3.530 \(\int \frac{\cos ^4(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=326 \[ -\frac{2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (6 a^2-b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt{a+b \cos (c+d x)}}{5 b^2 d \left (a^2-b^2\right )}-\frac{2 a \left (8 a^2-3 b^2\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{5 b^3 d \left (a^2-b^2\right )}-\frac{8 a \left (4 a^2+b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{5 b^4 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (-8 a^2 b^2+16 a^4-3 b^4\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{5 b^4 d \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]
[Out]
(2*(16*a^4 - 8*a^2*b^2 - 3*b^4)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(5*b^4*(a^2 -
b^2)*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (8*a*(4*a^2 + b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(
c + d*x)/2, (2*b)/(a + b)])/(5*b^4*d*Sqrt[a + b*Cos[c + d*x]]) - (2*a^2*Cos[c + d*x]^2*Sin[c + d*x])/(b*(a^2 -
b^2)*d*Sqrt[a + b*Cos[c + d*x]]) - (2*a*(8*a^2 - 3*b^2)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*b^3*(a^2 -
b^2)*d) + (2*(6*a^2 - b^2)*Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*b^2*(a^2 - b^2)*d)
________________________________________________________________________________________
Rubi [A] time = 0.513377, antiderivative size = 326, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used =
{2792, 3049, 3023, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (6 a^2-b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt{a+b \cos (c+d x)}}{5 b^2 d \left (a^2-b^2\right )}-\frac{2 a \left (8 a^2-3 b^2\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{5 b^3 d \left (a^2-b^2\right )}-\frac{8 a \left (4 a^2+b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{5 b^4 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (-8 a^2 b^2+16 a^4-3 b^4\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{5 b^4 d \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^4/(a + b*Cos[c + d*x])^(3/2),x]
[Out]
(2*(16*a^4 - 8*a^2*b^2 - 3*b^4)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(5*b^4*(a^2 -
b^2)*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (8*a*(4*a^2 + b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(
c + d*x)/2, (2*b)/(a + b)])/(5*b^4*d*Sqrt[a + b*Cos[c + d*x]]) - (2*a^2*Cos[c + d*x]^2*Sin[c + d*x])/(b*(a^2 -
b^2)*d*Sqrt[a + b*Cos[c + d*x]]) - (2*a*(8*a^2 - 3*b^2)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*b^3*(a^2 -
b^2)*d) + (2*(6*a^2 - b^2)*Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*b^2*(a^2 - b^2)*d)
Rule 2792
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(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*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Rule 3049
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] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Rule 3023
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 2752
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[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 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 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 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 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]
Rubi steps
\begin{align*} \int \frac{\cos ^4(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx &=-\frac{2 a^2 \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{2 \int \frac{\cos (c+d x) \left (2 a^2-\frac{1}{2} a b \cos (c+d x)-\frac{1}{2} \left (6 a^2-b^2\right ) \cos ^2(c+d x)\right )}{\sqrt{a+b \cos (c+d x)}} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (6 a^2-b^2\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac{4 \int \frac{-\frac{1}{2} a \left (6 a^2-b^2\right )+\frac{1}{4} b \left (2 a^2+3 b^2\right ) \cos (c+d x)+\frac{3}{4} a \left (8 a^2-3 b^2\right ) \cos ^2(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{5 b^2 \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{2 a \left (8 a^2-3 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac{2 \left (6 a^2-b^2\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac{8 \int \frac{-\frac{3}{8} a b \left (4 a^2+b^2\right )-\frac{3}{8} \left (16 a^4-8 a^2 b^2-3 b^4\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{15 b^3 \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{2 a \left (8 a^2-3 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac{2 \left (6 a^2-b^2\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac{\left (4 a \left (4 a^2+b^2\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{5 b^4}+\frac{\left (16 a^4-8 a^2 b^2-3 b^4\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{5 b^4 \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{2 a \left (8 a^2-3 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac{2 \left (6 a^2-b^2\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}+\frac{\left (\left (16 a^4-8 a^2 b^2-3 b^4\right ) \sqrt{a+b \cos (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}} \, dx}{5 b^4 \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (4 a \left (4 a^2+b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{5 b^4 \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 \left (16 a^4-8 a^2 b^2-3 b^4\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{5 b^4 \left (a^2-b^2\right ) d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{8 a \left (4 a^2+b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{5 b^4 d \sqrt{a+b \cos (c+d x)}}-\frac{2 a^2 \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{2 a \left (8 a^2-3 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac{2 \left (6 a^2-b^2\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}\\ \end{align*}
Mathematica [A] time = 1.27169, size = 242, normalized size = 0.74 \[ \frac{-b \sin (c+d x) \left (4 a b \left (a^2-b^2\right ) \cos (c+d x)+\left (b^4-a^2 b^2\right ) \cos (2 (c+d x))-7 a^2 b^2+16 a^4+b^4\right )-8 a \left (-3 a^2 b^2+4 a^4-b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+2 \left (-8 a^3 b^2-8 a^2 b^3+16 a^4 b+16 a^5-3 a b^4-3 b^5\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{5 b^4 d (a-b) (a+b) \sqrt{a+b \cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^4/(a + b*Cos[c + d*x])^(3/2),x]
[Out]
(2*(16*a^5 + 16*a^4*b - 8*a^3*b^2 - 8*a^2*b^3 - 3*a*b^4 - 3*b^5)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[
(c + d*x)/2, (2*b)/(a + b)] - 8*a*(4*a^4 - 3*a^2*b^2 - b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c +
d*x)/2, (2*b)/(a + b)] - b*(16*a^4 - 7*a^2*b^2 + b^4 + 4*a*b*(a^2 - b^2)*Cos[c + d*x] + (-(a^2*b^2) + b^4)*Cos
[2*(c + d*x)])*Sin[c + d*x])/(5*(a - b)*b^4*(a + b)*d*Sqrt[a + b*Cos[c + d*x]])
________________________________________________________________________________________
Maple [B] time = 4.046, size = 1285, normalized size = 3.9 \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*cos(d*x+c))^(3/2),x)
[Out]
-2/5*(-8*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*(a^2-b^2)*cos(1/2*d*x+1/2*c)*sin(1/2
*d*x+1/2*c)^6-8*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2*(a^3-a^2*b-a*b^2+b^3)*sin(1/2
*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+2*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(8*a^4+2*a^3
*b-4*a^2*b^2-2*a*b^3+b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-16*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*
d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF
(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^5+12*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-
2*b/(a-b))^(1/2))*a^3*b^2+4*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a
*b^4+16*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*
sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^5-16*(-2*b*sin(1/2*
d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(
a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^4*b-8*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*si
n(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*Ell
ipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b^2+8*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)
^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+
1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^3-3*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))
^(1/2))*a*b^4+3*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*
b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^5)/b^4/(a-b
)/(a+b)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*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*b+a+b)^(1/2)/d
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{4}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4/(a+b*cos(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
integrate(cos(d*x + c)^4/(b*cos(d*x + c) + a)^(3/2), x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4/(a+b*cos(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
integral(sqrt(b*cos(d*x + c) + a)*cos(d*x + c)^4/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2), x)
________________________________________________________________________________________
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*cos(d*x+c))**(3/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{4}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4/(a+b*cos(d*x+c))^(3/2),x, algorithm="giac")
[Out]
integrate(cos(d*x + c)^4/(b*cos(d*x + c) + a)^(3/2), x)