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 \[ \frac {2 B \sin (c+d x) \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-2 n-1);\frac {1}{4} (3-2 n);\cos ^2(c+d x)\right )}{d (2 n+1) \sqrt {\sin ^2(c+d x)}}-\frac {2 A \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (1-2 n);\frac {1}{4} (5-2 n);\cos ^2(c+d x)\right )}{d (1-2 n) \sqrt {\sin ^2(c+d x)} \sqrt {\sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 163, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.24, size = 140, normalized size = 0.86 \[ \frac {2 \sqrt {-\tan ^2(c+d x)} \csc (c+d x) (b \sec (c+d x))^n \left (A (2 n+3) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+1);\frac {1}{4} (2 n+5);\sec ^2(c+d x)\right )+B (2 n+1) \sec (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+3);\frac {1}{4} (2 n+7);\sec ^2(c+d x)\right )\right )}{d (2 n+1) (2 n+3) \sqrt {\sec (c+d x)}} \]

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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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt {\sec \left (d x + c\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt {\sec \left (d x + c\right )}\,{d x} \]

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)

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maple [F]  time = 1.57, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )\right ) \left (\sqrt {\sec }\left (d x +c \right )\right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt {\sec \left (d x + c\right )}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + B \sec {\left (c + d x \right )}\right ) \sqrt {\sec {\left (c + d x \right )}}\, dx \]

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]

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

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