3.4.61 \(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\) [361]

Optimal. Leaf size=147 \[ \frac {3 A \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}+\frac {3 B \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 (A+4 C) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{8 b^2 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

3/4*A*sin(d*x+c)/d/(b*cos(d*x+c))^(4/3)+3*B*hypergeom([-1/6, 1/2],[5/6],cos(d*x+c)^2)*sin(d*x+c)/b/d/(b*cos(d*
x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)-3/8*(A+4*C)*(b*cos(d*x+c))^(2/3)*hypergeom([1/3, 1/2],[4/3],cos(d*x+c)^2)*sin
(d*x+c)/b^2/d/(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 = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {16, 3100, 2827, 2722} \begin {gather*} -\frac {3 (A+4 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right )}{8 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 A \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}+\frac {3 B \sin (c+d x) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*Sin[c + d*x])/(4*d*(b*Cos[c + d*x])^(4/3)) + (3*B*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]*Sin[c
+ d*x])/(b*d*(b*Cos[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2]) - (3*(A + 4*C)*(b*Cos[c + d*x])^(2/3)*Hypergeometri
c2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*b^2*d*Sqrt[Sin[c + d*x]^2])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3100

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx &=b \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{7/3}} \, dx\\ &=\frac {3 A \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}+\frac {3 \int \frac {\frac {4 b^2 B}{3}+\frac {1}{3} b^2 (A+4 C) \cos (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx}{4 b^2}\\ &=\frac {3 A \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}+B \int \frac {1}{(b \cos (c+d x))^{4/3}} \, dx+\frac {(A+4 C) \int \frac {1}{\sqrt [3]{b \cos (c+d x)}} \, dx}{4 b}\\ &=\frac {3 A \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}+\frac {3 B \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 (A+4 C) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{8 b^2 d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(703\) vs. \(2(147)=294\).
time = 6.33, size = 703, normalized size = 4.78 \begin {gather*} \frac {\frac {\cos ^3(c+d x) \left (C+B \sec (c+d x)+A \sec ^2(c+d x)\right ) \left (\frac {6 B \csc (c) \sec (c)}{d}+\frac {3 A \sec (c) \sec ^2(c+d x) \sin (d x)}{2 d}+\frac {3 \sec (c) \sec (c+d x) (A \sin (c)+4 B \sin (d x))}{2 d}\right )}{\sqrt [3]{b \cos (c+d x)} (2 A+C+2 B \cos (c+d x)+C \cos (2 c+2 d x))}-\frac {A \cos ^{\frac {7}{3}}(c+d x) \cos (d x-\text {ArcTan}(\cot (c))) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {3}{2};\cos ^2(d x-\text {ArcTan}(\cot (c)))\right ) \left (C+B \sec (c+d x)+A \sec ^2(c+d x)\right ) \sin (d x-\text {ArcTan}(\cot (c)))}{2 d \sqrt [3]{b \cos (c+d x)} (2 A+C+2 B \cos (c+d x)+C \cos (2 c+2 d x)) \sqrt [3]{\cos (c) \cos (d x)-\sin (c) \sin (d x)} \sqrt [3]{\sin ^2(d x-\text {ArcTan}(\cot (c)))}}-\frac {2 C \cos ^{\frac {7}{3}}(c+d x) \cos (d x-\text {ArcTan}(\cot (c))) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {3}{2};\cos ^2(d x-\text {ArcTan}(\cot (c)))\right ) \left (C+B \sec (c+d x)+A \sec ^2(c+d x)\right ) \sin (d x-\text {ArcTan}(\cot (c)))}{d \sqrt [3]{b \cos (c+d x)} (2 A+C+2 B \cos (c+d x)+C \cos (2 c+2 d x)) \sqrt [3]{\cos (c) \cos (d x)-\sin (c) \sin (d x)} \sqrt [3]{\sin ^2(d x-\text {ArcTan}(\cot (c)))}}+\frac {4 B \cos ^{\frac {7}{3}}(c+d x) \csc (c) \left (C+B \sec (c+d x)+A \sec ^2(c+d x)\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {5}{6};\cos ^2(d x+\text {ArcTan}(\tan (c)))\right ) \sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {1+\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt [3]{\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {3 \cos ^2(c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}{2 \left (\cos ^2(c)+\sin ^2(c)\right )}}{\sqrt [3]{\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{d \sqrt [3]{b \cos (c+d x)} (2 A+C+2 B \cos (c+d x)+C \cos (2 c+2 d x))}}{b} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((Cos[c + d*x]^3*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*((6*B*Csc[c]*Sec[c])/d + (3*A*Sec[c]*Sec[c + d*x]^2*S
in[d*x])/(2*d) + (3*Sec[c]*Sec[c + d*x]*(A*Sin[c] + 4*B*Sin[d*x]))/(2*d)))/((b*Cos[c + d*x])^(1/3)*(2*A + C +
2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) - (A*Cos[c + d*x]^(7/3)*Cos[d*x - ArcTan[Cot[c]]]*Hypergeometric2F1[1/
2, 2/3, 3/2, Cos[d*x - ArcTan[Cot[c]]]^2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*Sin[d*x - ArcTan[Cot[c]]])/(
2*d*(b*Cos[c + d*x])^(1/3)*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])*(Cos[c]*Cos[d*x] - Sin[c]*Sin[d*x
])^(1/3)*(Sin[d*x - ArcTan[Cot[c]]]^2)^(1/3)) - (2*C*Cos[c + d*x]^(7/3)*Cos[d*x - ArcTan[Cot[c]]]*Hypergeometr
ic2F1[1/2, 2/3, 3/2, Cos[d*x - ArcTan[Cot[c]]]^2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*Sin[d*x - ArcTan[Cot
[c]]])/(d*(b*Cos[c + d*x])^(1/3)*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])*(Cos[c]*Cos[d*x] - Sin[c]*S
in[d*x])^(1/3)*(Sin[d*x - ArcTan[Cot[c]]]^2)^(1/3)) + (4*B*Cos[c + d*x]^(7/3)*Csc[c]*(C + B*Sec[c + d*x] + A*S
ec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/6}, {5/6}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]
*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*(Cos[c]*Cos[d*x + ArcTan[Tan
[c]]]*Sqrt[1 + Tan[c]^2])^(1/3)*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] +
(3*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(2*(Cos[c]^2 + Sin[c]^2)))/(Cos[c]*Cos[d*x + ArcTan
[Tan[c]]]*Sqrt[1 + Tan[c]^2])^(1/3)))/(d*(b*Cos[c + d*x])^(1/3)*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*
x])))/b

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Maple [F]
time = 0.29, size = 0, normalized size = 0.00 \[\int \frac {\left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \sec \left (d x +c \right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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