3.6.68 \(\int \cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\) [568]

Optimal. Leaf size=182 \[ \frac {2 \left (6 a A b+3 a^2 B+5 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (5 a^2 A+7 b (A b+2 a B)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 \left (5 a^2 A+7 b (A b+2 a B)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a (9 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x)) \sin (c+d x)}{7 d} \]

[Out]

2/5*(6*A*a*b+3*B*a^2+5*B*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))/d+2/21*(5*a^2*A+7*b*(A*b+2*B*a))*(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))/d+2/35*a*(9*A*b+7*B*a)*cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/7*a*A*cos(d*x+c)^(3/2)*(b+a*cos(d*x+c))*
sin(d*x+c)/d+2/21*(5*a^2*A+7*b*(A*b+2*B*a))*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]
time = 0.24, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3033, 3069, 3102, 2827, 2719, 2715, 2720} \begin {gather*} \frac {2 \left (3 a^2 B+6 a A b+5 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (5 a^2 A+7 b (2 a B+A b)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 \left (5 a^2 A+7 b (2 a B+A b)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d}+\frac {2 a (7 a B+9 A b) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{35 d}+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(6*a*A*b + 3*a^2*B + 5*b^2*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(5*a^2*A + 7*b*(A*b + 2*a*B))*EllipticF
[(c + d*x)/2, 2])/(21*d) + (2*(5*a^2*A + 7*b*(A*b + 2*a*B))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*a*(9*
A*b + 7*a*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(35*d) + (2*a*A*Cos[c + d*x]^(3/2)*(b + a*Cos[c + d*x])*Sin[c +
d*x])/(7*d)

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3033

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[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 3069

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*
x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f
*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c
- b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m
, 1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\int \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2 (B+A \cos (c+d x)) \, dx\\ &=\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x)) \sin (c+d x)}{7 d}+\frac {2}{7} \int \sqrt {\cos (c+d x)} \left (\frac {1}{2} b (3 a A+7 b B)+\frac {1}{2} \left (5 a^2 A+7 b (A b+2 a B)\right ) \cos (c+d x)+\frac {1}{2} a (9 A b+7 a B) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 a (9 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x)) \sin (c+d x)}{7 d}+\frac {4}{35} \int \sqrt {\cos (c+d x)} \left (\frac {7}{4} \left (6 a A b+3 a^2 B+5 b^2 B\right )+\frac {5}{4} \left (5 a^2 A+7 b (A b+2 a B)\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {2 a (9 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x)) \sin (c+d x)}{7 d}+\frac {1}{5} \left (6 a A b+3 a^2 B+5 b^2 B\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{7} \left (5 a^2 A+7 b (A b+2 a B)\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {2 \left (6 a A b+3 a^2 B+5 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (5 a^2 A+7 b (A b+2 a B)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a (9 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x)) \sin (c+d x)}{7 d}+\frac {1}{21} \left (5 a^2 A+7 b (A b+2 a B)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (6 a A b+3 a^2 B+5 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (5 a^2 A+7 b (A b+2 a B)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 \left (5 a^2 A+7 b (A b+2 a B)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a (9 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x)) \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]
time = 1.26, size = 139, normalized size = 0.76 \begin {gather*} \frac {42 \left (6 a A b+3 a^2 B+5 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (5 a^2 A+7 A b^2+14 a b B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sqrt {\cos (c+d x)} \left (42 a (2 A b+a B) \cos (c+d x)+5 \left (13 a^2 A+14 A b^2+28 a b B+3 a^2 A \cos (2 (c+d x))\right )\right ) \sin (c+d x)}{105 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(42*(6*a*A*b + 3*a^2*B + 5*b^2*B)*EllipticE[(c + d*x)/2, 2] + 10*(5*a^2*A + 7*A*b^2 + 14*a*b*B)*EllipticF[(c +
d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(42*a*(2*A*b + a*B)*Cos[c + d*x] + 5*(13*a^2*A + 14*A*b^2 + 28*a*b*B + 3*a^2*
A*Cos[2*(c + d*x)]))*Sin[c + d*x])/(105*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(547\) vs. \(2(218)=436\).
time = 2.21, size = 548, normalized size = 3.01

method	result	size
default	\(-\frac {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 (240 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}+\left (-360 a^{2} A -336 A b a -168 a^{2} B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (280 a^{2} A +336 A b a +140 A \,b^{2}+168 a^{2} B +280 B a b \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-80 a^{2} A -84 A b a -70 A \,b^{2}-42 a^{2} B -140 B a b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+25 a^{2} 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}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+35 A \,b^{2} \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}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-126 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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b +70 B a 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}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}-105 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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}\right )}{105 \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}\)	\(548\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(7/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8*
a^2+(-360*A*a^2-336*A*a*b-168*B*a^2)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(280*A*a^2+336*A*a*b+140*A*b^2+16
8*B*a^2+280*B*a*b)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-80*A*a^2-84*A*a*b-70*A*b^2-42*B*a^2-140*B*a*b)*si
n(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+25*a^2*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)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))+35*A*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))-126*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)*Ellipt
icE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b+70*B*a*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)*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))-63*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)*Elliptic
E(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-105*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)*Elliptic
E(cos(1/2*d*x+1/2*c),2^(1/2))*b^2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(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

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.74, size = 243, normalized size = 1.34 \begin {gather*} \frac {2 \, {\left (15 \, A a^{2} \cos \left (d x + c\right )^{2} + 25 \, A a^{2} + 70 \, B a b + 35 \, A b^{2} + 21 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, \sqrt {2} {\left (5 i \, A a^{2} + 14 i \, B a b + 7 i \, A b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, \sqrt {2} {\left (-5 i \, A a^{2} - 14 i \, B a b - 7 i \, A b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 \, \sqrt {2} {\left (-3 i \, B a^{2} - 6 i \, A a b - 5 i \, B b^{2}\right )} {\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 ) - 21 \, \sqrt {2} {\left (3 i \, B a^{2} + 6 i \, A a b + 5 i \, B b^{2}\right )} {\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 )}{105 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(2*(15*A*a^2*cos(d*x + c)^2 + 25*A*a^2 + 70*B*a*b + 35*A*b^2 + 21*(B*a^2 + 2*A*a*b)*cos(d*x + c))*sqrt(c
os(d*x + c))*sin(d*x + c) - 5*sqrt(2)*(5*I*A*a^2 + 14*I*B*a*b + 7*I*A*b^2)*weierstrassPInverse(-4, 0, cos(d*x
+ c) + I*sin(d*x + c)) - 5*sqrt(2)*(-5*I*A*a^2 - 14*I*B*a*b - 7*I*A*b^2)*weierstrassPInverse(-4, 0, cos(d*x +
c) - I*sin(d*x + c)) - 21*sqrt(2)*(-3*I*B*a^2 - 6*I*A*a*b - 5*I*B*b^2)*weierstrassZeta(-4, 0, weierstrassPInve
rse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*sqrt(2)*(3*I*B*a^2 + 6*I*A*a*b + 5*I*B*b^2)*weierstrassZeta(-4
, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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Mupad [B]
time = 3.17, size = 229, normalized size = 1.26 \begin {gather*} \frac {2\,A\,b^2\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,B\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a\,b\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}-\frac {2\,A\,a^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,A\,a\,b\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*A*b^2*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*B*b^2*ellipticE(c/2 + (d*
x)/2, 2))/d + (2*B*a*b*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d - (2*A*a
^2*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2)) -
(2*B*a^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(
1/2)) - (4*A*a*b*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*
x)^2)^(1/2))

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