3.112 \(\int \sec ^5(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=149 \[ -\frac {35 a^2}{96 d (a \sin (c+d x)+a)^{3/2}}-\frac {35 a}{64 d \sqrt {a \sin (c+d x)+a}}+\frac {35 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}+\frac {\sec ^4(c+d x) \sqrt {a \sin (c+d x)+a}}{4 d}+\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a \sin (c+d x)+a}} \]

[Out]

-35/96*a^2/d/(a+a*sin(d*x+c))^(3/2)+35/128*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^(1/2)
/d-35/64*a/d/(a+a*sin(d*x+c))^(1/2)+7/16*a*sec(d*x+c)^2/d/(a+a*sin(d*x+c))^(1/2)+1/4*sec(d*x+c)^4*(a+a*sin(d*x
+c))^(1/2)/d

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Rubi [A]  time = 0.20, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2675, 2687, 2667, 51, 63, 206} \[ -\frac {35 a^2}{96 d (a \sin (c+d x)+a)^{3/2}}-\frac {35 a}{64 d \sqrt {a \sin (c+d x)+a}}+\frac {35 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}+\frac {\sec ^4(c+d x) \sqrt {a \sin (c+d x)+a}}{4 d}+\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(35*Sqrt[a]*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(64*Sqrt[2]*d) - (35*a^2)/(96*d*(a + a*Sin[c
+ d*x])^(3/2)) - (35*a)/(64*d*Sqrt[a + a*Sin[c + d*x]]) + (7*a*Sec[c + d*x]^2)/(16*d*Sqrt[a + a*Sin[c + d*x]])
+ (Sec[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(4*d)

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2675

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p + 1)), x] + Dist[(a*(m + p + 1))/(g^2*(p + 1)), Int[(
g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ[m + 1/2, 2*p]

Rule 2687

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Simp[(b*(g*
Cos[e + f*x])^(p + 1))/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(a*(2*p + 1))/(2*g^2*(p + 1)), Int[
(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[p, -1] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \sec ^5(c+d x) \sqrt {a+a \sin (c+d x)} \, dx &=\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {1}{8} (7 a) \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {1}{32} \left (35 a^2\right ) \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {\left (35 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,a \sin (c+d x)\right )}{32 d}\\ &=-\frac {35 a^2}{96 d (a+a \sin (c+d x))^{3/2}}+\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {\left (35 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{64 d}\\ &=-\frac {35 a^2}{96 d (a+a \sin (c+d x))^{3/2}}-\frac {35 a}{64 d \sqrt {a+a \sin (c+d x)}}+\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {(35 a) \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{128 d}\\ &=-\frac {35 a^2}{96 d (a+a \sin (c+d x))^{3/2}}-\frac {35 a}{64 d \sqrt {a+a \sin (c+d x)}}+\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {(35 a) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{64 d}\\ &=\frac {35 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}-\frac {35 a^2}{96 d (a+a \sin (c+d x))^{3/2}}-\frac {35 a}{64 d \sqrt {a+a \sin (c+d x)}}+\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}\\ \end {align*}

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Mathematica [C]  time = 0.49, size = 179, normalized size = 1.20 \[ \frac {\sqrt {a (\sin (c+d x)+1)} \left (\frac {329 \sin (c+d x)+105 \sin (3 (c+d x))-70 \cos (2 (c+d x))-102}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}-(420-420 i) \sqrt [4]{-1} \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \sec \left (\frac {d x}{4}\right ) \left (\sin \left (\frac {1}{4} (2 c+d x)\right )+\cos \left (\frac {1}{4} (2 c+d x)\right )\right )\right )\right )}{768 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^5*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*((-420 + 420*I)*(-1)^(1/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*Sec[(d*x)/4]*(Cos[(2*c +
d*x)/4] + Sin[(2*c + d*x)/4])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + (-102 - 70*Cos[2*(c + d*x)] + 329*Si
n[c + d*x] + 105*Sin[3*(c + d*x)])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4))/(768*d*(Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2])^4)

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fricas [A]  time = 0.80, size = 121, normalized size = 0.81 \[ \frac {105 \, \sqrt {2} \sqrt {a} \cos \left (d x + c\right )^{4} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \, {\left (35 \, \cos \left (d x + c\right )^{2} - 7 \, {\left (15 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 8\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{768 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/768*(105*sqrt(2)*sqrt(a)*cos(d*x + c)^4*log(-(a*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) +
3*a)/(sin(d*x + c) - 1)) - 4*(35*cos(d*x + c)^2 - 7*(15*cos(d*x + c)^2 + 8)*sin(d*x + c) + 8)*sqrt(a*sin(d*x +
c) + a))/(d*cos(d*x + c)^4)

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giac [B]  time = 2.75, size = 442, normalized size = 2.97 \[ -\frac {\sqrt {2} {\left (420 \, \log \left (\frac {{\left | -\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}}{{\left | \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {3 \, {\left (\frac {24 \, {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} - \frac {210 \, {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{2}} - \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{2}} - \frac {72 \, {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} + \frac {3 \, {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{2}} + \frac {256 \, {\left (\frac {9 \, {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} + \frac {6 \, {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{2}} + 5 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )}}{{\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} + 1\right )}^{3}}\right )} \sqrt {a}}{3072 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-1/3072*sqrt(2)*(420*log(abs(-cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/abs(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1))*sgn
(cos(-1/4*pi + 1/2*d*x + 1/2*c)) + 3*(24*(cos(-1/4*pi + 1/2*d*x + 1/2*c) - 1)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*
c))/(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) - 210*(cos(-1/4*pi + 1/2*d*x + 1/2*c) - 1)^2*sgn(cos(-1/4*pi + 1/2*d*
x + 1/2*c))/(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)^2 - sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))*(cos(-1/4*pi + 1/2*d
*x + 1/2*c) + 1)^2/(cos(-1/4*pi + 1/2*d*x + 1/2*c) - 1)^2 - 72*(cos(-1/4*pi + 1/2*d*x + 1/2*c) - 1)*sgn(cos(-1
/4*pi + 1/2*d*x + 1/2*c))/(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) + 3*(cos(-1/4*pi + 1/2*d*x + 1/2*c) - 1)^2*sgn(
cos(-1/4*pi + 1/2*d*x + 1/2*c))/(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)^2 + 256*(9*(cos(-1/4*pi + 1/2*d*x + 1/2*c
) - 1)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) + 6*(cos(-1/4*pi + 1/2*d*x + 1
/2*c) - 1)^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)^2 + 5*sgn(cos(-1/4*pi +
1/2*d*x + 1/2*c)))/((cos(-1/4*pi + 1/2*d*x + 1/2*c) - 1)/(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) + 1)^3)*sqrt(a)/
d

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maple [A]  time = 0.33, size = 118, normalized size = 0.79 \[ -\frac {2 a^{5} \left (\frac {3}{16 a^{4} \sqrt {a +a \sin \left (d x +c \right )}}+\frac {1}{24 a^{3} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}\, a \left (11 \sin \left (d x +c \right )-15\right )}{8 \left (a \sin \left (d x +c \right )-a \right )^{2}}-\frac {35 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 \sqrt {a}}}{16 a^{4}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^(1/2),x)

[Out]

-2*a^5*(3/16/a^4/(a+a*sin(d*x+c))^(1/2)+1/24/a^3/(a+a*sin(d*x+c))^(3/2)+1/16/a^4*(1/8*(a+a*sin(d*x+c))^(1/2)*a
*(11*sin(d*x+c)-15)/(a*sin(d*x+c)-a)^2-35/16*2^(1/2)/a^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2
))))/d

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maxima [A]  time = 1.01, size = 168, normalized size = 1.13 \[ -\frac {105 \, \sqrt {2} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (105 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} a^{2} - 350 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a^{3} + 224 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{4} + 64 \, a^{5}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a + 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2}}}{768 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-1/768*(105*sqrt(2)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*
x + c) + a))) + 4*(105*(a*sin(d*x + c) + a)^3*a^2 - 350*(a*sin(d*x + c) + a)^2*a^3 + 224*(a*sin(d*x + c) + a)*
a^4 + 64*a^5)/((a*sin(d*x + c) + a)^(7/2) - 4*(a*sin(d*x + c) + a)^(5/2)*a + 4*(a*sin(d*x + c) + a)^(3/2)*a^2)
)/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+a\,\sin \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(1/2)/cos(c + d*x)^5,x)

[Out]

int((a + a*sin(c + d*x))^(1/2)/cos(c + d*x)^5, x)

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sympy [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(sec(d*x+c)**5*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Timed out

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