3.278 \(\int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt [3]{e \sec (c+d x)}} \, dx\)

Optimal. Leaf size=662 \[ \frac {\sqrt {2} 3^{3/4} a^2 \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt {\frac {\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt {3}\right )}{d e^{2/3} (a-a \sec (c+d x)) \sqrt {a \sec (c+d x)+a} \sqrt {\frac {\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}-\frac {3 \sqrt [4]{3} \sqrt {2-\sqrt {3}} a^2 \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt {\frac {\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} E\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt {3}\right )}{2 d e^{2/3} (a-a \sec (c+d x)) \sqrt {a \sec (c+d x)+a} \sqrt {\frac {\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}+\frac {3 a \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a} \sqrt [3]{e \sec (c+d x)}}+\frac {3 a \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )} \]

[Out]

3*a*tan(d*x+c)/d/(e*sec(d*x+c))^(1/3)/(a+a*sec(d*x+c))^(1/2)+3*a*tan(d*x+c)/d/(-(e*sec(d*x+c))^(1/3)+e^(1/3)*(
1+3^(1/2)))/(a+a*sec(d*x+c))^(1/2)+3^(3/4)*a^2*EllipticF((-(e*sec(d*x+c))^(1/3)+e^(1/3)*(1-3^(1/2)))/(-(e*sec(
d*x+c))^(1/3)+e^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(e^(1/3)-(e*sec(d*x+c))^(1/3))*2^(1/2)*((e^(2/3)+e^(1/3)*(e*
sec(d*x+c))^(1/3)+(e*sec(d*x+c))^(2/3))/(-(e*sec(d*x+c))^(1/3)+e^(1/3)*(1+3^(1/2)))^2)^(1/2)*tan(d*x+c)/d/e^(2
/3)/(a-a*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2)/(e^(1/3)*(e^(1/3)-(e*sec(d*x+c))^(1/3))/(-(e*sec(d*x+c))^(1/3)+e^(
1/3)*(1+3^(1/2)))^2)^(1/2)-3/2*3^(1/4)*a^2*EllipticE((-(e*sec(d*x+c))^(1/3)+e^(1/3)*(1-3^(1/2)))/(-(e*sec(d*x+
c))^(1/3)+e^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(e^(1/3)-(e*sec(d*x+c))^(1/3))*(1/2*6^(1/2)-1/2*2^(1/2))*((e^(2/
3)+e^(1/3)*(e*sec(d*x+c))^(1/3)+(e*sec(d*x+c))^(2/3))/(-(e*sec(d*x+c))^(1/3)+e^(1/3)*(1+3^(1/2)))^2)^(1/2)*tan
(d*x+c)/d/e^(2/3)/(a-a*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2)/(e^(1/3)*(e^(1/3)-(e*sec(d*x+c))^(1/3))/(-(e*sec(d*x
+c))^(1/3)+e^(1/3)*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]  time = 0.48, antiderivative size = 662, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3806, 51, 63, 303, 218, 1877} \[ \frac {\sqrt {2} 3^{3/4} a^2 \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt {\frac {\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt {3}\right )}{d e^{2/3} (a-a \sec (c+d x)) \sqrt {a \sec (c+d x)+a} \sqrt {\frac {\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}-\frac {3 \sqrt [4]{3} \sqrt {2-\sqrt {3}} a^2 \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt {\frac {\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} E\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt {3}\right )}{2 d e^{2/3} (a-a \sec (c+d x)) \sqrt {a \sec (c+d x)+a} \sqrt {\frac {\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}+\frac {3 a \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a} \sqrt [3]{e \sec (c+d x)}}+\frac {3 a \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*Tan[c + d*x])/(d*(e*Sec[c + d*x])^(1/3)*Sqrt[a + a*Sec[c + d*x]]) + (3*a*Tan[c + d*x])/(d*Sqrt[a + a*Sec[
c + d*x]]*((1 + Sqrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))) - (3*3^(1/4)*Sqrt[2 - Sqrt[3]]*a^2*EllipticE[ArcSi
n[((1 - Sqrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))/((1 + Sqrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))], -7 - 4*S
qrt[3]]*(e^(1/3) - (e*Sec[c + d*x])^(1/3))*Sqrt[(e^(2/3) + e^(1/3)*(e*Sec[c + d*x])^(1/3) + (e*Sec[c + d*x])^(
2/3))/((1 + Sqrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))^2]*Tan[c + d*x])/(2*d*e^(2/3)*(a - a*Sec[c + d*x])*Sqrt
[a + a*Sec[c + d*x]]*Sqrt[(e^(1/3)*(e^(1/3) - (e*Sec[c + d*x])^(1/3)))/((1 + Sqrt[3])*e^(1/3) - (e*Sec[c + d*x
])^(1/3))^2]) + (Sqrt[2]*3^(3/4)*a^2*EllipticF[ArcSin[((1 - Sqrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))/((1 + S
qrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))], -7 - 4*Sqrt[3]]*(e^(1/3) - (e*Sec[c + d*x])^(1/3))*Sqrt[(e^(2/3) +
e^(1/3)*(e*Sec[c + d*x])^(1/3) + (e*Sec[c + d*x])^(2/3))/((1 + Sqrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))^2]*
Tan[c + d*x])/(d*e^(2/3)*(a - a*Sec[c + d*x])*Sqrt[a + a*Sec[c + d*x]]*Sqrt[(e^(1/3)*(e^(1/3) - (e*Sec[c + d*x
])^(1/3)))/((1 + Sqrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))^2])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 303

Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Dist[(Sq
rt[2]*s)/(Sqrt[2 + Sqrt[3]]*r), Int[1/Sqrt[a + b*x^3], x], x] + Dist[1/r, Int[((1 - Sqrt[3])*s + r*x)/Sqrt[a +
b*x^3], x], x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 1877

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Simplify[((1 - Sqrt[3])*d)/c]]
, s = Denom[Simplify[((1 - Sqrt[3])*d)/c]]}, Simp[(2*d*s^3*Sqrt[a + b*x^3])/(a*r^2*((1 + Sqrt[3])*s + r*x)), x
] - Simp[(3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*Elli
pticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(r^2*Sqrt[a + b*x^3]*Sqrt[(s*(
s + r*x))/((1 + Sqrt[3])*s + r*x)^2]), x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && EqQ[b*c^3 - 2*(5 - 3*Sqrt[3
])*a*d^3, 0]

Rule 3806

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

Rubi steps

\begin {align*} \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt [3]{e \sec (c+d x)}} \, dx &=-\frac {\left (a^2 e \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{(e x)^{4/3} \sqrt {a-a x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {a-a \sec (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ &=\frac {3 a \tan (c+d x)}{d \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {\left (a^2 \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{e x} \sqrt {a-a x}} \, dx,x,\sec (c+d x)\right )}{2 d \sqrt {a-a \sec (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ &=\frac {3 a \tan (c+d x)}{d \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {\left (3 a^2 \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {a-\frac {a x^3}{e}}} \, dx,x,\sqrt [3]{e \sec (c+d x)}\right )}{2 d e \sqrt {a-a \sec (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ &=\frac {3 a \tan (c+d x)}{d \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {\left (3 a^2 \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {\left (1-\sqrt {3}\right ) \sqrt [3]{e}-x}{\sqrt {a-\frac {a x^3}{e}}} \, dx,x,\sqrt [3]{e \sec (c+d x)}\right )}{2 d e \sqrt {a-a \sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {\left (3 \sqrt {\frac {1}{2} \left (2-\sqrt {3}\right )} a^2 \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-\frac {a x^3}{e}}} \, dx,x,\sqrt [3]{e \sec (c+d x)}\right )}{d e^{2/3} \sqrt {a-a \sec (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ &=\frac {3 a \tan (c+d x)}{d \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {3 a \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}-\frac {3 \sqrt [4]{3} \sqrt {2-\sqrt {3}} a^2 E\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt {3}\right ) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt {\frac {e^{2/3}+\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} \tan (c+d x)}{2 d e^{2/3} (a-a \sec (c+d x)) \sqrt {a+a \sec (c+d x)} \sqrt {\frac {\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}+\frac {\sqrt {2} 3^{3/4} a^2 F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt {3}\right ) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt {\frac {e^{2/3}+\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} \tan (c+d x)}{d e^{2/3} (a-a \sec (c+d x)) \sqrt {a+a \sec (c+d x)} \sqrt {\frac {\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}\\ \end {align*}

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Mathematica [C]  time = 0.16, size = 71, normalized size = 0.11 \[ \frac {2 \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt [3]{\sec (c+d x)} \sqrt {a (\sec (c+d x)+1)} \, _2F_1\left (\frac {1}{2},\frac {4}{3};\frac {3}{2};1-\sec (c+d x)\right )}{d \sqrt [3]{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Hypergeometric2F1[1/2, 4/3, 3/2, 1 - Sec[c + d*x]]*Sec[c + d*x]^(1/3)*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d
*x)/2])/(d*(e*Sec[c + d*x])^(1/3))

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac {2}{3}}}{e \sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*sec(d*x + c) + a)*(e*sec(d*x + c))^(2/3)/(e*sec(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*sec(d*x + c) + a)/(e*sec(d*x + c))^(1/3), x)

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maple [F]  time = 1.31, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a +a \sec \left (d x +c \right )}}{\left (e \sec \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*sec(d*x + c) + a)/(e*sec(d*x + c))^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}{\sqrt [3]{e \sec {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**(1/2)/(e*sec(d*x+c))**(1/3),x)

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))/(e*sec(c + d*x))**(1/3), x)

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