Integrand size = 41, antiderivative size = 228 \[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\frac {3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (2+3 m) \sqrt [3]{b \sec (c+d x)}}+\frac {3 (C (1-3 m)-A (2+3 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (4-3 m) (2+3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d (1-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \] Output:
3*C*sec(d*x+c)^(1+m)*sin(d*x+c)/d/(2+3*m)/(b*sec(d*x+c))^(1/3)+3*(C*(1-3*m )-A*(2+3*m))*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)/d/(4-3*m)/(2+3*m)/(b*sec(d*x+c))^(1/3)/(sin(d*x+c)^2) ^(1/2)-3*B*hypergeom([1/2, 1/6-1/2*m],[7/6-1/2*m],cos(d*x+c)^2)*sec(d*x+c) ^m*sin(d*x+c)/d/(1-3*m)/(b*sec(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)
Time = 1.03 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\frac {3 b \csc (c+d x) \sec ^m(c+d x) \left (A \left (10+21 m+9 m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1+3 m),\frac {1}{6} (5+3 m),\sec ^2(c+d x)\right )+(-1+3 m) \left (B (5+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\sec ^2(c+d x)\right )+C (2+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\sec ^2(c+d x)\right )\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (-1+3 m) (2+3 m) (5+3 m) (b \sec (c+d x))^{4/3}} \] Input:
Integrate[(Sec[c + d*x]^m*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(b*Sec[ c + d*x])^(1/3),x]
Output:
(3*b*Csc[c + d*x]*Sec[c + d*x]^m*(A*(10 + 21*m + 9*m^2)*Hypergeometric2F1[ 1/2, (-1 + 3*m)/6, (5 + 3*m)/6, Sec[c + d*x]^2] + (-1 + 3*m)*(B*(5 + 3*m)* Cos[c + d*x]*Hypergeometric2F1[1/2, (2 + 3*m)/6, (8 + 3*m)/6, Sec[c + d*x] ^2] + C*(2 + 3*m)*Hypergeometric2F1[1/2, (5 + 3*m)/6, (11 + 3*m)/6, Sec[c + d*x]^2])*Sec[c + d*x]^2)*Sqrt[-Tan[c + d*x]^2])/(d*(-1 + 3*m)*(2 + 3*m)* (5 + 3*m)*(b*Sec[c + d*x])^(4/3))
Time = 0.98 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.98, 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.
| \(\displaystyle \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 2034 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \int \sec ^{m-\frac {1}{3}}(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )dx}{\sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {1}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{\sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\int \sec ^{m-\frac {1}{3}}(c+d x) \left (C \sec ^2(c+d x)+A\right )dx+B \int \sec ^{m+\frac {2}{3}}(c+d x)dx\right )}{\sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {1}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {2}{3}}dx\right )}{\sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {1}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \cos ^{m+\frac {2}{3}}(c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \int \cos ^{-m-\frac {2}{3}}(c+d x)dx\right )}{\sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {1}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \cos ^{m+\frac {2}{3}}(c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-m-\frac {2}{3}}dx\right )}{\sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {1}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx-\frac {3 B \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right )}{d (1-3 m) \sqrt {\sin ^2(c+d x)}}\right )}{\sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (-\frac {(C (1-3 m)-A (3 m+2)) \int \sec ^{m-\frac {1}{3}}(c+d x)dx}{3 m+2}-\frac {3 B \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right )}{d (1-3 m) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {2}{3}}(c+d x)}{d (3 m+2)}\right )}{\sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (-\frac {(C (1-3 m)-A (3 m+2)) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {1}{3}}dx}{3 m+2}-\frac {3 B \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right )}{d (1-3 m) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {2}{3}}(c+d x)}{d (3 m+2)}\right )}{\sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (-\frac {(C (1-3 m)-A (3 m+2)) \cos ^{m+\frac {2}{3}}(c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \int \cos ^{\frac {1}{3}-m}(c+d x)dx}{3 m+2}-\frac {3 B \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right )}{d (1-3 m) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {2}{3}}(c+d x)}{d (3 m+2)}\right )}{\sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (-\frac {(C (1-3 m)-A (3 m+2)) \cos ^{m+\frac {2}{3}}(c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{\frac {1}{3}-m}dx}{3 m+2}-\frac {3 B \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right )}{d (1-3 m) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {2}{3}}(c+d x)}{d (3 m+2)}\right )}{\sqrt [3]{b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\frac {3 (C (1-3 m)-A (3 m+2)) \sin (c+d x) \sec ^{m-\frac {4}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right )}{d (4-3 m) (3 m+2) \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right )}{d (1-3 m) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {2}{3}}(c+d x)}{d (3 m+2)}\right )}{\sqrt [3]{b \sec (c+d x)}}\) |
Input:
Int[(Sec[c + d*x]^m*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(b*Sec[c + d* x])^(1/3),x]
Output:
(Sec[c + d*x]^(1/3)*((3*C*Sec[c + d*x]^(2/3 + m)*Sin[c + d*x])/(d*(2 + 3*m )) + (3*(C*(1 - 3*m) - A*(2 + 3*m))*Hypergeometric2F1[1/2, (4 - 3*m)/6, (1 0 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-4/3 + m)*Sin[c + d*x])/(d*(4 - 3*m)*(2 + 3*m)*Sqrt[Sin[c + d*x]^2]) - (3*B*Hypergeometric2F1[1/2, (1 - 3* m)/6, (7 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-1/3 + m)*Sin[c + d*x])/( d*(1 - 3*m)*Sqrt[Sin[c + d*x]^2])))/(b*Sec[c + d*x])^(1/3)
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]
\[\int \frac {\sec \left (d x +c \right )^{m} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
Input:
int(sec(d*x+c)^m*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(1/3),x)
Output:
int(sec(d*x+c)^m*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(1/3),x)
\[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^m*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(1/3 ),x, algorithm="fricas")
Output:
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*sec(d*x + c)), x)
\[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{m}{\left (c + d x \right )}}{\sqrt [3]{b \sec {\left (c + d x \right )}}}\, dx \] Input:
integrate(sec(d*x+c)**m*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(b*sec(d*x+c))**( 1/3),x)
Output:
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**m/(b*sec(c + d*x))**(1/3), x)
\[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^m*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(1/3 ),x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^m/(b*sec(d* x + c))^(1/3), x)
\[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^m*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(1/3 ),x, algorithm="giac")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^m/(b*sec(d* x + c))^(1/3), x)
Timed out. \[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \] Input:
int(((1/cos(c + d*x))^m*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(b/cos(c + d*x))^(1/3),x)
Output:
int(((1/cos(c + d*x))^m*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(b/cos(c + d*x))^(1/3), x)
\[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\frac {\left (\int \frac {\sec \left (d x +c \right )^{m}}{\sec \left (d x +c \right )^{\frac {1}{3}}}d x \right ) a +\left (\int \sec \left (d x +c \right )^{m} \sec \left (d x +c \right )^{\frac {5}{3}}d x \right ) c +\left (\int \sec \left (d x +c \right )^{m} \sec \left (d x +c \right )^{\frac {2}{3}}d x \right ) b}{b^{\frac {1}{3}}} \] Input:
int(sec(d*x+c)^m*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(1/3),x)
Output:
(int(sec(c + d*x)**m/sec(c + d*x)**(1/3),x)*a + int((sec(c + d*x)**m*sec(c + d*x)**2)/sec(c + d*x)**(1/3),x)*c + int((sec(c + d*x)**m*sec(c + d*x))/ sec(c + d*x)**(1/3),x)*b)/b**(1/3)