\(\int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [42]

Optimal result

Integrand size = 31, antiderivative size = 163 \[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5-2 n),\frac {1}{4} (9-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (5-2 n) \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-2*A*hypergeom([1/2, 5/4-1/2*n],[9/4-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(5-2*n)/sec(d*x+c)^(5/
2)/(sin(d*x+c)^2)^(1/2)-2*B*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
/(3-2*n)/sec(d*x+c)^(3/2)/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {20, 3872, 3857, 2722} \[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 A \sin (c+d x) (b \sec (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5-2 n),\frac {1}{4} (9-2 n),\cos ^2(c+d x)\right )}{d (5-2 n) \sqrt {\sin ^2(c+d x)} \sec ^{\frac {5}{2}}(c+d x)}-\frac {2 B \sin (c+d x) (b \sec (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt {\sin ^2(c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \]

[In]

Int[((b*Sec[c + d*x])^n*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(3/2),x]

[Out]

(-2*A*Hypergeometric2F1[1/2, (5 - 2*n)/4, (9 - 2*n)/4, Cos[c + d*x]^2]*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(5
- 2*n)*Sec[c + d*x]^(5/2)*Sqrt[Sin[c + d*x]^2]) - (2*B*Hypergeometric2F1[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)*Sec[c + d*x]^(3/2)*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 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{-\frac {3}{2}+n}(c+d x) (A+B \sec (c+d x)) \, dx \\ & = \left (A \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{-\frac {3}{2}+n}(c+d x) \, 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 \\ & = \left (A \cos ^{\frac {1}{2}+n}(c+d x) \sqrt {\sec (c+d x)} (b \sec (c+d x))^n\right ) \int \cos ^{\frac {3}{2}-n}(c+d x) \, dx+\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 {2 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5-2 n),\frac {1}{4} (9-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (5-2 n) \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \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) \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.86 \[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \csc (c+d x) (b \sec (c+d x))^n \left (A (-1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\sec ^2(c+d x)\right )+B (-3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\sec ^2(c+d x)\right ) \sec (c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (-3+2 n) (-1+2 n) \sec ^{\frac {5}{2}}(c+d x)} \]

[In]

Integrate[((b*Sec[c + d*x])^n*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(3/2),x]

[Out]

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

Maple [F]

\[\int \frac {\left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )\right )}{\sec \left (d x +c \right )^{\frac {3}{2}}}d x\]

[In]

int((b*sec(d*x+c))^n*(A+B*sec(d*x+c))/sec(d*x+c)^(3/2),x)

[Out]

int((b*sec(d*x+c))^n*(A+B*sec(d*x+c))/sec(d*x+c)^(3/2),x)

Fricas [F]

\[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((b*sec(d*x+c))^n*(A+B*sec(d*x+c))/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

integral((B*sec(d*x + c) + A)*(b*sec(d*x + c))^n/sec(d*x + c)^(3/2), x)

Sympy [F]

\[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + B \sec {\left (c + d x \right )}\right )}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

[In]

integrate((b*sec(d*x+c))**n*(A+B*sec(d*x+c))/sec(d*x+c)**(3/2),x)

[Out]

Integral((b*sec(c + d*x))**n*(A + B*sec(c + d*x))/sec(c + d*x)**(3/2), x)

Maxima [F]

\[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((b*sec(d*x+c))^n*(A+B*sec(d*x+c))/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^n/sec(d*x + c)^(3/2), x)

Giac [F]

\[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((b*sec(d*x+c))^n*(A+B*sec(d*x+c))/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^n/sec(d*x + c)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int(((A + B/cos(c + d*x))*(b/cos(c + d*x))^n)/(1/cos(c + d*x))^(3/2),x)

[Out]

int(((A + B/cos(c + d*x))*(b/cos(c + d*x))^n)/(1/cos(c + d*x))^(3/2), x)