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

Optimal. Leaf size=221 \[ -\frac {2 (A (2 n+3)+2 C n+C) \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (1-2 n);\frac {1}{4} (5-2 n);\cos ^2(c+d x)\right )}{d (1-2 n) (2 n+3) \sqrt {\sin ^2(c+d x)} \sqrt {\sec (c+d x)}}+\frac {2 B \sin (c+d x) \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-2 n-1);\frac {1}{4} (3-2 n);\cos ^2(c+d x)\right )}{d (2 n+1) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n}{d (2 n+3)} \]

[Out]

2*C*sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(3+2*n)-2*(C+2*C*n+A*(3+2*n))*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/(-4*n^2-4*n+3)/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 = 0.18, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {20, 4047, 3772, 2643, 4046} \[ -\frac {2 (A (2 n+3)+2 C n+C) \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (1-2 n);\frac {1}{4} (5-2 n);\cos ^2(c+d x)\right )}{d (1-2 n) (2 n+3) \sqrt {\sin ^2(c+d x)} \sqrt {\sec (c+d x)}}+\frac {2 B \sin (c+d x) \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-2 n-1);\frac {1}{4} (3-2 n);\cos ^2(c+d x)\right )}{d (2 n+1) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n}{d (2 n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*C*Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(3 + 2*n)) - (2*(C + 2*C*n + A*(3 + 2*n))*Hypergeo
metric2F1[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)*(3 + 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 4046

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 4047

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*} \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \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+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 {3}{2}+n}(c+d x) \, dx\\ &=\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (3+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 {3}{2}-n}(c+d x) \, dx+\frac {\left (\left (C \left (\frac {1}{2}+n\right )+A \left (\frac {3}{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 {3}{2}+n}\\ &=\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (3+2 n)}+\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)}}+\frac {\left (\left (C \left (\frac {1}{2}+n\right )+A \left (\frac {3}{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 {3}{2}+n}\\ &=\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (3+2 n)}-\frac {2 (C+2 C n+A (3+2 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 ) (b \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3+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 [C]  time = 8.01, size = 492, normalized size = 2.23 \[ -\frac {i 2^{n+\frac {5}{2}} e^{-\frac {1}{2} i d (2 n+1) x} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{n+\frac {1}{2}} \left (1+e^{2 i (c+d x)}\right )^{n+\frac {1}{2}} \sec ^{-n-2}(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {e^{i c} \left (e^{i c} \left (\frac {2 (A+2 C) e^{\frac {1}{2} i d (2 n+5) x} \, _2F_1\left (n+\frac {5}{2},\frac {1}{4} (2 n+5);\frac {1}{4} (2 n+9);-e^{2 i (c+d x)}\right )}{2 n+5}+\frac {A e^{\frac {1}{2} i (4 c+d (2 n+9) x)} \, _2F_1\left (n+\frac {5}{2},\frac {1}{4} (2 n+9);\frac {1}{4} (2 n+13);-e^{2 i (c+d x)}\right )}{2 n+9}+\frac {2 B e^{\frac {1}{2} i (2 c+d (2 n+7) x)} \, _2F_1\left (n+\frac {5}{2},\frac {1}{4} (2 n+7);\frac {1}{4} (2 n+11);-e^{2 i (c+d x)}\right )}{2 n+7}\right )+\frac {2 B e^{\frac {1}{2} i d (2 n+3) x} \, _2F_1\left (n+\frac {5}{2},\frac {1}{4} (2 n+3);\frac {1}{4} (2 n+7);-e^{2 i (c+d x)}\right )}{2 n+3}\right )}{d}+\frac {A e^{\frac {1}{2} i x (2 d n+d)} \, _2F_1\left (n+\frac {5}{2},\frac {1}{4} (2 n+1);\frac {1}{4} (2 n+5);-e^{2 i (c+d x)}\right )}{2 d n+d}\right )}{A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*2^(5/2 + n)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(1/2 + n)*(1 + E^((2*I)*(c + d*x)))^(1/2 + n)*((
A*E^((I/2)*(d + 2*d*n)*x)*Hypergeometric2F1[5/2 + n, (1 + 2*n)/4, (5 + 2*n)/4, -E^((2*I)*(c + d*x))])/(d + 2*d
*n) + (E^(I*c)*((2*B*E^((I/2)*d*(3 + 2*n)*x)*Hypergeometric2F1[5/2 + n, (3 + 2*n)/4, (7 + 2*n)/4, -E^((2*I)*(c
+ d*x))])/(3 + 2*n) + E^(I*c)*((2*(A + 2*C)*E^((I/2)*d*(5 + 2*n)*x)*Hypergeometric2F1[5/2 + n, (5 + 2*n)/4, (
9 + 2*n)/4, -E^((2*I)*(c + d*x))])/(5 + 2*n) + (2*B*E^((I/2)*(2*c + d*(7 + 2*n)*x))*Hypergeometric2F1[5/2 + n,
(7 + 2*n)/4, (11 + 2*n)/4, -E^((2*I)*(c + d*x))])/(7 + 2*n) + (A*E^((I/2)*(4*c + d*(9 + 2*n)*x))*Hypergeometr
ic2F1[5/2 + n, (9 + 2*n)/4, (13 + 2*n)/4, -E^((2*I)*(c + d*x))])/(9 + 2*n))))/d)*Sec[c + d*x]^(-2 - n)*(b*Sec[
c + d*x])^n*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(E^((I/2)*d*(1 + 2*n)*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*
Cos[2*c + 2*d*x]))

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt {\sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt {\sec \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 1.29, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right ) \left (\sqrt {\sec }\left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt {\sec \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

Timed out

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