3.699 \(\int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx\)

Optimal. Leaf size=313 \[ -\frac {a \left (9 a^2+11 b^2\right ) \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{10 \sqrt {2} b^3 d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac {\left (18 a^2+25 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{20 \sqrt {2} b^3 d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{20 b^2 d}+\frac {3 \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{2/3}}{8 b d} \]

[Out]

-9/20*a*(a+b*sec(d*x+c))^(2/3)*tan(d*x+c)/b^2/d+3/8*sec(d*x+c)*(a+b*sec(d*x+c))^(2/3)*tan(d*x+c)/b/d+1/40*(18*
a^2+25*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^3/d/((a+b*sec(d*x+c))/(a+b))^(2/3)*2^(1/2)/(1+sec(d*x+c))^(1/2)-1/20*a*(9*a^2+11*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^3/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 = 0.49, antiderivative size = 313, normalized size of antiderivative = 1.00, 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 = {3865, 4082, 4007, 3834, 139, 138} \[ -\frac {a \left (9 a^2+11 b^2\right ) \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{10 \sqrt {2} b^3 d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac {\left (18 a^2+25 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{20 \sqrt {2} b^3 d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{20 b^2 d}+\frac {3 \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{2/3}}{8 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + b*Sec[c + d*x])^(1/3),x]

[Out]

(-9*a*(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(20*b^2*d) + (3*Sec[c + d*x]*(a + b*Sec[c + d*x])^(2/3)*Tan[c +
d*x])/(8*b*d) + ((18*a^2 + 25*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])/(20*Sqrt[2]*b^3*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*
x])/(a + b))^(2/3)) - (a*(9*a^2 + 11*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])/(10*Sqrt[2]*b^3*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 3865

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(d^3*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3))/(b*f*(m + n - 1)), x] + Dist[d^3/(b*(m + n
- 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 3)*Simp[a*(n - 3) + b*(m + n - 2)*Csc[e + f*x] - a*(n
- 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 3] && (Integer
Q[n] || IntegersQ[2*m, 2*n]) &&  !IGtQ[m, 2]

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 4082

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

Rubi steps

\begin {align*} \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx &=\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{8 b d}+\frac {3 \int \frac {\sec (c+d x) \left (a+\frac {5}{3} b \sec (c+d x)-2 a \sec ^2(c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{8 b}\\ &=-\frac {9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{20 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{8 b d}+\frac {9 \int \frac {\sec (c+d x) \left (\frac {a b}{3}+\frac {1}{9} \left (18 a^2+25 b^2\right ) \sec (c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{40 b^2}\\ &=-\frac {9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{20 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{8 b d}-\frac {\left (a \left (9 a^2+11 b^2\right )\right ) \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{20 b^3}+\frac {\left (18 a^2+25 b^2\right ) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \, dx}{40 b^3}\\ &=-\frac {9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{20 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{8 b d}+\frac {\left (a \left (9 a^2+11 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 )}{20 b^3 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {\left (\left (18 a^2+25 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 )}{40 b^3 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\\ &=-\frac {9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{20 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{8 b d}-\frac {\left (\left (18 a^2+25 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 )}{40 b^3 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 (9 a^2+11 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 )}{20 b^3 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ &=-\frac {9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{20 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{8 b d}+\frac {\left (18 a^2+25 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)}{20 \sqrt {2} b^3 d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {a \left (9 a^2+11 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)}{10 \sqrt {2} b^3 d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ \end {align*}

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Mathematica [B]  time = 26.60, size = 19015, normalized size = 60.75 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^4/(a + b*Sec[c + d*x])^(1/3),x]

[Out]

Result too large to show

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fricas [F]  time = 1.11, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*sec(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

integral(sec(d*x + c)^4/(b*sec(d*x + c) + a)^(1/3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*sec(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^4/(b*sec(d*x + c) + a)^(1/3), x)

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maple [F]  time = 0.82, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}\left (d x +c \right )}{\left (a +b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+b*sec(d*x+c))^(1/3),x)

[Out]

int(sec(d*x+c)^4/(a+b*sec(d*x+c))^(1/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*sec(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^4/(b*sec(d*x + c) + a)^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a + b/cos(c + d*x))^(1/3)),x)

[Out]

int(1/(cos(c + d*x)^4*(a + b/cos(c + d*x))^(1/3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}{\left (c + d x \right )}}{\sqrt [3]{a + b \sec {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+b*sec(d*x+c))**(1/3),x)

[Out]

Integral(sec(c + d*x)**4/(a + b*sec(c + d*x))**(1/3), x)

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