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

Optimal. Leaf size=90 \[ \frac {3 A b \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}}-\frac {3 (4 A+7 C) \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right )}{28 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}} \]

[Out]

-3/28*(4*A+7*C)*hypergeom([1/2, 2/3],[5/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(4/3)/(sin(d*x+c)^2)^(1/2
)+3/7*A*b*tan(d*x+c)/d/(b*sec(d*x+c))^(7/3)

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Rubi [A]  time = 0.09, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {16, 4045, 3772, 2643} \[ \frac {3 A b \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}}-\frac {3 (4 A+7 C) \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right )}{28 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(4*A + 7*C)*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(28*d*(b*Sec[c + d*x])^(4/3)*Sq
rt[Sin[c + d*x]^2]) + (3*A*b*Tan[c + d*x])/(7*d*(b*Sec[c + d*x])^(7/3))

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 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 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx &=b \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/3}} \, dx\\ &=\frac {3 A b \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}}+\frac {(4 A+7 C) \int \frac {1}{\sqrt [3]{b \sec (c+d x)}} \, dx}{7 b}\\ &=\frac {3 A b \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}}+\frac {\left ((4 A+7 C) \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \sqrt [3]{\frac {\cos (c+d x)}{b}} \, dx}{7 b}\\ &=-\frac {3 (4 A+7 C) \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{28 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 A b \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 92, normalized size = 1.02 \[ -\frac {3 \sqrt {-\tan ^2(c+d x)} \cot (c+d x) \left (A \cos ^2(c+d x) \, _2F_1\left (-\frac {7}{6},\frac {1}{2};-\frac {1}{6};\sec ^2(c+d x)\right )+7 C \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\sec ^2(c+d x)\right )\right )}{7 b d \sqrt [3]{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Cot[c + d*x]*(A*Cos[c + d*x]^2*Hypergeometric2F1[-7/6, 1/2, -1/6, Sec[c + d*x]^2] + 7*C*Hypergeometric2F1[
-1/6, 1/2, 5/6, Sec[c + d*x]^2])*Sqrt[-Tan[c + d*x]^2])/(7*b*d*(b*Sec[c + d*x])^(1/3))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}}{b^{2} \sec \left (d x + c\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.57, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x +c \right ) \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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