3.858 \(\int \sqrt [3]{a+b \sec (c+d x)} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=229 \[ \frac {\sqrt {2} (b B-a C) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}+\frac {\sqrt {2} C (a+b) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {4}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}} \]

[Out]

(a+b)*C*AppellF1(1/2,-4/3,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c))*(a+b*sec(d*x+c))^(1/3)*2^(1/2)*ta
n(d*x+c)/b/d/((a+b*sec(d*x+c))/(a+b))^(1/3)/(1+sec(d*x+c))^(1/2)+(B*b-C*a)*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))^(1/3)*2^(1/2)*tan(d*x+c)/b/d/((a+b*sec(d*x+c))/(a+b))^(1/3)
/(1+sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {4062, 12, 3834, 139, 138} \[ \frac {\sqrt {2} (b B-a C) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}+\frac {\sqrt {2} C (a+b) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {4}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^(1/3)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 4062

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

Rubi steps

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

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Mathematica [B]  time = 26.64, size = 21684, normalized size = 94.69 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Sec[c + d*x])^(1/3)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

Result too large to show

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fricas [F]  time = 1.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(1/3)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c))*(b*sec(d*x + c) + a)^(1/3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(1/3)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*(b*sec(d*x + c) + a)^(1/3), x)

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maple [F]  time = 1.21, size = 0, normalized size = 0.00 \[ \int \left (a +b \sec \left (d x +c \right )\right )^{\frac {1}{3}} \left (B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^(1/3)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

int((a+b*sec(d*x+c))^(1/3)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(1/3)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*(b*sec(d*x + c) + a)^(1/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(1/3),x)

[Out]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**(1/3)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

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

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