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

Optimal. Leaf size=77 \[ \frac {3 \sqrt {2} \tan (c+d x) (a \sec (c+d x)+a)^{2/3} F_1\left (\frac {7}{6};\frac {1}{2},1;\frac {13}{6};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{7 d \sqrt {1-\sec (c+d x)}} \]

[Out]

3/7*AppellF1(7/6,1,1/2,13/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*(a+a*sec(d*x+c))^(2/3)*2^(1/2)*tan(d*x+c)/d/(1-se
c(d*x+c))^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3779, 3778, 136} \[ \frac {3 \sqrt {2} \tan (c+d x) (a \sec (c+d x)+a)^{2/3} F_1\left (\frac {7}{6};\frac {1}{2},1;\frac {13}{6};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{7 d \sqrt {1-\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(2/3),x]

[Out]

(3*Sqrt[2]*AppellF1[7/6, 1/2, 1, 13/6, (1 + Sec[c + d*x])/2, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^(2/3)*Tan[
c + d*x])/(7*d*Sqrt[1 - Sec[c + d*x]])

Rule 136

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

Rule 3778

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[(a^n*Cot[c + d*x])/(d*Sqrt[1 + Csc[c + d*x]
]*Sqrt[1 - Csc[c + d*x]]), Subst[Int[(1 + (b*x)/a)^(n - 1/2)/(x*Sqrt[1 - (b*x)/a]), x], x, Csc[c + d*x]], x] /
; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 3779

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Csc[c + d*x])^FracPart
[n])/(1 + (b*Csc[c + d*x])/a)^FracPart[n], Int[(1 + (b*Csc[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^{2/3} \, dx &=\frac {(a+a \sec (c+d x))^{2/3} \int (1+\sec (c+d x))^{2/3} \, dx}{(1+\sec (c+d x))^{2/3}}\\ &=-\frac {\left ((a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt [6]{1+x}}{\sqrt {1-x} x} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac {3 \sqrt {2} F_1\left (\frac {7}{6};\frac {1}{2},1;\frac {13}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{7 d \sqrt {1-\sec (c+d x)}}\\ \end {align*}

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Mathematica [B]  time = 8.98, size = 694, normalized size = 9.01 \[ \frac {45 \tan (c+d x) (a (\sec (c+d x)+1))^{5/3} F_1\left (\frac {1}{2};\frac {2}{3},1;\frac {3}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (2 \tan ^2\left (\frac {1}{2} (c+d x)\right ) \left (2 F_1\left (\frac {3}{2};\frac {5}{3},1;\frac {5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-3 F_1\left (\frac {3}{2};\frac {2}{3},2;\frac {5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )+9 F_1\left (\frac {1}{2};\frac {2}{3},1;\frac {3}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )}{a d \left (135 F_1\left (\frac {1}{2};\frac {2}{3},1;\frac {3}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ){}^2 \left (\left (2 \tan ^2(c+d x)+3\right ) \sec (c+d x)+3\right )+40 \sin ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (3 F_1\left (\frac {3}{2};\frac {2}{3},2;\frac {5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 F_1\left (\frac {3}{2};\frac {5}{3},1;\frac {5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ){}^2-6 \sin ^2\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) F_1\left (\frac {1}{2};\frac {2}{3},1;\frac {3}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (24 \cos (c+d x) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \left (9 F_1\left (\frac {5}{2};\frac {2}{3},3;\frac {7}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-6 F_1\left (\frac {5}{2};\frac {5}{3},2;\frac {7}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+5 F_1\left (\frac {5}{2};\frac {8}{3},1;\frac {7}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )+10 (16 \cos (c+d x)-3 \cos (2 (c+d x))-7) F_1\left (\frac {3}{2};\frac {5}{3},1;\frac {5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+15 (-16 \cos (c+d x)+3 \cos (2 (c+d x))+7) F_1\left (\frac {3}{2};\frac {2}{3},2;\frac {5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(45*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(a*(1 + Sec[c + d*x]))^(5/3)*(9*Appell
F1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(-3*AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x)
/2]^2, -Tan[(c + d*x)/2]^2] + 2*AppellF1[3/2, 5/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c +
d*x)/2]^2)*Tan[c + d*x])/(a*d*(40*(3*AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*A
ppellF1[3/2, 5/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])^2*Sec[c + d*x]^2*Sin[(c + d*x)/2]^2*Tan[(c
+ d*x)/2]^2 - 6*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[c + d*x]^3*Sin[(c + d
*x)/2]^2*(10*AppellF1[3/2, 5/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(-7 + 16*Cos[c + d*x] - 3*Cos
[2*(c + d*x)]) + 15*AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(7 - 16*Cos[c + d*x] +
3*Cos[2*(c + d*x)]) + 24*(9*AppellF1[5/2, 2/3, 3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 6*AppellF1[
5/2, 5/3, 2, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 5*AppellF1[5/2, 8/3, 1, 7/2, Tan[(c + d*x)/2]^2,
-Tan[(c + d*x)/2]^2])*Cos[c + d*x]*Tan[(c + d*x)/2]^2) + 135*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -T
an[(c + d*x)/2]^2]^2*(3 + Sec[c + d*x]*(3 + 2*Tan[c + d*x]^2))))

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.79, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (d x +c \right )\right )^{\frac {2}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{2/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(2/3),x)

[Out]

int((a + a/cos(c + d*x))^(2/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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