\(\int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [81]

Optimal result

Integrand size = 41, antiderivative size = 222 \[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 C \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \sin (c+d x)}{d (1+2 n)}-\frac {2 (A-C (1-2 n)+2 A n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (1+2 n) \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1-2 n),\frac {1}{4} (5-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) \sqrt {\sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \]

[Out]

2*C*(b*sec(d*x+c))^n*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(1+2*n)-2*(A-C*(1-2*n)+2*A*n)*hypergeom([1/2, 3/4-1/2*n],[7
/4-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(-4*n^2+4*n+3)/sec(d*x+c)^(3/2)/(sin(d*x+c)^2)^(1/2)-2*B
*hypergeom([1/2, 1/4-1/2*n],[5/4-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(1-2*n)/sec(d*x+c)^(1/2)/(
sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {20, 4132, 3857, 2722, 4131} \[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=-\frac {2 (2 A n+A-C (1-2 n)) \sin (c+d x) (b \sec (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right )}{d (3-2 n) (2 n+1) \sqrt {\sin ^2(c+d x)} \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 B \sin (c+d x) (b \sec (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1-2 n),\frac {1}{4} (5-2 n),\cos ^2(c+d x)\right )}{d (1-2 n) \sqrt {\sin ^2(c+d x)} \sqrt {\sec (c+d x)}}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (b \sec (c+d x))^n}{d (2 n+1)} \]

[In]

Int[((b*Sec[c + d*x])^n*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[Sec[c + d*x]],x]

[Out]

(2*C*Sqrt[Sec[c + d*x]]*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(1 + 2*n)) - (2*(A - C*(1 - 2*n) + 2*A*n)*Hypergeo
metric2F1[1/2, (3 - 2*n)/4, (7 - 2*n)/4, Cos[c + d*x]^2]*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(3 - 2*n)*(1 + 2*
n)*Sec[c + d*x]^(3/2)*Sqrt[Sin[c + d*x]^2]) - (2*B*Hypergeometric2F1[1/2, (1 - 2*n)/4, (5 - 2*n)/4, Cos[c + d*
x]^2]*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(1 - 2*n)*Sqrt[Sec[c + d*x]]*Sqrt[Sin[c + d*x]^2])

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[b^IntPart[n]*((b*v)^FracPart[n]/(a^IntPart[n]
*(a*v)^FracPart[n])), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 2722

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]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3857

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]

Rule 4131

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] + Dist[(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]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[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]

Rubi steps \begin{align*} \text {integral}& = \left (\sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{-\frac {1}{2}+n}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx \\ & = \left (\sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{-\frac {1}{2}+n}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx+\left (B \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{\frac {1}{2}+n}(c+d x) \, dx \\ & = \frac {2 C \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \sin (c+d x)}{d (1+2 n)}+\left (B \cos ^{\frac {1}{2}+n}(c+d x) \sqrt {\sec (c+d x)} (b \sec (c+d x))^n\right ) \int \cos ^{-\frac {1}{2}-n}(c+d x) \, dx+\frac {\left (\left (C \left (-\frac {1}{2}+n\right )+A \left (\frac {1}{2}+n\right )\right ) \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{-\frac {1}{2}+n}(c+d x) \, dx}{\frac {1}{2}+n} \\ & = \frac {2 C \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \sin (c+d x)}{d (1+2 n)}-\frac {2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1-2 n),\frac {1}{4} (5-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) \sqrt {\sec (c+d x)} \sqrt {\sin ^2(c+d x)}}+\frac {\left (\left (C \left (-\frac {1}{2}+n\right )+A \left (\frac {1}{2}+n\right )\right ) \cos ^{\frac {1}{2}+n}(c+d x) \sqrt {\sec (c+d x)} (b \sec (c+d x))^n\right ) \int \cos ^{\frac {1}{2}-n}(c+d x) \, dx}{\frac {1}{2}+n} \\ & = \frac {2 C \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \sin (c+d x)}{d (1+2 n)}-\frac {2 (A-C (1-2 n)+2 A n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (1+2 n) \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1-2 n),\frac {1}{4} (5-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) \sqrt {\sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.08 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.92 \[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 \csc (c+d x) (b \sec (c+d x))^n \left (A \left (3+8 n+4 n^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\sec ^2(c+d x)\right )+(-1+2 n) \left (B (3+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\sec ^2(c+d x)\right )+C (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\sec ^2(c+d x)\right )\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (-1+2 n) (1+2 n) (3+2 n) \sec ^{\frac {3}{2}}(c+d x)} \]

[In]

Integrate[((b*Sec[c + d*x])^n*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[Sec[c + d*x]],x]

[Out]

(2*Csc[c + d*x]*(b*Sec[c + d*x])^n*(A*(3 + 8*n + 4*n^2)*Hypergeometric2F1[1/2, (-1 + 2*n)/4, (3 + 2*n)/4, Sec[
c + d*x]^2] + (-1 + 2*n)*(B*(3 + 2*n)*Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + 2*n)/4, (5 + 2*n)/4, Sec[c + d*
x]^2] + C*(1 + 2*n)*Hypergeometric2F1[1/2, (3 + 2*n)/4, (7 + 2*n)/4, Sec[c + d*x]^2])*Sec[c + d*x]^2)*Sqrt[-Ta
n[c + d*x]^2])/(d*(-1 + 2*n)*(1 + 2*n)*(3 + 2*n)*Sec[c + d*x]^(3/2))

Maple [F]

\[\int \frac {\left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )}{\sqrt {\sec \left (d x +c \right )}}d x\]

[In]

int((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x)

[Out]

int((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x)

Fricas [F]

\[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]

[In]

integrate((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^n/sqrt(sec(d*x + c)), x)

Sympy [F]

\[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\sqrt {\sec {\left (c + d x \right )}}}\, dx \]

[In]

integrate((b*sec(d*x+c))**n*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(1/2),x)

[Out]

Integral((b*sec(c + d*x))**n*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)/sqrt(sec(c + d*x)), x)

Maxima [F]

\[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]

[In]

integrate((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^n/sqrt(sec(d*x + c)), x)

Giac [F]

\[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]

[In]

integrate((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^n/sqrt(sec(d*x + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

[In]

int(((b/cos(c + d*x))^n*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(1/2),x)

[Out]

int(((b/cos(c + d*x))^n*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(1/2), x)