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

Optimal. Leaf size=356 \[ -\frac {a \left (30 a^2-373 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{220 \sqrt {2} b^2 d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {3 \left (15 a^2-64 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{440 b d}+\frac {\left (15 a^4-79 a^2 b^2+64 b^4\right ) \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{110 \sqrt {2} b^2 d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{88 b d} \]

[Out]

-3/440*(15*a^2-64*b^2)*(a+b*sec(d*x+c))^(2/3)*tan(d*x+c)/b/d-9/88*a*(a+b*sec(d*x+c))^(5/3)*tan(d*x+c)/b/d+3/11
*(a+b*sec(d*x+c))^(8/3)*tan(d*x+c)/b/d-1/440*a*(30*a^2-373*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^2/d/((a+b*sec(d*x+c))/(a+b))^(2/3)*2^(1/2)/(1+sec(d
*x+c))^(1/2)+1/220*(15*a^4-79*a^2*b^2+64*b^4)*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^2/d/(a+b*sec(d*x+c))^(1/3)*2^(1/2)/(1+sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.65, antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3840, 4002, 4007, 3834, 139, 138} \[ -\frac {a \left (30 a^2-373 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{220 \sqrt {2} b^2 d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {\left (-79 a^2 b^2+15 a^4+64 b^4\right ) \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{110 \sqrt {2} b^2 d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}-\frac {3 \left (15 a^2-64 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{440 b d}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{88 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(15*a^2 - 64*b^2)*(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(440*b*d) - (9*a*(a + b*Sec[c + d*x])^(5/3)*Tan
[c + d*x])/(88*b*d) + (3*(a + b*Sec[c + d*x])^(8/3)*Tan[c + d*x])/(11*b*d) - (a*(30*a^2 - 373*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])/(220*Sqrt[2]*b^2*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(2/3)) + ((15*a^4 - 79*a^2*b^2 +
64*b^4)*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])/(110*Sqrt[2]*b^2*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(1/3))

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])

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 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 3840

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

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]

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/3} \, dx &=\frac {3 (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{11 b d}+\frac {3 \int \sec (c+d x) \left (\frac {8 b}{3}-a \sec (c+d x)\right ) (a+b \sec (c+d x))^{5/3} \, dx}{11 b}\\ &=-\frac {9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{88 b d}+\frac {3 (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{11 b d}+\frac {9 \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \left (\frac {49 a b}{9}-\frac {1}{9} \left (15 a^2-64 b^2\right ) \sec (c+d x)\right ) \, dx}{88 b}\\ &=-\frac {3 \left (15 a^2-64 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{440 b d}-\frac {9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{88 b d}+\frac {3 (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{11 b d}+\frac {27 \int \frac {\sec (c+d x) \left (\frac {1}{27} b \left (215 a^2+128 b^2\right )-\frac {1}{27} a \left (30 a^2-373 b^2\right ) \sec (c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{440 b}\\ &=-\frac {3 \left (15 a^2-64 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{440 b d}-\frac {9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{88 b d}+\frac {3 (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{11 b d}+\frac {1}{440} \left (a \left (373-\frac {30 a^2}{b^2}\right )\right ) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \, dx+\frac {\left (15 a^4-79 a^2 b^2+64 b^4\right ) \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{220 b^2}\\ &=-\frac {3 \left (15 a^2-64 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{440 b d}-\frac {9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{88 b d}+\frac {3 (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{11 b d}-\frac {\left (a \left (373-\frac {30 a^2}{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 )}{440 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {\left (\left (15 a^4-79 a^2 b^2+64 b^4\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 )}{220 b^2 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\\ &=-\frac {3 \left (15 a^2-64 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{440 b d}-\frac {9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{88 b d}+\frac {3 (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{11 b d}-\frac {\left (a \left (373-\frac {30 a^2}{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 )}{440 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 (\left (15 a^4-79 a^2 b^2+64 b^4\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 )}{220 b^2 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ &=-\frac {3 \left (15 a^2-64 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{440 b d}-\frac {9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{88 b d}+\frac {3 (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{11 b d}+\frac {a \left (373-\frac {30 a^2}{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)}{220 \sqrt {2} d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {\left (15 a^4-79 a^2 b^2+64 b^4\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)}{110 \sqrt {2} b^2 d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ \end {align*}

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Mathematica [B]  time = 27.34, size = 21890, normalized size = 61.49 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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fricas [F]  time = 1.27, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (d x + c\right )^{4} + a \sec \left (d x + c\right )^{3}\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \sec \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(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)^3, x)

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maple [F]  time = 0.89, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{3}\left (d x +c \right )\right ) \left (a +b \sec \left (d x +c \right )\right )^{\frac {5}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \sec \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(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)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/3}}{{\cos \left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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