3.7.89 \(\int \sec ^2(c+d x) (a+b \sec (c+d x))^{2/3} \, dx\) [689]
Optimal. Leaf size=260 \[ \frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {2 \sqrt {2} a F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {2 \sqrt {2} \left (a^2-b^2\right ) F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{5 b d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \]
[Out]
3/5*(a+b*sec(d*x+c))^(2/3)*tan(d*x+c)/d+2/5*a*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)*2^(1/2)*tan(d*x+c)/b/d/((a+b*sec(d*x+c))/(a+b))^(2/3)/(1+sec(d*x+c))^(1/2)-2/5*(a^
2-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)*2^(1
/2)*tan(d*x+c)/b/d/(a+b*sec(d*x+c))^(1/3)/(1+sec(d*x+c))^(1/2)
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Rubi [A]
time = 0.23, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3920, 4092,
3919, 144, 143} \begin {gather*} -\frac {2 \sqrt {2} \left (a^2-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 )}{5 b d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac {2 \sqrt {2} a \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 )}{5 b d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^(2/3),x]
[Out]
(3*(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(5*d) + (2*Sqrt[2]*a*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])/(5*b*d*Sqrt[1 + Sec[c + d*x]]*
((a + b*Sec[c + d*x])/(a + b))^(2/3)) - (2*Sqrt[2]*(a^2 - 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])/(5*b*d*Sqrt[1 + Sec[c +
d*x]]*(a + b*Sec[c + d*x])^(1/3))
Rule 143
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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 144
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 3919
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 3920
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(
(a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[m/(m + 1), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(b + a
*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]
Rule 4092
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 ^2(c+d x) (a+b \sec (c+d x))^{2/3} \, dx &=\frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {2}{5} \int \frac {\sec (c+d x) (b+a \sec (c+d x))}{\sqrt [3]{a+b \sec (c+d x)}} \, dx\\ &=\frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {(2 a) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \, dx}{5 b}-\frac {\left (2 \left (a^2-b^2\right )\right ) \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{5 b}\\ &=\frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}-\frac {(2 a \tan (c+d x)) \text {Subst}\left (\int \frac {(a+b x)^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{5 b d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {\left (2 \left (a^2-b^2\right ) \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{a+b x}} \, dx,x,\sec (c+d x)\right )}{5 b d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\\ &=\frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}-\frac {\left (2 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {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 )}{5 b 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 (2 \left (a^2-b^2\right ) \sqrt [3]{-\frac {a+b \sec (c+d x)}{-a-b}} \tan (c+d x)\right ) \text {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 )}{5 b d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ &=\frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {2 \sqrt {2} a 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)}{5 b d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {2 \sqrt {2} \left (a^2-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)}{5 b d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(2505\) vs. \(2(260)=520\).
time = 48.32, size = 2505, normalized size = 9.63 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^(2/3),x]
[Out]
((a + b*Sec[c + d*x])^(2/3)*((3*a*Sin[c + d*x])/(5*b) + (3*Tan[c + d*x])/5))/d - ((-2*b + 3*a*Cos[c + d*x])*(a
+ b*Sec[c + d*x])^(2/3)*(3*a*(b + a*Cos[c + d*x])^(2/3)*Sqrt[1 - Cos[c + d*x]^2]*Sec[c + d*x]^(2/3) - (3*(b +
a*Cos[c + d*x])^(2/3)*Sqrt[(1 - Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 + a*Sqrt[b^(-2)])]*Sqrt[(1 + Sqrt[b^(-2)]*b*S
ec[c + d*x])/(1 - a*Sqrt[b^(-2)])]*(-5*(a^2 - b^2)*AppellF1[2/3, 1/2, 1/2, 5/3, -((a + b*Sec[c + d*x])/(-a + 1
/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])] + 2*a*AppellF1[5/3, 1/2, 1/2, 8/3, -((a + b*Sec[c
+ d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*(a + b*Sec[c + d*x])))/(5*b*Sqrt[1
- Cos[c + d*x]^2]*Sec[c + d*x]^(1/3))))/(5*b*d*((3*a*(b + a*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(Sqrt[1 - Cos[c
+ d*x]^2]*Sec[c + d*x]^(1/3)) - (2*a^2*Sqrt[1 - Cos[c + d*x]^2]*Sec[c + d*x]^(2/3)*Sin[c + d*x])/(b + a*Cos[c
+ d*x])^(1/3) + 2*a*(b + a*Cos[c + d*x])^(2/3)*Sqrt[1 - Cos[c + d*x]^2]*Sec[c + d*x]^(5/3)*Sin[c + d*x] - (3*S
qrt[b^(-2)]*(b + a*Cos[c + d*x])^(2/3)*Sec[c + d*x]^(5/3)*Sqrt[(1 - Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 + a*Sqrt[b
^(-2)])]*(-5*(a^2 - b^2)*AppellF1[2/3, 1/2, 1/2, 5/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Se
c[c + d*x])/(a + 1/Sqrt[b^(-2)])] + 2*a*AppellF1[5/3, 1/2, 1/2, 8/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2
)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*(a + b*Sec[c + d*x]))*Sin[c + d*x])/(10*(1 - a*Sqrt[b^(-2)])*
Sqrt[1 - Cos[c + d*x]^2]*Sqrt[(1 + Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 - a*Sqrt[b^(-2)])]) + (3*Sqrt[b^(-2)]*(b +
a*Cos[c + d*x])^(2/3)*Sec[c + d*x]^(5/3)*Sqrt[(1 + Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 - a*Sqrt[b^(-2)])]*(-5*(a^2
- b^2)*AppellF1[2/3, 1/2, 1/2, 5/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a +
1/Sqrt[b^(-2)])] + 2*a*AppellF1[5/3, 1/2, 1/2, 8/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[
c + d*x])/(a + 1/Sqrt[b^(-2)])]*(a + b*Sec[c + d*x]))*Sin[c + d*x])/(10*(1 + a*Sqrt[b^(-2)])*Sqrt[1 - Cos[c +
d*x]^2]*Sqrt[(1 - Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 + a*Sqrt[b^(-2)])]) + (3*(b + a*Cos[c + d*x])^(2/3)*Sqrt[(1
- Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 + a*Sqrt[b^(-2)])]*Sqrt[(1 + Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 - a*Sqrt[b^(-2)
])]*(-5*(a^2 - b^2)*AppellF1[2/3, 1/2, 1/2, 5/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c +
d*x])/(a + 1/Sqrt[b^(-2)])] + 2*a*AppellF1[5/3, 1/2, 1/2, 8/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])),
(a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*(a + b*Sec[c + d*x]))*Sin[c + d*x])/(5*b*(1 - Cos[c + d*x]^2)^(3/2
)*Sec[c + d*x]^(4/3)) + (2*a*Sqrt[(1 - Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 + a*Sqrt[b^(-2)])]*Sqrt[(1 + Sqrt[b^(-2
)]*b*Sec[c + d*x])/(1 - a*Sqrt[b^(-2)])]*(-5*(a^2 - b^2)*AppellF1[2/3, 1/2, 1/2, 5/3, -((a + b*Sec[c + d*x])/(
-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])] + 2*a*AppellF1[5/3, 1/2, 1/2, 8/3, -((a + b*
Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*(a + b*Sec[c + d*x]))*Sin[c +
d*x])/(5*b*(b + a*Cos[c + d*x])^(1/3)*Sqrt[1 - Cos[c + d*x]^2]*Sec[c + d*x]^(1/3)) + ((b + a*Cos[c + d*x])^(2
/3)*Sec[c + d*x]^(2/3)*Sqrt[(1 - Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 + a*Sqrt[b^(-2)])]*Sqrt[(1 + Sqrt[b^(-2)]*b*S
ec[c + d*x])/(1 - a*Sqrt[b^(-2)])]*(-5*(a^2 - b^2)*AppellF1[2/3, 1/2, 1/2, 5/3, -((a + b*Sec[c + d*x])/(-a + 1
/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])] + 2*a*AppellF1[5/3, 1/2, 1/2, 8/3, -((a + b*Sec[c
+ d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*(a + b*Sec[c + d*x]))*Sin[c + d*x])
/(5*b*Sqrt[1 - Cos[c + d*x]^2]) - (3*(b + a*Cos[c + d*x])^(2/3)*Sqrt[(1 - Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 + a*
Sqrt[b^(-2)])]*Sqrt[(1 + Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 - a*Sqrt[b^(-2)])]*(2*a*b*AppellF1[5/3, 1/2, 1/2, 8/3
, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*Sec[c + d*x]*Tan[c
+ d*x] - 5*(a^2 - b^2)*((b*AppellF1[5/3, 1/2, 3/2, 8/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b
*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*Sec[c + d*x]*Tan[c + d*x])/(5*(a + 1/Sqrt[b^(-2)])) - (b*AppellF1[5/3, 3/
2, 1/2, 8/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*Sec[c +
d*x]*Tan[c + d*x])/(5*(-a + 1/Sqrt[b^(-2)]))) + 2*a*(a + b*Sec[c + d*x])*((5*b*AppellF1[8/3, 1/2, 3/2, 11/3,
-((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*Sec[c + d*x]*Tan[c +
d*x])/(16*(a + 1/Sqrt[b^(-2)])) - (5*b*AppellF1[8/3, 3/2, 1/2, 11/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-
2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*Sec[c + d*x]*Tan[c + d*x])/(16*(-a + 1/Sqrt[b^(-2)])))))/(5*
b*Sqrt[1 - Cos[c + d*x]^2]*Sec[c + d*x]^(1/3))))
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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \left (\sec ^{2}\left (d x +c \right )\right ) \left (a +b \sec \left (d x +c \right )\right )^{\frac {2}{3}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(2/3),x)
[Out]
int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(2/3),x)
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(2/3),x, algorithm="maxima")
[Out]
integrate((b*sec(d*x + c) + a)^(2/3)*sec(d*x + c)^2, x)
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(2/3),x, algorithm="fricas")
[Out]
integral((b*sec(d*x + c) + a)^(2/3)*sec(d*x + c)^2, x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}} \sec ^{2}{\left (c + d x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)**2*(a+b*sec(d*x+c))**(2/3),x)
[Out]
Integral((a + b*sec(c + d*x))**(2/3)*sec(c + d*x)**2, x)
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(2/3),x, algorithm="giac")
[Out]
integrate((b*sec(d*x + c) + a)^(2/3)*sec(d*x + c)^2, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}}{{\cos \left (c+d\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b/cos(c + d*x))^(2/3)/cos(c + d*x)^2,x)
[Out]
int((a + b/cos(c + d*x))^(2/3)/cos(c + d*x)^2, x)
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