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

Optimal. Leaf size=95 \[ \frac{3 (13 A+10 C) \sin (c+d x) (b \sec (c+d x))^{7/3} \text{Hypergeometric2F1}\left (-\frac{7}{6},\frac{1}{2},-\frac{1}{6},\cos ^2(c+d x)\right )}{91 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{10/3}}{13 b^2 d} \]

[Out]

(3*(13*A + 10*C)*Hypergeometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(7/3)*Sin[c + d*x])/(91*b
*d*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(10/3)*Tan[c + d*x])/(13*b^2*d)

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Rubi [A]  time = 0.0857016, antiderivative size = 95, normalized size of antiderivative = 1., 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} \[ \frac{3 (13 A+10 C) \sin (c+d x) (b \sec (c+d x))^{7/3} \, _2F_1\left (-\frac{7}{6},\frac{1}{2};-\frac{1}{6};\cos ^2(c+d x)\right )}{91 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{10/3}}{13 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(13*A + 10*C)*Hypergeometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(7/3)*Sin[c + d*x])/(91*b
*d*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(10/3)*Tan[c + d*x])/(13*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 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]

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 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]

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{\int (b \sec (c+d x))^{10/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^2}\\ &=\frac{3 C (b \sec (c+d x))^{10/3} \tan (c+d x)}{13 b^2 d}+\frac{(13 A+10 C) \int (b \sec (c+d x))^{10/3} \, dx}{13 b^2}\\ &=\frac{3 C (b \sec (c+d x))^{10/3} \tan (c+d x)}{13 b^2 d}+\frac{\left ((13 A+10 C) \sqrt [3]{\frac{\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac{1}{\left (\frac{\cos (c+d x)}{b}\right )^{10/3}} \, dx}{13 b^2}\\ &=\frac{3 C (b \sec (c+d x))^{10/3} \tan (c+d x)}{13 b^2 d}+\frac{3 b (13 A+10 C) \, _2F_1\left (-\frac{7}{6},\frac{1}{2};-\frac{1}{6};\cos ^2(c+d x)\right ) \sec (c+d x) \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{91 d \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 1.74144, size = 236, normalized size = 2.48 \[ \frac{12 i e^{i (c+d x)} \cos ^3(c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \left ((13 A+10 C) \left (1+e^{2 i (c+d x)}\right )^{13/3} \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{2}{3},\frac{5}{3},-e^{2 i (c+d x)}\right )-13 A \left (5 e^{2 i (c+d x)}+2 e^{4 i (c+d x)}+1\right ) \left (1+e^{2 i (c+d x)}\right )^2-2 C \left (21 e^{2 i (c+d x)}+79 e^{4 i (c+d x)}+45 e^{6 i (c+d x)}+10 e^{8 i (c+d x)}+5\right )\right )}{91 d \left (1+e^{2 i (c+d x)}\right )^4 (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((12*I)/91)*E^(I*(c + d*x))*Cos[c + d*x]^3*(-13*A*(1 + E^((2*I)*(c + d*x)))^2*(1 + 5*E^((2*I)*(c + d*x)) + 2*
E^((4*I)*(c + d*x))) - 2*C*(5 + 21*E^((2*I)*(c + d*x)) + 79*E^((4*I)*(c + d*x)) + 45*E^((6*I)*(c + d*x)) + 10*
E^((8*I)*(c + d*x))) + (13*A + 10*C)*(1 + E^((2*I)*(c + d*x)))^(13/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -E^((2*
I)*(c + d*x))])*(b*Sec[c + d*x])^(4/3)*(A + C*Sec[c + d*x]^2))/(d*(1 + E^((2*I)*(c + d*x)))^4*(A + 2*C + A*Cos
[2*(c + d*x)]))

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Maple [F]  time = 0.136, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{2} \left ( b\sec \left ( dx+c \right ) \right ) ^{{\frac{4}{3}}} \left ( A+C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b \sec \left (d x + c\right )^{5} + A b \sec \left (d x + c\right )^{3}\right )} \left (b \sec \left (d x + c\right )\right )^{\frac{1}{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b*sec(d*x + c)^5 + A*b*sec(d*x + c)^3)*(b*sec(d*x + c))^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac{4}{3}} \sec \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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