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 \[ \frac {2 (A (2 n+7)+C (2 n+5)) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-2 n-3);\frac {1}{4} (1-2 n);\cos ^2(c+d x)\right )}{d (2 n+3) (2 n+7) \sqrt {\sin ^2(c+d x)}}+\frac {2 B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-2 n-5);\frac {1}{4} (-2 n-1);\cos ^2(c+d x)\right )}{d (2 n+5) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (b \sec (c+d x))^n}{d (2 n+7)} \]

[Out]

2*C*sec(d*x+c)^(7/2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(7+2*n)+2*(C*(5+2*n)+A*(7+2*n))*hypergeom([1/2, -3/4-1/2*n]
,[1/4-1/2*n],cos(d*x+c)^2)*sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(4*n^2+20*n+21)/(sin(d*x+c)^2)^(1/2)
+2*B*hypergeom([1/2, -5/4-1/2*n],[-1/4-1/2*n],cos(d*x+c)^2)*sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(5+
2*n)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 223, 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} \[ \frac {2 (A (2 n+7)+C (2 n+5)) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-2 n-3);\frac {1}{4} (1-2 n);\cos ^2(c+d x)\right )}{d (2 n+3) (2 n+7) \sqrt {\sin ^2(c+d x)}}+\frac {2 B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-2 n-5);\frac {1}{4} (-2 n-1);\cos ^2(c+d x)\right )}{d (2 n+5) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (b \sec (c+d x))^n}{d (2 n+7)} \]

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 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 \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*}

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Mathematica [F]  time = 180.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[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]

$Aborted

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{4} + B \sec \left (d x + c\right )^{3} + A \sec \left (d x + c\right )^{2}\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt {\sec \left (d x + c\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

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)

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maple [F]  time = 1.54, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{\frac {5}{2}}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

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 = 0.00, 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)^(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cos(c + d*x))^n*(1/cos(c + d*x))^(5/2)*(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))^(5/2)*(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(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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