3.22 \(\int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=146 \[ \frac {3 C \sin (c+d x) (b \sec (c+d x))^{2/3} \sec ^{m+1}(c+d x)}{d (3 m+5)}-\frac {3 (A (3 m+5)+C (3 m+2)) \sin (c+d x) (b \sec (c+d x))^{2/3} \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (1-3 m);\frac {1}{6} (7-3 m);\cos ^2(c+d x)\right )}{d (1-3 m) (3 m+5) \sqrt {\sin ^2(c+d x)}} \]

[Out]

3*C*sec(d*x+c)^(1+m)*(b*sec(d*x+c))^(2/3)*sin(d*x+c)/d/(5+3*m)-3*(C*(2+3*m)+A*(5+3*m))*hypergeom([1/2, 1/6-1/2
*m],[7/6-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-1+m)*(b*sec(d*x+c))^(2/3)*sin(d*x+c)/d/(-9*m^2-12*m+5)/(sin(d*x+c)^
2)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 146, 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} \[ \frac {3 C \sin (c+d x) (b \sec (c+d x))^{2/3} \sec ^{m+1}(c+d x)}{d (3 m+5)}-\frac {3 (A (3 m+5)+C (3 m+2)) \sin (c+d x) (b \sec (c+d x))^{2/3} \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (1-3 m);\frac {1}{6} (7-3 m);\cos ^2(c+d x)\right )}{d (1-3 m) (3 m+5) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*Sec[c + d*x]^(1 + m)*(b*Sec[c + d*x])^(2/3)*Sin[c + d*x])/(d*(5 + 3*m)) - (3*(C*(2 + 3*m) + A*(5 + 3*m))*
Hypergeometric2F1[1/2, (1 - 3*m)/6, (7 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*(b*Sec[c + d*x])^(2/3)*
Sin[c + d*x])/(d*(1 - 3*m)*(5 + 3*m)*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 ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {(b \sec (c+d x))^{2/3} \int \sec ^{\frac {2}{3}+m}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx}{\sec ^{\frac {2}{3}}(c+d x)}\\ &=\frac {3 C \sec ^{1+m}(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (5+3 m)}+\frac {\left (\left (C \left (\frac {2}{3}+m\right )+A \left (\frac {5}{3}+m\right )\right ) (b \sec (c+d x))^{2/3}\right ) \int \sec ^{\frac {2}{3}+m}(c+d x) \, dx}{\left (\frac {5}{3}+m\right ) \sec ^{\frac {2}{3}}(c+d x)}\\ &=\frac {3 C \sec ^{1+m}(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (5+3 m)}+\frac {\left (\left (C \left (\frac {2}{3}+m\right )+A \left (\frac {5}{3}+m\right )\right ) \cos ^{\frac {2}{3}+m}(c+d x) \sec ^m(c+d x) (b \sec (c+d x))^{2/3}\right ) \int \cos ^{-\frac {2}{3}-m}(c+d x) \, dx}{\frac {5}{3}+m}\\ &=\frac {3 C \sec ^{1+m}(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (5+3 m)}-\frac {3 (C (2+3 m)+A (5+3 m)) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (1-3 m);\frac {1}{6} (7-3 m);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (1-3 m) (5+3 m) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 3.23, size = 303, normalized size = 2.08 \[ -\frac {3 i 2^{m+\frac {5}{3}} e^{-\frac {1}{3} i (3 m+5) (c+d x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+\frac {5}{3}} (b \sec (c+d x))^{2/3} \left (A+C \sec ^2(c+d x)\right ) \left (\frac {2 (A+2 C) e^{\frac {1}{3} i (3 m+8) (c+d x)} \, _2F_1\left (1,\frac {1}{6} (-3 m-2);\frac {m}{2}+\frac {7}{3};-e^{2 i (c+d x)}\right )}{3 m+8}+\frac {A e^{\frac {1}{3} i (3 m+2) (c+d x)} \, _2F_1\left (1,\frac {1}{6} (-3 m-8);\frac {1}{6} (3 m+8);-e^{2 i (c+d x)}\right )}{3 m+2}+\frac {A e^{\frac {1}{3} i (3 m+14) (c+d x)} \, _2F_1\left (1,\frac {1}{6} (4-3 m);\frac {1}{6} (3 m+20);-e^{2 i (c+d x)}\right )}{3 m+14}\right )}{d \sec ^{\frac {8}{3}}(c+d x) (A \cos (2 c+2 d x)+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-3*I)*2^(5/3 + m)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(5/3 + m)*((A*E^((I/3)*(2 + 3*m)*(c + d*x))*Hy
pergeometric2F1[1, (-8 - 3*m)/6, (8 + 3*m)/6, -E^((2*I)*(c + d*x))])/(2 + 3*m) + (2*(A + 2*C)*E^((I/3)*(8 + 3*
m)*(c + d*x))*Hypergeometric2F1[1, (-2 - 3*m)/6, 7/3 + m/2, -E^((2*I)*(c + d*x))])/(8 + 3*m) + (A*E^((I/3)*(14
+ 3*m)*(c + d*x))*Hypergeometric2F1[1, (4 - 3*m)/6, (20 + 3*m)/6, -E^((2*I)*(c + d*x))])/(14 + 3*m))*(b*Sec[c
+ d*x])^(2/3)*(A + C*Sec[c + d*x]^2))/(d*E^((I/3)*(5 + 3*m)*(c + d*x))*(A + 2*C + A*Cos[2*c + 2*d*x])*Sec[c +
d*x]^(8/3))

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.44, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{m}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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