3.1.36 \(\int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n (A+C \sec ^2(c+d x)) \, dx\) [36]

Optimal. Leaf size=140 \[ \frac {2 C \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (3+2 n)}-\frac {2 (C+2 C n+A (3+2 n)) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (1-2 n);\frac {1}{4} (5-2 n);\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3+2 n) \sqrt {\sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \]

[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)

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 140, 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, 4131, 3857, 2722} \begin {gather*} \frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n}{d (2 n+3)}-\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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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 4131

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 \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \left (A+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\\ &=\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 {\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 {\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)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 2.16, size = 336, normalized size = 2.40 \begin {gather*} -\frac {i 2^{\frac {5}{2}+n} e^{-\frac {1}{2} i d (1+2 n) x} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{\frac {1}{2}+n} \left (1+e^{2 i (c+d x)}\right )^{\frac {1}{2}+n} \left (\frac {A e^{\frac {1}{2} i (d+2 d n) x} \, _2F_1\left (\frac {5}{2}+n,\frac {1}{4} (1+2 n);\frac {1}{4} (5+2 n);-e^{2 i (c+d x)}\right )}{d+2 d n}+\frac {e^{\frac {1}{2} i (4 c+d (5+2 n) x)} \left (2 (A+2 C) (9+2 n) \, _2F_1\left (\frac {5}{2}+n,\frac {1}{4} (5+2 n);\frac {1}{4} (9+2 n);-e^{2 i (c+d x)}\right )+A e^{2 i (c+d x)} (5+2 n) \, _2F_1\left (\frac {5}{2}+n,\frac {1}{4} (9+2 n);\frac {1}{4} (13+2 n);-e^{2 i (c+d x)}\right )\right )}{d (5+2 n) (9+2 n)}\right ) \sec ^{-2-n}(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right )}{A+2 C+A \cos (2 c+2 d x)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[Sec[c + d*x]]*(b*Sec[c + d*x])^n*(A + 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/2)*(4*c + d*(5 + 2*n)*x))*(2*(A + 2*C)*(9 + 2*n)*Hypergeometric2F1[5/2 + n, (5 + 2*n)/4, (9 + 2*n
)/4, -E^((2*I)*(c + d*x))] + A*E^((2*I)*(c + d*x))*(5 + 2*n)*Hypergeometric2F1[5/2 + n, (9 + 2*n)/4, (13 + 2*n
)/4, -E^((2*I)*(c + d*x))]))/(d*(5 + 2*n)*(9 + 2*n)))*Sec[c + d*x]^(-2 - n)*(b*Sec[c + d*x])^n*(A + C*Sec[c +
d*x]^2))/(E^((I/2)*d*(1 + 2*n)*x)*(A + 2*C + A*Cos[2*c + 2*d*x]))

________________________________________________________________________________________

Maple [F]
time = 0.42, size = 0, normalized size = 0.00 \[\int \left (b \sec \left (d x +c \right )\right )^{n} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right ) \left (\sqrt {\sec }\left (d x +c \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \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 \end {gather*}

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)

________________________________________________________________________________________