3.849 \(\int \cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=303 \[ \frac {2 \left (5 a^2+9 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{21 d}+\frac {2 b \left (29 a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{21 a d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}+\frac {2 \left (5 a^4-2 a^2 b^2-3 b^4\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{21 a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {6 a b \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d} \]

[Out]

2/21*(5*a^4-2*a^2*b^2-3*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1
/2)*(a/(a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)/a/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+6/7*a*b*cos(d*
x+c)^(3/2)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/d+2/7*a^2*cos(d*x+c)^(5/2)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/d+2/
21*(5*a^2+9*b^2)*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/d+2/21*b*(29*a^2+3*b^2)*(cos(1/2*d*x+1/2*c
)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*cos(d*x+c)^(1/2)*(a+b*sec(
d*x+c))^(1/2)/a/d/((b+a*cos(d*x+c))/(a+b))^(1/2)

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Rubi [A]  time = 1.01, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4264, 3841, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac {2 \left (5 a^2+9 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{21 d}+\frac {2 \left (-2 a^2 b^2+5 a^4-3 b^4\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{21 a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 b \left (29 a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{21 a d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}+\frac {6 a b \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(7/2)*(a + b*Sec[c + d*x])^(5/2),x]

[Out]

(2*(5*a^4 - 2*a^2*b^2 - 3*b^4)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(21*a
*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*b*(29*a^2 + 3*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)
/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(21*a*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (2*(5*a^2 + 9*b^2)*
Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(21*d) + (6*a*b*Cos[c + d*x]^(3/2)*Sqrt[a + b*Sec[c
+ d*x]]*Sin[c + d*x])/(7*d) + (2*a^2*Cos[c + d*x]^(5/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(7*d)

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 3841

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2*(n + 1))*Csc[e + f*x] - b*(b^2*
n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2]
&& ((IntegerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))

Rule 3856

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3858

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4035

Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]

Rule 4104

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

Rule 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

\begin {align*} \int \cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \sec (c+d x))^{5/2}}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {1}{7} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {15 a^2 b}{2}+\frac {1}{2} a \left (5 a^2+21 b^2\right ) \sec (c+d x)+\frac {1}{2} b \left (4 a^2+7 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {6 a b \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {2 a^2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d}-\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {5}{4} a^2 \left (5 a^2+9 b^2\right )-\frac {5}{4} a b \left (13 a^2+7 b^2\right ) \sec (c+d x)-\frac {15}{2} a^2 b^2 \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx}{35 a}\\ &=\frac {2 \left (5 a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{21 d}+\frac {6 a b \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {2 a^2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {5}{8} a^2 b \left (29 a^2+3 b^2\right )+\frac {5}{8} a^3 \left (5 a^2+27 b^2\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{105 a^2}\\ &=\frac {2 \left (5 a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{21 d}+\frac {6 a b \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {2 a^2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {\left (b \left (29 a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{21 a}+\frac {\left (\left (5 a^4-2 a^2 b^2-3 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{21 a}\\ &=\frac {2 \left (5 a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{21 d}+\frac {6 a b \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {2 a^2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {\left (\left (5 a^4-2 a^2 b^2-3 b^4\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{21 a \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (b \left (29 a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{21 a \sqrt {b+a \cos (c+d x)}}\\ &=\frac {2 \left (5 a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{21 d}+\frac {6 a b \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {2 a^2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {\left (\left (5 a^4-2 a^2 b^2-3 b^4\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{21 a \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (b \left (29 a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{21 a \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}\\ &=\frac {2 \left (5 a^4-2 a^2 b^2-3 b^4\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{21 a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 b \left (29 a^2+3 b^2\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{21 a d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 \left (5 a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{21 d}+\frac {6 a b \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {2 a^2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [C]  time = 12.14, size = 419, normalized size = 1.38 \[ \frac {\cos ^{\frac {5}{2}}(c+d x) \left (\frac {1}{42} \left (23 a^2+36 b^2\right ) \sin (c+d x)+\frac {1}{14} a^2 \sin (3 (c+d x))+\frac {3}{7} a b \sin (2 (c+d x))\right ) (a+b \sec (c+d x))^{5/2}}{d (a \cos (c+d x)+b)^2}+\frac {2 \cos ^{\frac {3}{2}}(c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} (a+b \sec (c+d x))^{5/2} \left (b \left (29 a^2+3 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} (a \cos (c+d x)+b)-i a \left (5 a^3+29 a^2 b+27 a b^2+3 b^3\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} F\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+i b \left (29 a^3+29 a^2 b+3 a b^2+3 b^3\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{21 a d \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+b)^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[c + d*x]^(7/2)*(a + b*Sec[c + d*x])^(5/2),x]

[Out]

(Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(5/2)*(((23*a^2 + 36*b^2)*Sin[c + d*x])/42 + (3*a*b*Sin[2*(c + d*x)])
/7 + (a^2*Sin[3*(c + d*x)])/14))/(d*(b + a*Cos[c + d*x])^2) + (2*Cos[c + d*x]^(3/2)*(Cos[(c + d*x)/2]^2*Sec[c
+ d*x])^(3/2)*(a + b*Sec[c + d*x])^(5/2)*(I*b*(29*a^3 + 29*a^2*b + 3*a*b^2 + 3*b^3)*EllipticE[I*ArcSinh[Tan[(c
+ d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - I*
a*(5*a^3 + 29*a^2*b + 27*a*b^2 + 3*b^3)*EllipticF[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)
/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + b*(29*a^2 + 3*b^2)*(b + a*Cos[c + d*x])*(Sec[(
c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]))/(21*a*d*(b + a*Cos[c + d*x])^3*Sec[c + d*x]^(5/2))

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fricas [F]  time = 1.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right )^{3} \sec \left (d x + c\right ) + a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((b^2*cos(d*x + c)^3*sec(d*x + c)^2 + 2*a*b*cos(d*x + c)^3*sec(d*x + c) + a^2*cos(d*x + c)^3)*sqrt(b*s
ec(d*x + c) + a)*sqrt(cos(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(7/2), x)

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maple [B]  time = 1.28, size = 2040, normalized size = 6.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(7/2)*(a+b*sec(d*x+c))^(5/2),x)

[Out]

-2/21/d*(-3*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*Ellip
ticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^4+5*cos(d*x+c)*sin(d*x+c)*Elliptic
F((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))
^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*a^4-5*a^3*b*((a-b)/(a+b))^(1/2)-29*a^2*b^2*((a-b)/(a+b))^(1/2)-9*a*b^3*((a-b)/
(a+b))^(1/2)+18*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*b^2+22*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*b+12*cos(d*x+
c)^2*((a-b)/(a+b))^(1/2)*a*b^3-29*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*b+11*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b
^2-3*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^3+12*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^3*b+3*cos(d*x+c)*((a-b)/(a+b))
^(1/2)*b^4+3*cos(d*x+c)^5*((a-b)/(a+b))^(1/2)*a^4+2*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^4-5*a^4*((a-b)/(a+b))^(
1/2)*cos(d*x+c)+5*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^4*((b+a*cos
(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-3*EllipticE((-1+cos(d*x+c))*((a-b)/(a
+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)
))^(1/2)*sin(d*x+c)-3*b^4*((a-b)/(a+b))^(1/2)-29*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a
+b)/(a-b))^(1/2))*a^3*b*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+27*E
llipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^2*((b+a*cos(d*x+c))/(1+cos
(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-3*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(
d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^3*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(
d*x+c)+29*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3*b*((b+a*cos(d*x+c
))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-29*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^
(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^2*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c))
)^(1/2)*sin(d*x+c)+3*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^3*((b+
a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-29*cos(d*x+c)*sin(d*x+c)*Ellipti
cF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b)
)^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*a^3*b+27*cos(d*x+c)*sin(d*x+c)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/
sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*a^2*b^
2-3*cos(d*x+c)*sin(d*x+c)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a
*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*a*b^3+29*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x
+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*
x+c),(-(a+b)/(a-b))^(1/2))*a^3*b-29*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+
cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^2+3*co
s(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d
*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^3)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x
+c)^(1/2)/(b+a*cos(d*x+c))/sin(d*x+c)/a/((a-b)/(a+b))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^{7/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(7/2)*(a + b/cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^(7/2)*(a + b/cos(c + d*x))^(5/2), 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(cos(d*x+c)**(7/2)*(a+b*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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