3.1.25 \(\int \frac {\sec ^m(c+d x) (A+C \sec ^2(c+d x))}{(b \sec (c+d x))^{2/3}} \, dx\) [25]

Optimal. Leaf size=145 \[ \frac {3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+3 m) (b \sec (c+d x))^{2/3}}-\frac {3 (A-C (2-3 m)+3 A m) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (5-3 m);\frac {1}{6} (11-3 m);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (5-3 m) (1+3 m) (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}} \]

[Out]

3*C*sec(d*x+c)^(1+m)*sin(d*x+c)/d/(1+3*m)/(b*sec(d*x+c))^(2/3)-3*(A-C*(2-3*m)+3*A*m)*hypergeom([1/2, 5/6-1/2*m
],[11/6-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-1+m)*sin(d*x+c)/d/(-9*m^2+12*m+5)/(b*sec(d*x+c))^(2/3)/(sin(d*x+c)^2
)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 145, 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 {3 C \sin (c+d x) \sec ^{m+1}(c+d x)}{d (3 m+1) (b \sec (c+d x))^{2/3}}-\frac {3 (3 A m+A-C (2-3 m)) \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (5-3 m);\frac {1}{6} (11-3 m);\cos ^2(c+d x)\right )}{d (5-3 m) (3 m+1) \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*Sec[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(1 + 3*m)*(b*Sec[c + d*x])^(2/3)) - (3*(A - C*(2 - 3*m) + 3*A*m)*Hy
pergeometric2F1[1/2, (5 - 3*m)/6, (11 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*Sin[c + d*x])/(d*(5 - 3*
m)*(1 + 3*m)*(b*Sec[c + d*x])^(2/3)*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 \frac {\sec ^m(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx &=\frac {\sec ^{\frac {2}{3}}(c+d x) \int \sec ^{-\frac {2}{3}+m}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx}{(b \sec (c+d x))^{2/3}}\\ &=\frac {3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+3 m) (b \sec (c+d x))^{2/3}}+\frac {\left (\left (C \left (-\frac {2}{3}+m\right )+A \left (\frac {1}{3}+m\right )\right ) \sec ^{\frac {2}{3}}(c+d x)\right ) \int \sec ^{-\frac {2}{3}+m}(c+d x) \, dx}{\left (\frac {1}{3}+m\right ) (b \sec (c+d x))^{2/3}}\\ &=\frac {3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+3 m) (b \sec (c+d x))^{2/3}}+\frac {\left (\left (C \left (-\frac {2}{3}+m\right )+A \left (\frac {1}{3}+m\right )\right ) \cos ^{\frac {1}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{\frac {2}{3}-m}(c+d x) \, dx}{\left (\frac {1}{3}+m\right ) (b \sec (c+d x))^{2/3}}\\ &=\frac {3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+3 m) (b \sec (c+d x))^{2/3}}-\frac {3 (A-C (2-3 m)+3 A m) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (5-3 m);\frac {1}{6} (11-3 m);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (5-3 m) (1+3 m) (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 4.39, size = 311, normalized size = 2.14 \begin {gather*} -\frac {3 i 2^{\frac {1}{3}+m} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-\frac {2}{3}+m} \left (1+e^{2 i (c+d x)}\right )^{-\frac {2}{3}+m} \left (A e^{4 i (c+d x)} \left (-8+6 m+9 m^2\right ) \, _2F_1\left (\frac {5}{3}+\frac {m}{2},\frac {4}{3}+m;\frac {8}{3}+\frac {m}{2};-e^{2 i (c+d x)}\right )+(10+3 m) \left (A (4+3 m) \, _2F_1\left (\frac {4}{3}+m,\frac {1}{6} (-2+3 m);\frac {1}{6} (4+3 m);-e^{2 i (c+d x)}\right )+2 (A+2 C) e^{2 i (c+d x)} (-2+3 m) \, _2F_1\left (\frac {4}{3}+m,\frac {1}{6} (4+3 m);\frac {5}{3}+\frac {m}{2};-e^{2 i (c+d x)}\right )\right )\right ) \left (A+C \sec ^2(c+d x)\right )}{d (-2+3 m) (4+3 m) (10+3 m) (A+2 C+A \cos (2 c+2 d x)) \sec ^{\frac {4}{3}}(c+d x) (b \sec (c+d x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*I)*2^(1/3 + m)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(-2/3 + m)*(1 + E^((2*I)*(c + d*x)))^(-2/3 + m
)*(A*E^((4*I)*(c + d*x))*(-8 + 6*m + 9*m^2)*Hypergeometric2F1[5/3 + m/2, 4/3 + m, 8/3 + m/2, -E^((2*I)*(c + d*
x))] + (10 + 3*m)*(A*(4 + 3*m)*Hypergeometric2F1[4/3 + m, (-2 + 3*m)/6, (4 + 3*m)/6, -E^((2*I)*(c + d*x))] + 2
*(A + 2*C)*E^((2*I)*(c + d*x))*(-2 + 3*m)*Hypergeometric2F1[4/3 + m, (4 + 3*m)/6, 5/3 + m/2, -E^((2*I)*(c + d*
x))]))*(A + C*Sec[c + d*x]^2))/(d*(-2 + 3*m)*(4 + 3*m)*(10 + 3*m)*(A + 2*C + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^
(4/3)*(b*Sec[c + d*x])^(2/3))

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Maple [F]
time = 0.39, size = 0, normalized size = 0.00 \[\int \frac {\left (\sec ^{m}\left (d x +c \right )\right ) \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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