\(\int \frac {(a+b \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [872]

Optimal result

Integrand size = 42, antiderivative size = 192 \[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (a^3 B+9 a b^2 B+9 a^2 b C+b^3 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (7 b B+3 a C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (a B-b C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

-2*(3*B*a^2*b-B*b^3+C*a^3-3*C*a*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+2/3*(B*a^3+9*B*a*b^2+9*C*a^2*b+C*b^3)*(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+2/3*a*B*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/3*a^2*(7*B*b+3*C*a)*
sin(d*x+c)/d/cos(d*x+c)^(1/2)-2/3*b^2*(B*a-C*b)*sin(d*x+c)*cos(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3108, 3068, 3110, 3102, 2827, 2720, 2719} \[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a^2 (3 a C+7 b B) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {2 \left (a^3 B+9 a^2 b C+9 a b^2 B+b^3 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {2 \left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 b^2 (a B-b C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

Int[((a + b*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]

[Out]

(-2*(3*a^2*b*B - b^3*B + a^3*C - 3*a*b^2*C)*EllipticE[(c + d*x)/2, 2])/d + (2*(a^3*B + 9*a*b^2*B + 9*a^2*b*C +
b^3*C)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^2*(7*b*B + 3*a*C)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]) - (2*
b^2*(a*B - b*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d) + (2*a*B*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d*Cos[
c + d*x]^(3/2))

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 3068

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*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1
)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Si
n[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c -
(A*b + a*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*
d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^
2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 1] && LtQ[n, -1]

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]

Rule 3108

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

Rule 3110

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {(a+b \cos (c+d x))^3 (B+C \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{2} a (7 b B+3 a C)+\frac {1}{2} \left (a^2 B+3 b^2 B+6 a b C\right ) \cos (c+d x)-\frac {3}{2} b (a B-b C) \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (7 b B+3 a C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {4}{3} \int \frac {-\frac {1}{4} a \left (a^2 B+10 b^2 B+9 a b C\right )+\frac {3}{4} \left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \cos (c+d x)+\frac {3}{4} b^2 (a B-b C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 a^2 (7 b B+3 a C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (a B-b C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {8}{9} \int \frac {-\frac {3}{8} \left (a^3 B+9 a b^2 B+9 a^2 b C+b^3 C\right )+\frac {9}{8} \left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 a^2 (7 b B+3 a C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (a B-b C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \int \sqrt {\cos (c+d x)} \, dx-\frac {1}{3} \left (-a^3 B-9 a b^2 B-9 a^2 b C-b^3 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {2 \left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (a^3 B+9 a b^2 B+9 a^2 b C+b^3 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (7 b B+3 a C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (a B-b C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.73 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {-6 \left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \left (a^3 B+9 a b^2 B+9 a^2 b C+b^3 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+18 a^2 b B \sin (c+d x)+6 a^3 C \sin (c+d x)+b^3 C \sin (2 (c+d x))+2 a^3 B \tan (c+d x)}{3 d \sqrt {\cos (c+d x)}} \]

[In]

Integrate[((a + b*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]

[Out]

(-6*(3*a^2*b*B - b^3*B + a^3*C - 3*a*b^2*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 2*(a^3*B + 9*a*b^2*
B + 9*a^2*b*C + b^3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 18*a^2*b*B*Sin[c + d*x] + 6*a^3*C*Sin[c
+ d*x] + b^3*C*Sin[2*(c + d*x)] + 2*a^3*B*Tan[c + d*x])/(3*d*Sqrt[Cos[c + d*x]])

Maple [C] (verified)

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

Time = 9.31 (sec) , antiderivative size = 771, normalized size of antiderivative = 4.02

method	result	size
parts	\(\text {Expression too large to display}\)	\(771\)
default	\(\text {Expression too large to display}\)	\(1210\)

[In]

int((a+cos(d*x+c)*b)^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

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

Time = 0.12 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.59 \[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {\sqrt {2} {\left (-i \, B a^{3} - 9 i \, C a^{2} b - 9 i \, B a b^{2} - i \, C b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, B a^{3} + 9 i \, C a^{2} b + 9 i \, B a b^{2} + i \, C b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (i \, C a^{3} + 3 i \, B a^{2} b - 3 i \, C a b^{2} - i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (-i \, C a^{3} - 3 i \, B a^{2} b + 3 i \, C a b^{2} + i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (C b^{3} \cos \left (d x + c\right )^{2} + B a^{3} + 3 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{2}} \]

[In]

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

1/3*(sqrt(2)*(-I*B*a^3 - 9*I*C*a^2*b - 9*I*B*a*b^2 - I*C*b^3)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*
x + c) + I*sin(d*x + c)) + sqrt(2)*(I*B*a^3 + 9*I*C*a^2*b + 9*I*B*a*b^2 + I*C*b^3)*cos(d*x + c)^2*weierstrassP
Inverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*(I*C*a^3 + 3*I*B*a^2*b - 3*I*C*a*b^2 - I*B*b^3)*cos(
d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(-I*C
*a^3 - 3*I*B*a^2*b + 3*I*C*a*b^2 + I*B*b^3)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c
os(d*x + c) - I*sin(d*x + c))) + 2*(C*b^3*cos(d*x + c)^2 + B*a^3 + 3*(C*a^3 + 3*B*a^2*b)*cos(d*x + c))*sqrt(co
s(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*cos(d*x+c))**3*(B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^3/cos(d*x + c)^(7/2), x)

Giac [F]

\[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^3/cos(d*x + c)^(7/2), x)

Mupad [B] (verification not implemented)

Time = 4.89 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.33 \[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2\,\left (B\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,b^3+3\,B\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,b^2\right )}{d}+\frac {C\,b^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {6\,C\,a\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a^2\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,B\,a^2\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^3)/cos(c + d*x)^(7/2),x)

[Out]

(2*(B*b^3*ellipticE(c/2 + (d*x)/2, 2) + 3*B*a*b^2*ellipticF(c/2 + (d*x)/2, 2)))/d + (C*b^3*((2*cos(c + d*x)^(1
/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (6*C*a*b^2*ellipticE(c/2 + (d*x)/2, 2))/d + (6*C
*a^2*b*ellipticF(c/2 + (d*x)/2, 2))/d + (2*B*a^3*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*
d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (2*C*a^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^
2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (6*B*a^2*b*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c
+ d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))