3.2.73 \(\int \frac {(A+C \cos ^2(c+d x)) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\) [173]

Optimal. Leaf size=90 \[ \frac {3 A \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}-\frac {3 (A+4 C) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{8 b^2 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

3/4*A*sin(d*x+c)/d/(b*cos(d*x+c))^(4/3)-3/8*(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 = 0.06, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {16, 3091, 2722} \begin {gather*} \frac {3 A \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}-\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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + 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*(A + 4*C)*(b*Cos[c + d*x])^(2/3)*Hypergeometric2F1[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 3091

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

Rubi steps

\begin {align*} \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx &=b \int \frac {A+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 {(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 (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 [A]
time = 0.65, size = 104, normalized size = 1.16 \begin {gather*} \frac {-\left ((A+4 C) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {3}{2};\cos ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sin (2 d x-2 \text {ArcTan}(\cot (c)))\right )+6 A \sqrt [3]{\sin ^2(d x-\text {ArcTan}(\cot (c)))} \tan (c+d x)}{8 b d \sqrt [3]{b \cos (c+d x)} \sqrt [3]{\sin ^2(d x-\text {ArcTan}(\cot (c)))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((A + 4*C)*Hypergeometric2F1[1/2, 2/3, 3/2, Cos[d*x - ArcTan[Cot[c]]]^2]*Sin[2*d*x - 2*ArcTan[Cot[c]]]) + 6*
A*(Sin[d*x - ArcTan[Cot[c]]]^2)^(1/3)*Tan[c + d*x])/(8*b*d*(b*Cos[c + d*x])^(1/3)*(Sin[d*x - ArcTan[Cot[c]]]^2
)^(1/3))

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Maple [F]
time = 0.27, size = 0, normalized size = 0.00 \[\int \frac {\left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \sec \left (d x +c \right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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