3.35 \(\int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=142 \[ \frac {2 (A (2 n+5)+C (2 n+3)) \sin (c+d x) \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-2 n-1);\frac {1}{4} (3-2 n);\cos ^2(c+d x)\right )}{d (2 n+1) (2 n+5) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n}{d (2 n+5)} \]

[Out]

2*C*sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(5+2*n)+2*(C*(3+2*n)+A*(5+2*n))*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/(4*n^2+12*n+5)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 142, 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} \[ \frac {2 (A (2 n+5)+C (2 n+3)) \sin (c+d x) \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-2 n-1);\frac {1}{4} (3-2 n);\cos ^2(c+d x)\right )}{d (2 n+1) (2 n+5) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n}{d (2 n+5)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2),x]

[Out]

(2*C*Sec[c + d*x]^(5/2)*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(5 + 2*n)) + (2*(C*(3 + 2*n) + A*(5 + 2*n))*Hyperg
eometric2F1[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)*(5 + 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]

Rubi steps

\begin {align*} \int \sec ^{\frac {3}{2}}(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 {3}{2}+n}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 C \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (5+2 n)}+\frac {\left (\left (C \left (\frac {3}{2}+n\right )+A \left (\frac {5}{2}+n\right )\right ) \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{\frac {3}{2}+n}(c+d x) \, dx}{\frac {5}{2}+n}\\ &=\frac {2 C \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (5+2 n)}+\frac {\left (\left (C \left (\frac {3}{2}+n\right )+A \left (\frac {5}{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 {3}{2}-n}(c+d x) \, dx}{\frac {5}{2}+n}\\ &=\frac {2 C \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (5+2 n)}+\frac {2 (C (3+2 n)+A (5+2 n)) \, _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) (5+2 n) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 2.37, size = 303, normalized size = 2.13 \[ -\frac {i 2^{n+\frac {7}{2}} e^{-\frac {1}{2} i (2 n+5) (c+d x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{n+\frac {5}{2}} \sec ^{-n-2}(c+d x) \left (A+C \sec ^2(c+d x)\right ) (b \sec (c+d x))^n \left (\frac {2 (A+2 C) e^{\frac {1}{2} i (2 n+7) (c+d x)} \, _2F_1\left (1,\frac {1}{4} (-2 n-3);\frac {1}{4} (2 n+11);-e^{2 i (c+d x)}\right )}{2 n+7}+\frac {A e^{\frac {1}{2} i (2 n+3) (c+d x)} \, _2F_1\left (1,\frac {1}{4} (-2 n-7);\frac {1}{4} (2 n+7);-e^{2 i (c+d x)}\right )}{2 n+3}+\frac {A e^{\frac {1}{2} i (2 n+11) (c+d x)} \, _2F_1\left (1,\frac {1}{4} (1-2 n);\frac {1}{4} (2 n+15);-e^{2 i (c+d x)}\right )}{2 n+11}\right )}{d (A \cos (2 c+2 d x)+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2),x]

[Out]

((-I)*2^(7/2 + n)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(5/2 + n)*((A*E^((I/2)*(3 + 2*n)*(c + d*x))*Hype
rgeometric2F1[1, (-7 - 2*n)/4, (7 + 2*n)/4, -E^((2*I)*(c + d*x))])/(3 + 2*n) + (2*(A + 2*C)*E^((I/2)*(7 + 2*n)
*(c + d*x))*Hypergeometric2F1[1, (-3 - 2*n)/4, (11 + 2*n)/4, -E^((2*I)*(c + d*x))])/(7 + 2*n) + (A*E^((I/2)*(1
1 + 2*n)*(c + d*x))*Hypergeometric2F1[1, (1 - 2*n)/4, (15 + 2*n)/4, -E^((2*I)*(c + d*x))])/(11 + 2*n))*Sec[c +
d*x]^(-2 - n)*(b*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2))/(d*E^((I/2)*(5 + 2*n)*(c + d*x))*(A + 2*C + A*Cos[2*
c + 2*d*x]))

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{3} + A \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt {\sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^3 + A*sec(d*x + c))*(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 \[ \int {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^n*sec(d*x + c)^(3/2), x)

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maple [F]  time = 1.52, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{\frac {3}{2}}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{n} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x)

[Out]

int(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^n*sec(d*x + c)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]

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))^(3/2),x)

[Out]

int((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^n*(1/cos(c + d*x))^(3/2), 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(sec(d*x+c)**(3/2)*(b*sec(d*x+c))**n*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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