3.696 \(\int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/3} \, dx\)

Optimal. Leaf size=299 \[ \frac{\left (2 a^2+5 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{4 \sqrt{2} b d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac{a \left (a^2-b^2\right ) \tan (c+d x) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{2 \sqrt{2} b d \sqrt{\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac{3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 d}+\frac{3 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{8 d} \]

[Out]

(3*a*(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(8*d) + (3*(a + b*Sec[c + d*x])^(5/3)*Tan[c + d*x])/(8*d) + ((2*
a^2 + 5*b^2)*AppellF1[1/2, 1/2, -2/3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*(a + b*Sec[c
+ d*x])^(2/3)*Tan[c + d*x])/(4*Sqrt[2]*b*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(2/3)) - (a*(
a^2 - b^2)*AppellF1[1/2, 1/2, 1/3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*((a + b*Sec[c +
d*x])/(a + b))^(1/3)*Tan[c + d*x])/(2*Sqrt[2]*b*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(1/3))

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Rubi [A]  time = 0.456605, antiderivative size = 299, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3835, 4002, 4007, 3834, 139, 138} \[ \frac{\left (2 a^2+5 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{4 \sqrt{2} b d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac{a \left (a^2-b^2\right ) \tan (c+d x) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{2 \sqrt{2} b d \sqrt{\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac{3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 d}+\frac{3 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^(5/3),x]

[Out]

(3*a*(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(8*d) + (3*(a + b*Sec[c + d*x])^(5/3)*Tan[c + d*x])/(8*d) + ((2*
a^2 + 5*b^2)*AppellF1[1/2, 1/2, -2/3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*(a + b*Sec[c
+ d*x])^(2/3)*Tan[c + d*x])/(4*Sqrt[2]*b*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(2/3)) - (a*(
a^2 - b^2)*AppellF1[1/2, 1/2, 1/3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*((a + b*Sec[c +
d*x])/(a + b))^(1/3)*Tan[c + d*x])/(2*Sqrt[2]*b*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(1/3))

Rule 3835

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

Rule 4002

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 4007

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

Rule 3834

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[Cot[e + f*x]/(f*Sqr
t[1 + Csc[e + f*x]]*Sqrt[1 - Csc[e + f*x]]), Subst[Int[(a + b*x)^m/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Csc[e + f
*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*m]

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/3} \, dx &=\frac{3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 d}+\frac{5}{8} \int \sec (c+d x) (b+a \sec (c+d x)) (a+b \sec (c+d x))^{2/3} \, dx\\ &=\frac{3 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{8 d}+\frac{3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 d}+\frac{3}{8} \int \frac{\sec (c+d x) \left (\frac{7 a b}{3}+\frac{1}{3} \left (2 a^2+5 b^2\right ) \sec (c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}} \, dx\\ &=\frac{3 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{8 d}+\frac{3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 d}-\frac{\left (a \left (a^2-b^2\right )\right ) \int \frac{\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{4 b}+\frac{\left (2 a^2+5 b^2\right ) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \, dx}{8 b}\\ &=\frac{3 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{8 d}+\frac{3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 d}+\frac{\left (a \left (a^2-b^2\right ) \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} \sqrt{1+x} \sqrt [3]{a+b x}} \, dx,x,\sec (c+d x)\right )}{4 b d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}-\frac{\left (\left (2 a^2+5 b^2\right ) \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{2/3}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{8 b d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}\\ &=\frac{3 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{8 d}+\frac{3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 d}-\frac{\left (\left (2 a^2+5 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^{2/3}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{8 b d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)} \left (-\frac{a+b \sec (c+d x)}{-a-b}\right )^{2/3}}+\frac{\left (a \left (a^2-b^2\right ) \sqrt [3]{-\frac{a+b \sec (c+d x)}{-a-b}} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} \sqrt{1+x} \sqrt [3]{-\frac{a}{-a-b}-\frac{b x}{-a-b}}} \, dx,x,\sec (c+d x)\right )}{4 b d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ &=\frac{3 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{8 d}+\frac{3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 d}+\frac{\left (2 a^2+5 b^2\right ) F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{4 \sqrt{2} b d \sqrt{1+\sec (c+d x)} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac{a \left (a^2-b^2\right ) F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{2 \sqrt{2} b d \sqrt{1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 26.6131, size = 19016, normalized size = 63.6 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^(5/3),x]

[Out]

Result too large to show

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Maple [F]  time = 0.1, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{2} \left ( a+b\sec \left ( dx+c \right ) \right ) ^{{\frac{5}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/3),x)

[Out]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{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*(a+b*sec(d*x+c))^(5/3),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^(5/3)*sec(d*x + c)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sec \left (d x + c\right )^{3} + a \sec \left (d x + c\right )^{2}\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/3),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c)^3 + a*sec(d*x + c)^2)*(b*sec(d*x + c) + a)^(2/3), 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*(a+b*sec(d*x+c))**(5/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{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*(a+b*sec(d*x+c))^(5/3),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^(5/3)*sec(d*x + c)^2, x)