\(\int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/3} \, dx\) [696]

Optimal result

Integrand size = 23, antiderivative size = 299 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=\frac {3 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{8 d}+\frac {3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 d}+\frac {\left (2 a^2+5 b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{4 \sqrt {2} b d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {a \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{2 \sqrt {2} b d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \]

[Out]

3/8*a*(a+b*sec(d*x+c))^(2/3)*tan(d*x+c)/d+3/8*(a+b*sec(d*x+c))^(5/3)*tan(d*x+c)/d+1/8*(2*a^2+5*b^2)*AppellF1(1
/2,-2/3,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c))*(a+b*sec(d*x+c))^(2/3)*tan(d*x+c)/b/d/((a+b*sec(d*x
+c))/(a+b))^(2/3)*2^(1/2)/(1+sec(d*x+c))^(1/2)-1/4*a*(a^2-b^2)*AppellF1(1/2,1/3,1/2,3/2,b*(1-sec(d*x+c))/(a+b)
,1/2-1/2*sec(d*x+c))*((a+b*sec(d*x+c))/(a+b))^(1/3)*tan(d*x+c)/b/d/(a+b*sec(d*x+c))^(1/3)*2^(1/2)/(1+sec(d*x+c
))^(1/2)

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3920, 4087, 4092, 3919, 144, 143} \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=\frac {\left (2 a^2+5 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{4 \sqrt {2} b d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {a \left (a^2-b^2\right ) \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{2 \sqrt {2} b d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 d}+\frac {3 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{8 d} \]

[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 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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])

Rule 144

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 3919

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 3920

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
*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 4087

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] /;
FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 4092

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]

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \operatorname {AppellF1}\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 ) \operatorname {AppellF1}\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] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(7809\) vs. \(2(299)=598\).

Time = 45.02 (sec) , antiderivative size = 7809, normalized size of antiderivative = 26.12 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

Result too large to show

Maple [F]

\[\int \sec \left (d x +c \right )^{2} \left (a +b \sec \left (d x +c \right )\right )^{\frac {5}{3}}d x\]

[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)

Fricas [F]

\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \sec \left (d x + c\right )^{2} \,d x } \]

[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)

Sympy [F]

\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{3}} \sec ^{2}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \sec \left (d x + c\right )^{2} \,d x } \]

[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)

Giac [F]

\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \sec \left (d x + c\right )^{2} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/3}}{{\cos \left (c+d\,x\right )}^2} \,d x \]

[In]

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

[Out]

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