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

Optimal. Leaf size=95 \[ \frac {3 C \tan (c+d x) (b \sec (c+d x))^{2/3}}{5 b^2 d}-\frac {3 (5 A+2 C) \sin (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(c+d x)\right )}{5 b d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 95, 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 = {16, 4046, 3772, 2643} \[ \frac {3 C \tan (c+d x) (b \sec (c+d x))^{2/3}}{5 b^2 d}-\frac {3 (5 A+2 C) \sin (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(c+d x)\right )}{5 b d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4046

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

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Mathematica [C]  time = 1.19, size = 165, normalized size = 1.74 \[ \frac {3 \left (A+C \sec ^2(c+d x)\right ) \left (2 C \sin (c+d x) \sec ^{\frac {5}{3}}(c+d x)-i 2^{2/3} (5 A+2 C) \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left (1+e^{2 i (c+d x)}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-e^{2 i (c+d x)}\right )\right )}{5 d \sec ^{\frac {2}{3}}(c+d x) (b \sec (c+d x))^{4/3} (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(A + C*Sec[c + d*x]^2)*((-I)*2^(2/3)*(5*A + 2*C)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(2/3)*(1 + E^(
(2*I)*(c + d*x)))^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -E^((2*I)*(c + d*x))] + 2*C*Sec[c + d*x]^(5/3)*Sin[c
+ d*x]))/(5*d*(A + 2*C + A*Cos[2*(c + d*x)])*Sec[c + d*x]^(2/3)*(b*Sec[c + d*x])^(4/3))

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}}{b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [F]  time = 1.26, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sec ^{2}\left (d x +c \right )\right ) \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(sec(d*x+c)^2*(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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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