*b*(C*(4 +3*m)+A*(7+3*m))*hypergeom([1/2, -1/6-1/2*m],[5/6-1/2*m],cos(d*x+c)^2)*sec (d*x+c)^m*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(9*m^2+24*m+7)/(sin(d*x+c)^2)^ (1/2)
Time = 0.62 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.98 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b \csc (c+d x) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A (10+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4+3 m),\frac {5}{3}+\frac {m}{2},\sec ^2(c+d x)\right )+C (4+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3}+\frac {m}{2},\frac {8}{3}+\frac {m}{2},\sec ^2(c+d x)\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (4+3 m) (10+3 m)} \]
(3*b*Csc[c + d*x]*Sec[c + d*x]^m*(b*Sec[c + d*x])^(1/3)*(A*(10 + 3*m)*Hype rgeometric2F1[1/2, (4 + 3*m)/6, 5/3 + m/2, Sec[c + d*x]^2] + C*(4 + 3*m)*H ypergeometric2F1[1/2, 5/3 + m/2, 8/3 + m/2, Sec[c + d*x]^2]*Sec[c + d*x]^2 )*Sqrt[-Tan[c + d*x]^2])/(d*(4 + 3*m)*(10 + 3*m))
Time = 0.54 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {2034, 3042, 4534, 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.
\(\displaystyle \int (b \sec (c+d x))^{4/3} \sec ^m(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \frac {b \sqrt [3]{b \sec (c+d x)} \int \sec ^{m+\frac {4}{3}}(c+d x) \left (C \sec ^2(c+d x)+A\right )dx}{\sqrt [3]{\sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \sqrt [3]{b \sec (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {4}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx}{\sqrt [3]{\sec (c+d x)}}\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {b \sqrt [3]{b \sec (c+d x)} \left (\frac {(A (3 m+7)+C (3 m+4)) \int \sec ^{m+\frac {4}{3}}(c+d x)dx}{3 m+7}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {7}{3}}(c+d x)}{d (3 m+7)}\right )}{\sqrt [3]{\sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \sqrt [3]{b \sec (c+d x)} \left (\frac {(A (3 m+7)+C (3 m+4)) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {4}{3}}dx}{3 m+7}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {7}{3}}(c+d x)}{d (3 m+7)}\right )}{\sqrt [3]{\sec (c+d x)}}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {b \sqrt [3]{b \sec (c+d x)} \left (\frac {(A (3 m+7)+C (3 m+4)) \cos ^{m+\frac {1}{3}}(c+d x) \sec ^{m+\frac {1}{3}}(c+d x) \int \cos ^{-m-\frac {4}{3}}(c+d x)dx}{3 m+7}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {7}{3}}(c+d x)}{d (3 m+7)}\right )}{\sqrt [3]{\sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \sqrt [3]{b \sec (c+d x)} \left (\frac {(A (3 m+7)+C (3 m+4)) \cos ^{m+\frac {1}{3}}(c+d x) \sec ^{m+\frac {1}{3}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-m-\frac {4}{3}}dx}{3 m+7}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {7}{3}}(c+d x)}{d (3 m+7)}\right )}{\sqrt [3]{\sec (c+d x)}}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {b \sqrt [3]{b \sec (c+d x)} \left (\frac {3 (A (3 m+7)+C (3 m+4)) \sin (c+d x) \sec ^{m+\frac {1}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-3 m-1),\frac {1}{6} (5-3 m),\cos ^2(c+d x)\right )}{d (3 m+1) (3 m+7) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {7}{3}}(c+d x)}{d (3 m+7)}\right )}{\sqrt [3]{\sec (c+d x)}}\) |
(b*(b*Sec[c + d*x])^(1/3)*((3*C*Sec[c + d*x]^(7/3 + m)*Sin[c + d*x])/(d*(7 + 3*m)) + (3*(C*(4 + 3*m) + A*(7 + 3*m))*Hypergeometric2F1[1/2, (-1 - 3*m )/6, (5 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(1/3 + m)*Sin[c + d*x])/(d* (1 + 3*m)*(7 + 3*m)*Sqrt[Sin[c + d*x]^2])))/Sec[c + d*x]^(1/3)
3.1.21.3.1 Defintions of rubi rules used
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[b^IntPart [n]*((b*v)^FracPart[n]/(a^IntPart[n]*(a*v)^FracPart[n])) Int[(a*v)^(m + n )*Fx, x], x] /; FreeQ[{a, b, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[m + n]
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[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
\[\int \sec \left (d x +c \right )^{m} \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +C \sec \left (d x +c \right )^{2}\right )d x\]
\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{m} \,d x } \]
Timed out. \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{m} \,d x } \]
\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{m} \,d x } \]
Timed out. \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m \,d x \]