\(\int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx\) [467]

Optimal result

Integrand size = 33, antiderivative size = 160 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {4 a^2 B \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {4 a^2 (3 A+2 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^2 (3 A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 B \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]

[Out]

2/3*B*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/3*a^2*(3*A-B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+4*a^2*B
*(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))*cos(d*x+c)^(1/2)*sec(d*
x+c)^(1/2)/d+4/3*a^2*(3*A+2*B)*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3039, 4102, 4082, 3872, 3856, 2719, 2720} \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {2 a^2 (3 A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^2 (3 A+2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 B \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d \sqrt {\sec (c+d x)}}+\frac {4 a^2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d} \]

[In]

Int[(a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x])*Sec[c + d*x]^(3/2),x]

[Out]

(4*a^2*B*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (4*a^2*(3*A + 2*B)*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*a^2*(3*A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/
(3*d) + (2*B*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3039

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

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

Rule 4082

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

Rule 4102

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {(a+a \sec (c+d x))^2 (B+A \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 B \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2}{3} \int \frac {(a+a \sec (c+d x)) \left (\frac {1}{2} a (3 A+5 B)+\frac {1}{2} a (3 A-B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 a^2 (3 A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 B \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {4}{3} \int \frac {\frac {3 a^2 B}{2}+\frac {1}{2} a^2 (3 A+2 B) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 a^2 (3 A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 B \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\left (2 a^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (2 a^2 (3 A+2 B)\right ) \int \sqrt {\sec (c+d x)} \, dx \\ & = \frac {2 a^2 (3 A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 B \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\left (2 a^2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (2 a^2 (3 A+2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {4 a^2 B \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {4 a^2 (3 A+2 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^2 (3 A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 B \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 2.33 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.95 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {2 a^2 \left (12 i B \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )-4 i (3 A+2 B) e^{i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )+\sqrt {1+e^{2 i (c+d x)}} (-6 i B+B \sin (c+d x)+3 A \tan (c+d x))\right )}{3 d \sqrt {1+e^{2 i (c+d x)}} \sqrt {\sec (c+d x)}} \]

[In]

Integrate[(a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x])*Sec[c + d*x]^(3/2),x]

[Out]

(2*a^2*((12*I)*B*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] - (4*I)*(3*A + 2*B)*E^(I*(c + d*x))*H
ypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))] + Sqrt[1 + E^((2*I)*(c + d*x))]*((-6*I)*B + B*Sin[c + d*
x] + 3*A*Tan[c + d*x])))/(3*d*Sqrt[1 + E^((2*I)*(c + d*x))]*Sqrt[Sec[c + d*x]])

Maple [A] (verified)

Time = 9.71 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.52

method	result	size
default	\(\frac {4 a^{2} \left (-2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(244\)
parts	\(-\frac {2 A \,a^{2} \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}+\frac {2 \left (A \,a^{2}+2 B \,a^{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}-\frac {2 \left (2 A \,a^{2}+B \,a^{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}-\frac {2 B \,a^{2} \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(657\)

[In]

int((a+cos(d*x+c)*a)^2*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

4/3*a^2*(-2*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+3*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*A*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+B*cos(1/2*d*x+1/2*
c)*sin(1/2*d*x+1/2*c)^2-2*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*
x+1/2*c),2^(1/2))+3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*
c),2^(1/2)))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.04 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {2 \, {\left (i \, \sqrt {2} {\left (3 \, A + 2 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {2} {\left (3 \, A + 2 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} B a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} B a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (B a^{2} \cos \left (d x + c\right ) + 3 \, A a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{3 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-2/3*(I*sqrt(2)*(3*A + 2*B)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - I*sqrt(2)*(3*A + 2
*B)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*B*a^2*weierstrassZeta(-4, 0, w
eierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*B*a^2*weierstrassZeta(-4, 0, weierstra
ssPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (B*a^2*cos(d*x + c) + 3*A*a^2)*sin(d*x + c)/sqrt(cos(d*x +
c)))/d

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**2*(A+B*cos(d*x+c))*sec(d*x+c)**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^2*sec(d*x + c)^(3/2), x)

Giac [F]

\[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^2*sec(d*x + c)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2 \,d x \]

[In]

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

[Out]

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