3.4.54 \(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx\) [354]

Optimal. Leaf size=149 \[ \frac {3 A \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}}-\frac {3 B (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{2 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 (2 A-C) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{5 b^2 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

3*A*sin(d*x+c)/d/(b*cos(d*x+c))^(1/3)-3/2*B*(b*cos(d*x+c))^(2/3)*hypergeom([1/3, 1/2],[4/3],cos(d*x+c)^2)*sin(
d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)+3/5*(2*A-C)*(b*cos(d*x+c))^(5/3)*hypergeom([1/2, 5/6],[11/6],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 = 149, 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 (2 A-C) \sin (c+d x) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};\cos ^2(c+d x)\right )}{5 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 A \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}}-\frac {3 B \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 )}{2 b d \sqrt {\sin ^2(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])^(1/3),x]

[Out]

(3*A*Sin[c + d*x])/(d*(b*Cos[c + d*x])^(1/3)) - (3*B*(b*Cos[c + d*x])^(2/3)*Hypergeometric2F1[1/3, 1/2, 4/3, C
os[c + d*x]^2]*Sin[c + d*x])/(2*b*d*Sqrt[Sin[c + d*x]^2]) + (3*(2*A - C)*(b*Cos[c + d*x])^(5/3)*Hypergeometric
2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*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)}{\sqrt [3]{b \cos (c+d x)}} \, dx &=b \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\\ &=\frac {3 A \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}}+\frac {3 \int \frac {\frac {b^2 B}{3}-\frac {1}{3} b^2 (2 A-C) \cos (c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx}{b^2}\\ &=\frac {3 A \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}}+B \int \frac {1}{\sqrt [3]{b \cos (c+d x)}} \, dx-\frac {(2 A-C) \int (b \cos (c+d x))^{2/3} \, dx}{b}\\ &=\frac {3 A \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}}-\frac {3 B (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)}{2 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 (2 A-C) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{5 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. \(779\) vs. \(2(149)=298\).
time = 6.32, size = 779, normalized size = 5.23 \begin {gather*} \frac {\cos ^2(c+d x) (B+C \cos (c+d x)+A \sec (c+d x)) \left (-\frac {3 (-4 A+C+C \cos (2 c)) \csc (c) \sec (c)}{2 d}+\frac {6 A \sec (c) \sec (c+d x) \sin (d x)}{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 {2 B \cos ^{\frac {4}{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 ) (B+C \cos (c+d x)+A \sec (c+d x)) \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 A \cos ^{\frac {4}{3}}(c+d x) \csc (c) (B+C \cos (c+d x)+A \sec (c+d x)) \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))}-\frac {2 C \cos ^{\frac {4}{3}}(c+d x) \csc (c) (B+C \cos (c+d x)+A \sec (c+d x)) \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))} \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])^(1/3),x]

[Out]

(Cos[c + d*x]^2*(B + C*Cos[c + d*x] + A*Sec[c + d*x])*((-3*(-4*A + C + C*Cos[2*c])*Csc[c]*Sec[c])/(2*d) + (6*A
*Sec[c]*Sec[c + d*x]*Sin[d*x])/d))/((b*Cos[c + d*x])^(1/3)*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x]))
- (2*B*Cos[c + d*x]^(4/3)*Cos[d*x - ArcTan[Cot[c]]]*Hypergeometric2F1[1/2, 2/3, 3/2, Cos[d*x - ArcTan[Cot[c]]]
^2]*(B + C*Cos[c + d*x] + A*Sec[c + d*x])*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]*Sin[d*x])^(1/3)*(Sin[d*x - ArcTan[Cot[c]]]^2)^(1/
3)) + (4*A*Cos[c + d*x]^(4/3)*Csc[c]*(B + C*Cos[c + d*x] + A*Sec[c + d*x])*((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 + Ta
n[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])) - (2*C*Cos[c + d*x]^(4/3)*Csc[c]*(B + C*Cos[c
+ d*x] + A*Sec[c + d*x])*((HypergeometricPFQ[{-1/2, -1/6}, {5/6}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcT
an[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]))

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Maple [F]
time = 0.28, size = 0, normalized size = 0.00 \[\int \frac {\left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \sec \left (d x +c \right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx\]

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))^(1/3),x)

[Out]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(b*cos(d*x+c))^(1/3),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))^(1/3),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))^(1/3), 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))^(1/3),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*cos(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 + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\sqrt [3]{b \cos {\left (c + d x \right )}}}\, dx \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))**(1/3),x)

[Out]

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sec(c + d*x)/(b*cos(c + d*x))**(1/3), 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((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(b*cos(d*x+c))^(1/3),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))^(1/3), 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 )}^{1/3}} \,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))^(1/3)),x)

[Out]

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

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