\(\int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n (A+C \sec ^2(c+d x)) \, dx\) [34]

Optimal result

Integrand size = 33, antiderivative size = 142 \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 C \sec ^{\frac {7}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (7+2 n)}+\frac {2 (C (5+2 n)+A (7+2 n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3-2 n),\frac {1}{4} (1-2 n),\cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (3+2 n) (7+2 n) \sqrt {\sin ^2(c+d x)}} \]

[Out]

2*C*sec(d*x+c)^(7/2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(7+2*n)+2*(C*(5+2*n)+A*(7+2*n))*hypergeom([1/2, -3/4-1/2*n]
,[1/4-1/2*n],cos(d*x+c)^2)*sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(4*n^2+20*n+21)/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {20, 4131, 3857, 2722} \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (A (2 n+7)+C (2 n+5)) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-2 n-3),\frac {1}{4} (1-2 n),\cos ^2(c+d x)\right )}{d (2 n+3) (2 n+7) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (b \sec (c+d x))^n}{d (2 n+7)} \]

[In]

Int[Sec[c + d*x]^(5/2)*(b*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2),x]

[Out]

(2*C*Sec[c + d*x]^(7/2)*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(7 + 2*n)) + (2*(C*(5 + 2*n) + A*(7 + 2*n))*Hyperg
eometric2F1[1/2, (-3 - 2*n)/4, (1 - 2*n)/4, Cos[c + d*x]^2]*Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^n*Sin[c + d*x]
)/(d*(3 + 2*n)*(7 + 2*n)*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]

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

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00 \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \csc (c+d x) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \left (A (9+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5+2 n),\frac {1}{4} (9+2 n),\sec ^2(c+d x)\right )+C (5+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (9+2 n),\frac {1}{4} (13+2 n),\sec ^2(c+d x)\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (5+2 n) (9+2 n)} \]

[In]

Integrate[Sec[c + d*x]^(5/2)*(b*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2),x]

[Out]

(2*Csc[c + d*x]*Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^n*(A*(9 + 2*n)*Hypergeometric2F1[1/2, (5 + 2*n)/4, (9 + 2*
n)/4, Sec[c + d*x]^2] + C*(5 + 2*n)*Hypergeometric2F1[1/2, (9 + 2*n)/4, (13 + 2*n)/4, Sec[c + d*x]^2]*Sec[c +
d*x]^2)*Sqrt[-Tan[c + d*x]^2])/(d*(5 + 2*n)*(9 + 2*n))

Maple [F]

\[\int \sec \left (d x +c \right )^{\frac {5}{2}} \left (b \sec \left (d x +c \right )\right )^{n} \left (A +C \sec \left (d x +c \right )^{2}\right )d x\]

[In]

int(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x)

[Out]

int(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x)

Fricas [F]

\[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^4 + A*sec(d*x + c)^2)*(b*sec(d*x + c))^n*sqrt(sec(d*x + c)), x)

Sympy [F(-1)]

Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**(5/2)*(b*sec(d*x+c))**n*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [F]

\[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^n*sec(d*x + c)^(5/2), x)

Giac [F]

\[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^n*sec(d*x + c)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

[In]

int((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^n*(1/cos(c + d*x))^(5/2),x)

[Out]

int((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^n*(1/cos(c + d*x))^(5/2), x)