3.78 \(\int \sec ^{\frac{5}{2}}(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=223 \[ \frac{2 (A (2 n+7)+C (2 n+5)) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (b \sec (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{4} (-2 n-3),\frac{1}{4} (1-2 n),\cos ^2(c+d x)\right )}{d (2 n+3) (2 n+7) \sqrt{\sin ^2(c+d x)}}+\frac{2 B \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (b \sec (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{4} (-2 n-5),\frac{1}{4} (-2 n-1),\cos ^2(c+d x)\right )}{d (2 n+5) \sqrt{\sin ^2(c+d x)}}+\frac{2 C \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) (b \sec (c+d x))^n}{d (2 n+7)} \]

[Out]

(2*C*Sec[c + d*x]^(7/2)*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(7 + 2*n)) + (2*(C*(5 + 2*n) + A*(7 + 2*n))*Hyperg
eometric2F1[1/2, (-3 - 2*n)/4, (1 - 2*n)/4, Cos[c + d*x]^2]*Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^n*Sin[c + d*x]
)/(d*(3 + 2*n)*(7 + 2*n)*Sqrt[Sin[c + d*x]^2]) + (2*B*Hypergeometric2F1[1/2, (-5 - 2*n)/4, (-1 - 2*n)/4, Cos[c
+ d*x]^2]*Sec[c + d*x]^(5/2)*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(5 + 2*n)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.195954, antiderivative size = 223, normalized size of antiderivative = 1., 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} \[ \frac{2 (A (2 n+7)+C (2 n+5)) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (-2 n-3);\frac{1}{4} (1-2 n);\cos ^2(c+d x)\right )}{d (2 n+3) (2 n+7) \sqrt{\sin ^2(c+d x)}}+\frac{2 B \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (-2 n-5);\frac{1}{4} (-2 n-1);\cos ^2(c+d x)\right )}{d (2 n+5) \sqrt{\sin ^2(c+d x)}}+\frac{2 C \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) (b \sec (c+d x))^n}{d (2 n+7)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(5/2)*(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]^(7/2)*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(7 + 2*n)) + (2*(C*(5 + 2*n) + A*(7 + 2*n))*Hyperg
eometric2F1[1/2, (-3 - 2*n)/4, (1 - 2*n)/4, Cos[c + d*x]^2]*Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^n*Sin[c + d*x]
)/(d*(3 + 2*n)*(7 + 2*n)*Sqrt[Sin[c + d*x]^2]) + (2*B*Hypergeometric2F1[1/2, (-5 - 2*n)/4, (-1 - 2*n)/4, Cos[c
+ d*x]^2]*Sec[c + d*x]^(5/2)*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(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 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]

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 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 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{5}{2}}(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{5}{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{5}{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{7}{2}+n}(c+d x) \, dx\\ &=\frac{2 C \sec ^{\frac{7}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (7+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{7}{2}-n}(c+d x) \, dx+\frac{\left (\left (C \left (\frac{5}{2}+n\right )+A \left (\frac{7}{2}+n\right )\right ) \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{\frac{5}{2}+n}(c+d x) \, dx}{\frac{7}{2}+n}\\ &=\frac{2 C \sec ^{\frac{7}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (7+2 n)}+\frac{2 B \, _2F_1\left (\frac{1}{2},\frac{1}{4} (-5-2 n);\frac{1}{4} (-1-2 n);\cos ^2(c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (5+2 n) \sqrt{\sin ^2(c+d x)}}+\frac{\left (\left (C \left (\frac{5}{2}+n\right )+A \left (\frac{7}{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{5}{2}-n}(c+d x) \, dx}{\frac{7}{2}+n}\\ &=\frac{2 C \sec ^{\frac{7}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (7+2 n)}+\frac{2 (C (5+2 n)+A (7+2 n)) \, _2F_1\left (\frac{1}{2},\frac{1}{4} (-3-2 n);\frac{1}{4} (1-2 n);\cos ^2(c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (3+2 n) (7+2 n) \sqrt{\sin ^2(c+d x)}}+\frac{2 B \, _2F_1\left (\frac{1}{2},\frac{1}{4} (-5-2 n);\frac{1}{4} (-1-2 n);\cos ^2(c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (5+2 n) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 7.87678, size = 493, normalized size = 2.21 \[ -\frac{i 2^{n+\frac{9}{2}} e^{2 i c-\frac{1}{2} i d (2 n+1) x} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{n+\frac{1}{2}} \left (1+e^{2 i (c+d x)}\right )^{n+\frac{1}{2}} \sec ^{-n-2}(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (e^{2 i c} \left (\frac{2 (A+2 C) e^{\frac{1}{2} i d (2 n+9) x} \text{Hypergeometric2F1}\left (n+\frac{9}{2},\frac{1}{4} (2 n+9),\frac{1}{4} (2 n+13),-e^{2 i (c+d x)}\right )}{2 n+9}+\frac{A e^{\frac{1}{2} i (4 c+d (2 n+13) x)} \text{Hypergeometric2F1}\left (n+\frac{9}{2},\frac{1}{4} (2 n+13),\frac{1}{4} (2 n+17),-e^{2 i (c+d x)}\right )}{2 n+13}+\frac{2 B e^{\frac{1}{2} i (2 c+d (2 n+11) x)} \text{Hypergeometric2F1}\left (n+\frac{9}{2},\frac{1}{4} (2 n+11),\frac{1}{4} (2 n+15),-e^{2 i (c+d x)}\right )}{2 n+11}\right )+\frac{A e^{\frac{1}{2} i d (2 n+5) x} \text{Hypergeometric2F1}\left (n+\frac{9}{2},\frac{1}{4} (2 n+5),\frac{1}{4} (2 n+9),-e^{2 i (c+d x)}\right )}{2 n+5}+\frac{2 B e^{\frac{1}{2} i (2 c+d (2 n+7) x)} \text{Hypergeometric2F1}\left (n+\frac{9}{2},\frac{1}{4} (2 n+7),\frac{1}{4} (2 n+11),-e^{2 i (c+d x)}\right )}{2 n+7}\right )}{d (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^(5/2)*(b*Sec[c + d*x])^n*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((-I)*2^(9/2 + n)*E^((2*I)*c - (I/2)*d*(1 + 2*n)*x)*(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*(5 + 2*n)*x)*Hypergeometric2F1[9/2 + 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[9/2 + n, (7 + 2*n
)/4, (11 + 2*n)/4, -E^((2*I)*(c + d*x))])/(7 + 2*n) + E^((2*I)*c)*((2*(A + 2*C)*E^((I/2)*d*(9 + 2*n)*x)*Hyperg
eometric2F1[9/2 + n, (9 + 2*n)/4, (13 + 2*n)/4, -E^((2*I)*(c + d*x))])/(9 + 2*n) + (2*B*E^((I/2)*(2*c + d*(11
+ 2*n)*x))*Hypergeometric2F1[9/2 + n, (11 + 2*n)/4, (15 + 2*n)/4, -E^((2*I)*(c + d*x))])/(11 + 2*n) + (A*E^((I
/2)*(4*c + d*(13 + 2*n)*x))*Hypergeometric2F1[9/2 + n, (13 + 2*n)/4, (17 + 2*n)/4, -E^((2*I)*(c + d*x))])/(13
+ 2*n)))*Sec[c + d*x]^(-2 - n)*(b*Sec[c + d*x])^n*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*C
os[c + d*x] + A*Cos[2*c + 2*d*x]))

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Maple [F]  time = 0.236, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}} \left ( b\sec \left ( dx+c \right ) \right ) ^{n} \left ( A+B\sec \left ( dx+c \right ) +C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

int(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \sec \left (d x + c\right )^{4} + B \sec \left (d x + c\right )^{3} + A \sec \left (d x + c\right )^{2}\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt{\sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^4 + B*sec(d*x + c)^3 + A*sec(d*x + c)^2)*(b*sec(d*x + c))^n*sqrt(sec(d*x + c)), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(5/2)*(b*sec(d*x+c))**n*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \sec \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^n*sec(d*x + c)^(5/2), x)