3.552 \(\int \cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=530 \[ \frac{\sqrt{a+b} \left (284 a^2 b+72 a^3+118 a b^2+15 b^3\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{192 a d}+\frac{b \left (284 a^2+15 b^2\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{192 a d}+\frac{\left (36 a^2+59 b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt{a+b \sec (c+d x)}}{96 d}+\frac{(a-b) \sqrt{a+b} \left (284 a^2+15 b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{192 a d}-\frac{\sqrt{a+b} \left (120 a^2 b^2+48 a^4-5 b^4\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{64 a^2 d}+\frac{a^2 \sin (c+d x) \cos ^3(c+d x) \sqrt{a+b \sec (c+d x)}}{4 d}+\frac{17 a b \sin (c+d x) \cos ^2(c+d x) \sqrt{a+b \sec (c+d x)}}{24 d} \]

[Out]

((a - b)*Sqrt[a + b]*(284*a^2 + 15*b^2)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (
a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(192*a*d) + (Sqr
t[a + b]*(72*a^3 + 284*a^2*b + 118*a*b^2 + 15*b^3)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt
[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(192*
a*d) - (Sqrt[a + b]*(48*a^4 + 120*a^2*b^2 - 5*b^4)*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c
+ d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a -
b))])/(64*a^2*d) + (b*(284*a^2 + 15*b^2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(192*a*d) + ((36*a^2 + 59*b^2
)*Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(96*d) + (17*a*b*Cos[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]
*Sin[c + d*x])/(24*d) + (a^2*Cos[c + d*x]^3*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(4*d)

________________________________________________________________________________________

Rubi [A]  time = 1.30013, antiderivative size = 530, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3841, 4104, 4058, 3921, 3784, 3832, 4004} \[ \frac{b \left (284 a^2+15 b^2\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{192 a d}+\frac{\left (36 a^2+59 b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt{a+b \sec (c+d x)}}{96 d}+\frac{\sqrt{a+b} \left (284 a^2 b+72 a^3+118 a b^2+15 b^3\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{192 a d}+\frac{(a-b) \sqrt{a+b} \left (284 a^2+15 b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{192 a d}-\frac{\sqrt{a+b} \left (120 a^2 b^2+48 a^4-5 b^4\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{64 a^2 d}+\frac{a^2 \sin (c+d x) \cos ^3(c+d x) \sqrt{a+b \sec (c+d x)}}{4 d}+\frac{17 a b \sin (c+d x) \cos ^2(c+d x) \sqrt{a+b \sec (c+d x)}}{24 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^(5/2),x]

[Out]

((a - b)*Sqrt[a + b]*(284*a^2 + 15*b^2)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (
a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(192*a*d) + (Sqr
t[a + b]*(72*a^3 + 284*a^2*b + 118*a*b^2 + 15*b^3)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt
[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(192*
a*d) - (Sqrt[a + b]*(48*a^4 + 120*a^2*b^2 - 5*b^4)*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c
+ d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a -
b))])/(64*a^2*d) + (b*(284*a^2 + 15*b^2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(192*a*d) + ((36*a^2 + 59*b^2
)*Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(96*d) + (17*a*b*Cos[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]
*Sin[c + d*x])/(24*d) + (a^2*Cos[c + d*x]^3*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(4*d)

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 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 4058

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

Rule 3921

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

Rule 3784

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*Sqrt[(b*(1 - Csc[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Csc[c + d*x]))/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[c + d*x]]/Rt[a
+ b, 2]], (a + b)/(a - b)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rubi steps

\begin{align*} \int \cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} \, dx &=\frac{a^2 \cos ^3(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{1}{4} \int \frac{\cos ^3(c+d x) \left (\frac{17 a^2 b}{2}+3 a \left (a^2+4 b^2\right ) \sec (c+d x)+\frac{1}{2} b \left (5 a^2+8 b^2\right ) \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{17 a b \cos ^2(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a^2 \cos ^3(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 d}-\frac{\int \frac{\cos ^2(c+d x) \left (-\frac{1}{4} a^2 \left (36 a^2+59 b^2\right )-\frac{1}{2} a b \left (49 a^2+24 b^2\right ) \sec (c+d x)-\frac{51}{4} a^2 b^2 \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{12 a}\\ &=\frac{\left (36 a^2+59 b^2\right ) \cos (c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{96 d}+\frac{17 a b \cos ^2(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a^2 \cos ^3(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{\int \frac{\cos (c+d x) \left (\frac{1}{8} a^2 b \left (284 a^2+15 b^2\right )+\frac{1}{4} a^3 \left (36 a^2+161 b^2\right ) \sec (c+d x)+\frac{1}{8} a^2 b \left (36 a^2+59 b^2\right ) \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{24 a^2}\\ &=\frac{b \left (284 a^2+15 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{192 a d}+\frac{\left (36 a^2+59 b^2\right ) \cos (c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{96 d}+\frac{17 a b \cos ^2(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a^2 \cos ^3(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 d}-\frac{\int \frac{-\frac{3}{16} a^2 \left (48 a^4+120 a^2 b^2-5 b^4\right )-\frac{1}{8} a^3 b \left (36 a^2+59 b^2\right ) \sec (c+d x)+\frac{1}{16} a^2 b^2 \left (284 a^2+15 b^2\right ) \sec ^2(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{24 a^3}\\ &=\frac{b \left (284 a^2+15 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{192 a d}+\frac{\left (36 a^2+59 b^2\right ) \cos (c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{96 d}+\frac{17 a b \cos ^2(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a^2 \cos ^3(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 d}-\frac{\int \frac{-\frac{3}{16} a^2 \left (48 a^4+120 a^2 b^2-5 b^4\right )+\left (-\frac{1}{16} a^2 b^2 \left (284 a^2+15 b^2\right )-\frac{1}{8} a^3 b \left (36 a^2+59 b^2\right )\right ) \sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{24 a^3}-\frac{\left (b^2 \left (284 a^2+15 b^2\right )\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{384 a}\\ &=\frac{(a-b) \sqrt{a+b} \left (284 a^2+15 b^2\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{192 a d}+\frac{b \left (284 a^2+15 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{192 a d}+\frac{\left (36 a^2+59 b^2\right ) \cos (c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{96 d}+\frac{17 a b \cos ^2(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a^2 \cos ^3(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{\left (b \left (72 a^3+284 a^2 b+118 a b^2+15 b^3\right )\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{384 a}+\frac{\left (48 a^4+120 a^2 b^2-5 b^4\right ) \int \frac{1}{\sqrt{a+b \sec (c+d x)}} \, dx}{128 a}\\ &=\frac{(a-b) \sqrt{a+b} \left (284 a^2+15 b^2\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{192 a d}+\frac{\sqrt{a+b} \left (72 a^3+284 a^2 b+118 a b^2+15 b^3\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{192 a d}-\frac{\sqrt{a+b} \left (48 a^4+120 a^2 b^2-5 b^4\right ) \cot (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{64 a^2 d}+\frac{b \left (284 a^2+15 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{192 a d}+\frac{\left (36 a^2+59 b^2\right ) \cos (c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{96 d}+\frac{17 a b \cos ^2(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a^2 \cos ^3(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [C]  time = 17.0028, size = 1688, normalized size = 3.18 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^(5/2),x]

[Out]

(Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*((17*a*b*Sin[c + d*x])/96 + ((48*a^2 + 59*b^2)*Sin[2*(c + d*x)])/19
2 + (17*a*b*Sin[3*(c + d*x)])/96 + (a^2*Sin[4*(c + d*x)])/32))/(d*(b + a*Cos[c + d*x])^2) + ((a + b*Sec[c + d*
x])^(5/2)*(-284*a^3*b*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2] - 284*a^2*b^2*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*
x)/2] - 15*a*b^3*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2] - 15*b^4*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2] + 56
8*a^3*b*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^3 + 30*a*b^3*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^3 - 284*a
^3*b*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 + 284*a^2*b^2*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 - 15*a*
b^3*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 + 15*b^4*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 + (288*I)*a^4
*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 -
Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (720*I)*a^2*b^2*Elli
pticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(
c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - (30*I)*b^4*EllipticPi[-((
a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(c + d*x)/2
]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (288*I)*a^4*EllipticPi[-((a + b)/(a
- b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(
c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (720*I)*a^2*b^2*EllipticP
i[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Tan[(c + d*x)/2]^2*
Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - (30*I)*b^4*
EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Tan[(c + d
*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + I*
b*(284*a^3 - 284*a^2*b + 15*a*b^2 - 15*b^3)*EllipticE[I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a +
b)/(a - b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[
(c + d*x)/2]^2)/(a + b)] - (2*I)*(72*a^4 - 36*a^3*b + 38*a^2*b^2 - 59*a*b^3 - 15*b^4)*EllipticF[I*ArcSinh[Sqrt
[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*S
qrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)]))/(192*a*Sqrt[(-a + b)/(a + b)]*d*(b + a*Co
s[c + d*x])^(5/2)*Sec[c + d*x]^(5/2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(-1 + Tan[(c + d*x)/2]^2)*(1 + Tan[(c
+ d*x)/2]^2)^(3/2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)])

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Maple [B]  time = 0.398, size = 2330, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+b*sec(d*x+c))^(5/2),x)

[Out]

-1/192/d/a*(-1+cos(d*x+c))^2*(48*cos(d*x+c)^6*a^4-72*cos(d*x+c)^2*a^4-15*cos(d*x+c)*b^4+15*(cos(d*x+c)/(cos(d*
x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+
b))^(1/2))*sin(d*x+c)*cos(d*x+c)*b^4+172*cos(d*x+c)^3*a^3*b+288*cos(d*x+c)*sin(d*x+c)*EllipticPi((-1+cos(d*x+c
))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*a^4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x
+c)+1))^(1/2)+24*cos(d*x+c)^4*a^4+15*cos(d*x+c)^2*b^4+133*cos(d*x+c)^3*a*b^3+30*cos(d*x+c)^2*a^2*b^2-118*cos(d
*x+c)*a*b^3+184*cos(d*x+c)^5*a^3*b+254*cos(d*x+c)^4*a^2*b^2-30*cos(d*x+c)*sin(d*x+c)*EllipticPi((-1+cos(d*x+c)
)/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*b^4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+
c)+1))^(1/2)-144*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((
-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*a^4+284*a^3*(cos(d*x+c)/(cos(d*x+c)+1))^(
1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))
*b*sin(d*x+c)+284*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(
(-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2*sin(d*x+c)+15*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+
b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3*sin(
d*x+c)+720*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticPi((-1+co
s(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*a^2*b^2*sin(d*x+c)+15*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*
(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*co
s(d*x+c)*a*b^3+720*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticP
i((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*a^2*b^2+72*(cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(
1/2))*sin(d*x+c)*cos(d*x+c)*a^3*b-644*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+
1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*a^2*b^2+118*(cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c)
,((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*a*b^3+284*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+
c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*a^3*
b+284*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+
c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*a^2*b^2-284*cos(d*x+c)^2*a^3*b-15*cos(d*x+c)^2*a*b^3
-72*cos(d*x+c)*a^3*b-284*cos(d*x+c)*a^2*b^2+15*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(co
s(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^4*sin(d*x+c)+288*sin(d*x+c)*Ell
ipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*a^4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a
*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)-30*sin(d*x+c)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))
*b^4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)-144*sin(d*x+c)*Elliptic
F((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+
c))/(cos(d*x+c)+1))^(1/2)+72*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)
*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b*sin(d*x+c)-644*(cos(d*x+c)/(cos(d*x+c)+1))^(1
/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*
a^2*b^2*sin(d*x+c)+118*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellip
ticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3*a*sin(d*x+c))*(cos(d*x+c)+1)^2*((b+a*cos(d*x+c))/cos(
d*x+c))^(1/2)/(b+a*cos(d*x+c))/sin(d*x+c)^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(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)^4, x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+b*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(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)^4, x)