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

Optimal. Leaf size=141 \[ \frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (b \sec (c+d x))^n}{d (2 n+1)}-\frac {2 (2 A n+A-C (1-2 n)) \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (3-2 n);\frac {1}{4} (7-2 n);\cos ^2(c+d x)\right )}{d (3-2 n) (2 n+1) \sqrt {\sin ^2(c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

2*C*(b*sec(d*x+c))^n*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(1+2*n)-2*(A-C*(1-2*n)+2*A*n)*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/(-4*n^2+4*n+3)/sec(d*x+c)^(3/2)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {20, 4046, 3772, 2643} \[ \frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (b \sec (c+d x))^n}{d (2 n+1)}-\frac {2 (2 A n+A-C (1-2 n)) \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (3-2 n);\frac {1}{4} (7-2 n);\cos ^2(c+d x)\right )}{d (3-2 n) (2 n+1) \sqrt {\sin ^2(c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*C*Sqrt[Sec[c + d*x]]*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(1 + 2*n)) - (2*(A - C*(1 - 2*n) + 2*A*n)*Hypergeo
metric2F1[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)*(1 + 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 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]

Rubi steps

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

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Mathematica [C]  time = 4.87, size = 311, normalized size = 2.21 \[ -\frac {i 2^{n+\frac {3}{2}} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{n-\frac {1}{2}} \left (1+e^{2 i (c+d x)}\right )^{n-\frac {1}{2}} \sec ^{-n-2}(c+d x) \left (A+C \sec ^2(c+d x)\right ) (b \sec (c+d x))^n \left ((2 n-1) e^{2 i (c+d x)} \left (2 (2 n+7) (A+2 C) \, _2F_1\left (n+\frac {3}{2},\frac {1}{4} (2 n+3);\frac {1}{4} (2 n+7);-e^{2 i (c+d x)}\right )+A (2 n+3) e^{2 i (c+d x)} \, _2F_1\left (n+\frac {3}{2},\frac {1}{4} (2 n+7);\frac {1}{4} (2 n+11);-e^{2 i (c+d x)}\right )\right )+A \left (4 n^2+20 n+21\right ) \, _2F_1\left (n+\frac {3}{2},\frac {1}{4} (2 n-1);\frac {1}{4} (2 n+3);-e^{2 i (c+d x)}\right )\right )}{d (2 n-1) (2 n+3) (2 n+7) (A \cos (2 c+2 d x)+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*2^(3/2 + 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*(21 + 20*n + 4*n^2)*Hypergeometric2F1[3/2 + n, (-1 + 2*n)/4, (3 + 2*n)/4, -E^((2*I)*(c + d*x))] + E^((2*I)*
(c + d*x))*(-1 + 2*n)*(2*(A + 2*C)*(7 + 2*n)*Hypergeometric2F1[3/2 + n, (3 + 2*n)/4, (7 + 2*n)/4, -E^((2*I)*(c
+ d*x))] + A*E^((2*I)*(c + d*x))*(3 + 2*n)*Hypergeometric2F1[3/2 + n, (7 + 2*n)/4, (11 + 2*n)/4, -E^((2*I)*(c
+ d*x))]))*Sec[c + d*x]^(-2 - n)*(b*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2))/(d*(-1 + 2*n)*(3 + 2*n)*(7 + 2*n)
*(A + 2*C + A*Cos[2*c + 2*d*x]))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sqrt {\sec \left (d x + c\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.76, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \sec \left (d x +c \right )\right )^{n} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )}{\sqrt {\sec \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*sec(d*x + c)^2 + 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 \[ \int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*sec(c + d*x))**n*(A + C*sec(c + d*x)**2)/sqrt(sec(c + d*x)), x)

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