3.1.21 \(\int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} (A+C \sec ^2(c+d x)) \, dx\) [21]

Optimal. Leaf size=146 \[ \frac {3 b C \sec ^{2+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (7+3 m)}+\frac {3 b (C (4+3 m)+A (7+3 m)) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (-1-3 m);\frac {1}{6} (5-3 m);\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (1+3 m) (7+3 m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

3*b*C*sec(d*x+c)^(2+m)*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(7+3*m)+3*b*(C*(4+3*m)+A*(7+3*m))*hypergeom([1/2, -1/
6-1/2*m],[5/6-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^m*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(9*m^2+24*m+7)/(sin(d*x+c)^2
)^(1/2)

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Rubi [A]
time = 0.09, 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, 4131, 3857, 2722} \begin {gather*} \frac {3 b (A (3 m+7)+C (3 m+4)) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (-3 m-1);\frac {1}{6} (5-3 m);\cos ^2(c+d x)\right )}{d (3 m+1) (3 m+7) \sqrt {\sin ^2(c+d x)}}+\frac {3 b C \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m+2}(c+d x)}{d (3 m+7)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^m*(b*Sec[c + d*x])^(4/3)*(A + C*Sec[c + d*x]^2),x]

[Out]

(3*b*C*Sec[c + d*x]^(2 + m)*(b*Sec[c + d*x])^(1/3)*Sin[c + d*x])/(d*(7 + 3*m)) + (3*b*(C*(4 + 3*m) + A*(7 + 3*
m))*Hypergeometric2F1[1/2, (-1 - 3*m)/6, (5 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^m*(b*Sec[c + d*x])^(1/3)*Si
n[c + d*x])/(d*(1 + 3*m)*(7 + 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 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 \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {\left (b \sqrt [3]{b \sec (c+d x)}\right ) \int \sec ^{\frac {4}{3}+m}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx}{\sqrt [3]{\sec (c+d x)}}\\ &=\frac {3 b C \sec ^{2+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (7+3 m)}+\frac {\left (b \left (C \left (\frac {4}{3}+m\right )+A \left (\frac {7}{3}+m\right )\right ) \sqrt [3]{b \sec (c+d x)}\right ) \int \sec ^{\frac {4}{3}+m}(c+d x) \, dx}{\left (\frac {7}{3}+m\right ) \sqrt [3]{\sec (c+d x)}}\\ &=\frac {3 b C \sec ^{2+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (7+3 m)}+\frac {\left (b \left (C \left (\frac {4}{3}+m\right )+A \left (\frac {7}{3}+m\right )\right ) \cos ^{\frac {1}{3}+m}(c+d x) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)}\right ) \int \cos ^{-\frac {4}{3}-m}(c+d x) \, dx}{\frac {7}{3}+m}\\ &=\frac {3 b C \sec ^{2+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (7+3 m)}+\frac {3 b (C (4+3 m)+A (7+3 m)) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (-1-3 m);\frac {1}{6} (5-3 m);\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (1+3 m) (7+3 m) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 3.56, size = 333, normalized size = 2.28 \begin {gather*} -\frac {3 i 2^{\frac {7}{3}+m} e^{-\frac {1}{3} i d (4+3 m) x} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{\frac {4}{3}+m} \left (1+e^{2 i (c+d x)}\right )^{\frac {4}{3}+m} \left (\frac {2 (A+2 C) e^{\frac {1}{3} i (6 c+d (10+3 m) x)} \, _2F_1\left (\frac {5}{3}+\frac {m}{2},\frac {10}{3}+m;\frac {8}{3}+\frac {m}{2};-e^{2 i (c+d x)}\right )}{10+3 m}+\frac {A e^{4 i c+\frac {1}{3} i d (16+3 m) x} \, _2F_1\left (\frac {8}{3}+\frac {m}{2},\frac {10}{3}+m;\frac {1}{6} (22+3 m);-e^{2 i (c+d x)}\right )}{16+3 m}+\frac {A e^{\frac {1}{3} i d (4+3 m) x} \, _2F_1\left (\frac {10}{3}+m,\frac {1}{6} (4+3 m);\frac {5}{3}+\frac {m}{2};-e^{2 i (c+d x)}\right )}{4+3 m}\right ) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right )}{d (A+2 C+A \cos (2 c+2 d x)) \sec ^{\frac {10}{3}}(c+d x)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^m*(b*Sec[c + d*x])^(4/3)*(A + C*Sec[c + d*x]^2),x]

[Out]

((-3*I)*2^(7/3 + m)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(4/3 + m)*(1 + E^((2*I)*(c + d*x)))^(4/3 + m)*
((2*(A + 2*C)*E^((I/3)*(6*c + d*(10 + 3*m)*x))*Hypergeometric2F1[5/3 + m/2, 10/3 + m, 8/3 + m/2, -E^((2*I)*(c
+ d*x))])/(10 + 3*m) + (A*E^((4*I)*c + (I/3)*d*(16 + 3*m)*x)*Hypergeometric2F1[8/3 + m/2, 10/3 + m, (22 + 3*m)
/6, -E^((2*I)*(c + d*x))])/(16 + 3*m) + (A*E^((I/3)*d*(4 + 3*m)*x)*Hypergeometric2F1[10/3 + m, (4 + 3*m)/6, 5/
3 + m/2, -E^((2*I)*(c + d*x))])/(4 + 3*m))*(b*Sec[c + d*x])^(4/3)*(A + C*Sec[c + d*x]^2))/(d*E^((I/3)*d*(4 + 3
*m)*x)*(A + 2*C + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(10/3))

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Maple [F]
time = 0.50, size = 0, normalized size = 0.00 \[\int \left (\sec ^{m}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^m*(b*sec(d*x+c))^(4/3)*(A+C*sec(d*x+c)^2),x)

[Out]

int(sec(d*x+c)^m*(b*sec(d*x+c))^(4/3)*(A+C*sec(d*x+c)^2),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*(b*sec(d*x+c))^(4/3)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^(4/3)*sec(d*x + c)^m, 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*(b*sec(d*x+c))^(4/3)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**m*(b*sec(d*x+c))**(4/3)*(A+C*sec(d*x+c)**2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6190 deep

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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*(b*sec(d*x+c))^(4/3)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^(4/3)*sec(d*x + c)^m, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^(4/3)*(1/cos(c + d*x))^m,x)

[Out]

int((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^(4/3)*(1/cos(c + d*x))^m, x)

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