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

Optimal. Leaf size=147 \[ \frac {3 (C (1-3 m)-A (3 m+2)) \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (4-3 m);\frac {1}{6} (10-3 m);\cos ^2(c+d x)\right )}{d (4-3 m) (3 m+2) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac {3 C \sin (c+d x) \sec ^{m+1}(c+d x)}{d (3 m+2) \sqrt [3]{b \sec (c+d x)}} \]

[Out]

3*C*sec(d*x+c)^(1+m)*sin(d*x+c)/d/(2+3*m)/(b*sec(d*x+c))^(1/3)+3*(C*(1-3*m)-A*(2+3*m))*hypergeom([1/2, 2/3-1/2
*m],[5/3-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-1+m)*sin(d*x+c)/d/(-9*m^2+6*m+8)/(b*sec(d*x+c))^(1/3)/(sin(d*x+c)^2
)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 147, 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} \[ \frac {3 (C (1-3 m)-A (3 m+2)) \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (4-3 m);\frac {1}{6} (10-3 m);\cos ^2(c+d x)\right )}{d (4-3 m) (3 m+2) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac {3 C \sin (c+d x) \sec ^{m+1}(c+d x)}{d (3 m+2) \sqrt [3]{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^m*(A + C*Sec[c + d*x]^2))/(b*Sec[c + d*x])^(1/3),x]

[Out]

(3*C*Sec[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + 3*m)*(b*Sec[c + d*x])^(1/3)) + (3*(C*(1 - 3*m) - A*(2 + 3*m))*
Hypergeometric2F1[1/2, (4 - 3*m)/6, (10 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*Sin[c + d*x])/(d*(4 -
3*m)*(2 + 3*m)*(b*Sec[c + d*x])^(1/3)*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 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 \frac {\sec ^m(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx &=\frac {\sqrt [3]{\sec (c+d x)} \int \sec ^{-\frac {1}{3}+m}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx}{\sqrt [3]{b \sec (c+d x)}}\\ &=\frac {3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (2+3 m) \sqrt [3]{b \sec (c+d x)}}+\frac {\left (\left (C \left (-\frac {1}{3}+m\right )+A \left (\frac {2}{3}+m\right )\right ) \sqrt [3]{\sec (c+d x)}\right ) \int \sec ^{-\frac {1}{3}+m}(c+d x) \, dx}{\left (\frac {2}{3}+m\right ) \sqrt [3]{b \sec (c+d x)}}\\ &=\frac {3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (2+3 m) \sqrt [3]{b \sec (c+d x)}}+\frac {\left (\left (C \left (-\frac {1}{3}+m\right )+A \left (\frac {2}{3}+m\right )\right ) \cos ^{\frac {2}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{\frac {1}{3}-m}(c+d x) \, dx}{\left (\frac {2}{3}+m\right ) \sqrt [3]{b \sec (c+d x)}}\\ &=\frac {3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (2+3 m) \sqrt [3]{b \sec (c+d x)}}+\frac {3 (C (1-3 m)-A (2+3 m)) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (4-3 m);\frac {1}{6} (10-3 m);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (4-3 m) (2+3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 8.54, size = 311, normalized size = 2.12 \[ -\frac {3 i 2^{m+\frac {2}{3}} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m-\frac {1}{3}} \left (1+e^{2 i (c+d x)}\right )^{m-\frac {1}{3}} \left (A+C \sec ^2(c+d x)\right ) \left ((3 m-1) e^{2 i (c+d x)} \left (2 (3 m+11) (A+2 C) \, _2F_1\left (m+\frac {5}{3},\frac {1}{6} (3 m+5);\frac {1}{6} (3 m+11);-e^{2 i (c+d x)}\right )+A (3 m+5) e^{2 i (c+d x)} \, _2F_1\left (m+\frac {5}{3},\frac {1}{6} (3 m+11);\frac {1}{6} (3 m+17);-e^{2 i (c+d x)}\right )\right )+A \left (9 m^2+48 m+55\right ) \, _2F_1\left (m+\frac {5}{3},\frac {1}{6} (3 m-1);\frac {1}{6} (3 m+5);-e^{2 i (c+d x)}\right )\right )}{d (3 m-1) (3 m+5) (3 m+11) \sec ^{\frac {5}{3}}(c+d x) \sqrt [3]{b \sec (c+d x)} (A \cos (2 c+2 d x)+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^m*(A + C*Sec[c + d*x]^2))/(b*Sec[c + d*x])^(1/3),x]

[Out]

((-3*I)*2^(2/3 + m)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(-1/3 + m)*(1 + E^((2*I)*(c + d*x)))^(-1/3 + m
)*(A*(55 + 48*m + 9*m^2)*Hypergeometric2F1[5/3 + m, (-1 + 3*m)/6, (5 + 3*m)/6, -E^((2*I)*(c + d*x))] + E^((2*I
)*(c + d*x))*(-1 + 3*m)*(2*(A + 2*C)*(11 + 3*m)*Hypergeometric2F1[5/3 + m, (5 + 3*m)/6, (11 + 3*m)/6, -E^((2*I
)*(c + d*x))] + A*E^((2*I)*(c + d*x))*(5 + 3*m)*Hypergeometric2F1[5/3 + m, (11 + 3*m)/6, (17 + 3*m)/6, -E^((2*
I)*(c + d*x))]))*(A + C*Sec[c + d*x]^2))/(d*(-1 + 3*m)*(5 + 3*m)*(11 + 3*m)*(A + 2*C + A*Cos[2*c + 2*d*x])*Sec
[c + d*x]^(5/3)*(b*Sec[c + d*x])^(1/3))

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

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))^(1/3),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}}\,{d x} \]

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))^(1/3),x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sec ^{m}\left (d x +c \right )\right ) \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx \]

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))^(1/3),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}}\,{d x} \]

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))^(1/3),x, algorithm="maxima")

[Out]

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

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

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))^(1/3),x)

[Out]

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

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

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))**(1/3),x)

[Out]

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

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