Integrand size = 41, antiderivative size = 234 \[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 C \sec ^m(c+d x) \sin (c+d x)}{b d (1-3 m) \sqrt [3]{b \sec (c+d x)}}-\frac {3 (A+C (4-3 m)-3 A m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7-3 m),\frac {1}{6} (13-3 m),\cos ^2(c+d x)\right ) \sec ^{-2+m}(c+d x) \sin (c+d x)}{b d (1-3 m) (7-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{b d (4-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \]
-3*C*sec(d*x+c)^m*sin(d*x+c)/b/d/(1-3*m)/(b*sec(d*x+c))^(1/3)-3*(A+C*(4-3* m)-3*A*m)*hypergeom([1/2, 7/6-1/2*m],[13/6-1/2*m],cos(d*x+c)^2)*sec(d*x+c) ^(-2+m)*sin(d*x+c)/b/d/(9*m^2-24*m+7)/(b*sec(d*x+c))^(1/3)/(sin(d*x+c)^2)^ (1/2)-3*B*hypergeom([1/2, 2/3-1/2*m],[5/3-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^ (-1+m)*sin(d*x+c)/b/d/(4-3*m)/(b*sec(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)
Time = 1.34 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.88 \[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\frac {3 b \csc (c+d x) \sec ^m(c+d x) \left (A \left (-2+3 m+9 m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-4+3 m),\frac {1}{6} (2+3 m),\sec ^2(c+d x)\right )+(-4+3 m) \left (B (2+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1+3 m),\frac {1}{6} (5+3 m),\sec ^2(c+d x)\right )+C (-1+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\sec ^2(c+d x)\right )\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (-4+3 m) (-1+3 m) (2+3 m) (b \sec (c+d x))^{7/3}} \]
(3*b*Csc[c + d*x]*Sec[c + d*x]^m*(A*(-2 + 3*m + 9*m^2)*Hypergeometric2F1[1 /2, (-4 + 3*m)/6, (2 + 3*m)/6, Sec[c + d*x]^2] + (-4 + 3*m)*(B*(2 + 3*m)*C os[c + d*x]*Hypergeometric2F1[1/2, (-1 + 3*m)/6, (5 + 3*m)/6, Sec[c + d*x] ^2] + C*(-1 + 3*m)*Hypergeometric2F1[1/2, (2 + 3*m)/6, (8 + 3*m)/6, Sec[c + d*x]^2])*Sec[c + d*x]^2)*Sqrt[-Tan[c + d*x]^2])/(d*(-4 + 3*m)*(-1 + 3*m) *(2 + 3*m)*(b*Sec[c + d*x])^(7/3))
Time = 0.98 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.96, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.293, Rules used = {2034, 3042, 4535, 3042, 4259, 3042, 3122, 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.
| \(\displaystyle \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx\) |
| \(\Big \downarrow \) 2034 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \int \sec ^{m-\frac {4}{3}}(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )dx}{b \sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {4}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{b \sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\int \sec ^{m-\frac {4}{3}}(c+d x) \left (C \sec ^2(c+d x)+A\right )dx+B \int \sec ^{m-\frac {1}{3}}(c+d x)dx\right )}{b \sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {4}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {1}{3}}dx\right )}{b \sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {4}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \cos ^{m+\frac {2}{3}}(c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \int \cos ^{\frac {1}{3}-m}(c+d x)dx\right )}{b \sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {4}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \cos ^{m+\frac {2}{3}}(c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{\frac {1}{3}-m}dx\right )}{b \sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {4}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx-\frac {3 B \sin (c+d x) \sec ^{m-\frac {4}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right )}{d (4-3 m) \sqrt {\sin ^2(c+d x)}}\right )}{b \sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\frac {(A (1-3 m)+C (4-3 m)) \int \sec ^{m-\frac {4}{3}}(c+d x)dx}{1-3 m}-\frac {3 B \sin (c+d x) \sec ^{m-\frac {4}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right )}{d (4-3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 C \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x)}{d (1-3 m)}\right )}{b \sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\frac {(A (1-3 m)+C (4-3 m)) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {4}{3}}dx}{1-3 m}-\frac {3 B \sin (c+d x) \sec ^{m-\frac {4}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right )}{d (4-3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 C \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x)}{d (1-3 m)}\right )}{b \sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\frac {(A (1-3 m)+C (4-3 m)) \cos ^{m+\frac {2}{3}}(c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \int \cos ^{\frac {4}{3}-m}(c+d x)dx}{1-3 m}-\frac {3 B \sin (c+d x) \sec ^{m-\frac {4}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right )}{d (4-3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 C \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x)}{d (1-3 m)}\right )}{b \sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\frac {(A (1-3 m)+C (4-3 m)) \cos ^{m+\frac {2}{3}}(c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{\frac {4}{3}-m}dx}{1-3 m}-\frac {3 B \sin (c+d x) \sec ^{m-\frac {4}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right )}{d (4-3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 C \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x)}{d (1-3 m)}\right )}{b \sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (-\frac {3 (A (1-3 m)+C (4-3 m)) \sin (c+d x) \sec ^{m-\frac {7}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7-3 m),\frac {1}{6} (13-3 m),\cos ^2(c+d x)\right )}{d (1-3 m) (7-3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) \sec ^{m-\frac {4}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right )}{d (4-3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 C \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x)}{d (1-3 m)}\right )}{b \sqrt [3]{b \sec (c+d x)}}\) |
(Sec[c + d*x]^(1/3)*((-3*C*Sec[c + d*x]^(-1/3 + m)*Sin[c + d*x])/(d*(1 - 3 *m)) - (3*(A*(1 - 3*m) + C*(4 - 3*m))*Hypergeometric2F1[1/2, (7 - 3*m)/6, (13 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-7/3 + m)*Sin[c + d*x])/(d*(1 - 3*m)*(7 - 3*m)*Sqrt[Sin[c + d*x]^2]) - (3*B*Hypergeometric2F1[1/2, (4 - 3*m)/6, (10 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-4/3 + m)*Sin[c + d*x] )/(d*(4 - 3*m)*Sqrt[Sin[c + d*x]^2])))/(b*(b*Sec[c + d*x])^(1/3))
3.1.70.3.1 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[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
\[\int \frac {\sec \left (d x +c \right )^{m} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}d x\]
\[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
integrate(sec(d*x+c)^m*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(4/3 ),x, algorithm="fricas")
integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^(2/3)*se c(d*x + c)^m/(b^2*sec(d*x + c)^2), x)
\[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{m}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**m/(b*sec(c + d*x))**(4/3), x)
\[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
integrate(sec(d*x+c)^m*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(4/3 ),x, algorithm="maxima")
\[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
integrate(sec(d*x+c)^m*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(4/3 ),x, algorithm="giac")
Timed out. \[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}} \,d x \]