3.40 \(\int \sqrt{\sec (c+d x)} (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=163 \[ \frac{2 B \sin (c+d x) \sqrt{\sec (c+d x)} (b \sec (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{4} (-2 n-1),\frac{1}{4} (3-2 n),\cos ^2(c+d x)\right )}{d (2 n+1) \sqrt{\sin ^2(c+d x)}}-\frac{2 A \sin (c+d x) (b \sec (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{4} (1-2 n),\frac{1}{4} (5-2 n),\cos ^2(c+d x)\right )}{d (1-2 n) \sqrt{\sin ^2(c+d x)} \sqrt{\sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.111319, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {20, 3787, 3772, 2643} \[ \frac{2 B \sin (c+d x) \sqrt{\sec (c+d x)} (b \sec (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (-2 n-1);\frac{1}{4} (3-2 n);\cos ^2(c+d x)\right )}{d (2 n+1) \sqrt{\sin ^2(c+d x)}}-\frac{2 A \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (1-2 n);\frac{1}{4} (5-2 n);\cos ^2(c+d x)\right )}{d (1-2 n) \sqrt{\sin ^2(c+d x)} \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*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]) + (2*B*Hypergeometric2F1[1/2, (-1 - 2*n)/4, (3 - 2*n)/4, Cos[c
+ d*x]^2]*Sqrt[Sec[c + d*x]]*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(1 + 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 3787

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]

Rule 3772

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 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin{align*} \int \sqrt{\sec (c+d x)} (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx &=\left (\sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{\frac{1}{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{1}{2}+n}(c+d x) \, dx+\left (B \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{\frac{3}{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{1}{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{3}{2}-n}(c+d x) \, dx\\ &=-\frac{2 A \, _2F_1\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{2 B \, _2F_1\left (\frac{1}{2},\frac{1}{4} (-1-2 n);\frac{1}{4} (3-2 n);\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} (b \sec (c+d x))^n \sin (c+d x)}{d (1+2 n) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.232687, size = 140, normalized size = 0.86 \[ \frac{2 \sqrt{-\tan ^2(c+d x)} \csc (c+d x) (b \sec (c+d x))^n \left (A (2 n+3) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{4} (2 n+1),\frac{1}{4} (2 n+5),\sec ^2(c+d x)\right )+B (2 n+1) \sec (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{4} (2 n+3),\frac{1}{4} (2 n+7),\sec ^2(c+d x)\right )\right )}{d (2 n+1) (2 n+3) \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.179, size = 0, normalized size = 0. \begin{align*} \int \left ( b\sec \left ( dx+c \right ) \right ) ^{n} \left ( A+B\sec \left ( dx+c \right ) \right ) \sqrt{\sec \left ( dx+c \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (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 )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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