3.500 \(\int \cos ^3(c+d x) (a+b \cos (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=371 \[ \frac{2 \left (8 a^2+81 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{693 b^2 d}+\frac{2 a \left (8 a^2+67 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{693 b^2 d}+\frac{2 \left (57 a^2 b^2+8 a^4+135 b^4\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{693 b^2 d}-\frac{2 \left (49 a^4 b^2+78 a^2 b^4+8 a^6-135 b^6\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{693 b^3 d \sqrt{a+b \cos (c+d x)}}+\frac{2 a \left (51 a^2 b^2+8 a^4+741 b^4\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{693 b^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{8 a \sin (c+d x) (a+b \cos (c+d x))^{7/2}}{99 b^2 d}+\frac{2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{7/2}}{11 b d} \]

[Out]

(2*a*(8*a^4 + 51*a^2*b^2 + 741*b^4)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(693*b^3*d
*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(8*a^6 + 49*a^4*b^2 + 78*a^2*b^4 - 135*b^6)*Sqrt[(a + b*Cos[c + d*x]
)/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(693*b^3*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(8*a^4 + 57*a^2*b^
2 + 135*b^4)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(693*b^2*d) + (2*a*(8*a^2 + 67*b^2)*(a + b*Cos[c + d*x])^(
3/2)*Sin[c + d*x])/(693*b^2*d) + (2*(8*a^2 + 81*b^2)*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(693*b^2*d) - (8
*a*(a + b*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(99*b^2*d) + (2*Cos[c + d*x]*(a + b*Cos[c + d*x])^(7/2)*Sin[c + d*
x])/(11*b*d)

________________________________________________________________________________________

Rubi [A]  time = 0.629158, antiderivative size = 371, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {2793, 3023, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (8 a^2+81 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{693 b^2 d}+\frac{2 a \left (8 a^2+67 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{693 b^2 d}+\frac{2 \left (57 a^2 b^2+8 a^4+135 b^4\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{693 b^2 d}-\frac{2 \left (49 a^4 b^2+78 a^2 b^4+8 a^6-135 b^6\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{693 b^3 d \sqrt{a+b \cos (c+d x)}}+\frac{2 a \left (51 a^2 b^2+8 a^4+741 b^4\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{693 b^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{8 a \sin (c+d x) (a+b \cos (c+d x))^{7/2}}{99 b^2 d}+\frac{2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{7/2}}{11 b d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(2*a*(8*a^4 + 51*a^2*b^2 + 741*b^4)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(693*b^3*d
*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(8*a^6 + 49*a^4*b^2 + 78*a^2*b^4 - 135*b^6)*Sqrt[(a + b*Cos[c + d*x]
)/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(693*b^3*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(8*a^4 + 57*a^2*b^
2 + 135*b^4)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(693*b^2*d) + (2*a*(8*a^2 + 67*b^2)*(a + b*Cos[c + d*x])^(
3/2)*Sin[c + d*x])/(693*b^2*d) + (2*(8*a^2 + 81*b^2)*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(693*b^2*d) - (8
*a*(a + b*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(99*b^2*d) + (2*Cos[c + d*x]*(a + b*Cos[c + d*x])^(7/2)*Sin[c + d*
x])/(11*b*d)

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !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 2753

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

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 \cos ^3(c+d x) (a+b \cos (c+d x))^{5/2} \, dx &=\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{11 b d}+\frac{2 \int (a+b \cos (c+d x))^{5/2} \left (a+\frac{9}{2} b \cos (c+d x)-2 a \cos ^2(c+d x)\right ) \, dx}{11 b}\\ &=-\frac{8 a (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{99 b^2 d}+\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{11 b d}+\frac{4 \int (a+b \cos (c+d x))^{5/2} \left (-\frac{5 a b}{2}+\frac{1}{4} \left (8 a^2+81 b^2\right ) \cos (c+d x)\right ) \, dx}{99 b^2}\\ &=\frac{2 \left (8 a^2+81 b^2\right ) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{693 b^2 d}-\frac{8 a (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{99 b^2 d}+\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{11 b d}+\frac{8 \int (a+b \cos (c+d x))^{3/2} \left (-\frac{15}{8} b \left (2 a^2-27 b^2\right )+\frac{5}{8} a \left (8 a^2+67 b^2\right ) \cos (c+d x)\right ) \, dx}{693 b^2}\\ &=\frac{2 a \left (8 a^2+67 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{693 b^2 d}+\frac{2 \left (8 a^2+81 b^2\right ) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{693 b^2 d}-\frac{8 a (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{99 b^2 d}+\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{11 b d}+\frac{16 \int \sqrt{a+b \cos (c+d x)} \left (-\frac{15}{8} a b \left (a^2-101 b^2\right )+\frac{15}{16} \left (8 a^4+57 a^2 b^2+135 b^4\right ) \cos (c+d x)\right ) \, dx}{3465 b^2}\\ &=\frac{2 \left (8 a^4+57 a^2 b^2+135 b^4\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{693 b^2 d}+\frac{2 a \left (8 a^2+67 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{693 b^2 d}+\frac{2 \left (8 a^2+81 b^2\right ) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{693 b^2 d}-\frac{8 a (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{99 b^2 d}+\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{11 b d}+\frac{32 \int \frac{\frac{15}{32} b \left (2 a^4+663 a^2 b^2+135 b^4\right )+\frac{15}{32} a \left (8 a^4+51 a^2 b^2+741 b^4\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{10395 b^2}\\ &=\frac{2 \left (8 a^4+57 a^2 b^2+135 b^4\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{693 b^2 d}+\frac{2 a \left (8 a^2+67 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{693 b^2 d}+\frac{2 \left (8 a^2+81 b^2\right ) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{693 b^2 d}-\frac{8 a (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{99 b^2 d}+\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{11 b d}+\frac{\left (a \left (8 a^4+51 a^2 b^2+741 b^4\right )\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{693 b^3}-\frac{\left (8 a^6+49 a^4 b^2+78 a^2 b^4-135 b^6\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{693 b^3}\\ &=\frac{2 \left (8 a^4+57 a^2 b^2+135 b^4\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{693 b^2 d}+\frac{2 a \left (8 a^2+67 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{693 b^2 d}+\frac{2 \left (8 a^2+81 b^2\right ) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{693 b^2 d}-\frac{8 a (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{99 b^2 d}+\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{11 b d}+\frac{\left (a \left (8 a^4+51 a^2 b^2+741 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}{693 b^3 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (\left (8 a^6+49 a^4 b^2+78 a^2 b^4-135 b^6\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}{693 b^3 \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 a \left (8 a^4+51 a^2 b^2+741 b^4\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{693 b^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 \left (8 a^6+49 a^4 b^2+78 a^2 b^4-135 b^6\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 )}{693 b^3 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (8 a^4+57 a^2 b^2+135 b^4\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{693 b^2 d}+\frac{2 a \left (8 a^2+67 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{693 b^2 d}+\frac{2 \left (8 a^2+81 b^2\right ) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{693 b^2 d}-\frac{8 a (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{99 b^2 d}+\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{11 b d}\\ \end{align*}

Mathematica [A]  time = 1.18883, size = 268, normalized size = 0.72 \[ \frac{16 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left (b \left (663 a^2 b^3+2 a^4 b+135 b^5\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+a \left (51 a^2 b^2+8 a^4+741 b^4\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )\right )-b \left (\left (-3732 a^2 b^2+64 a^4-2610 b^4\right ) \sin (c+d x)-b \left (4 \left (6 a^3+619 a b^2\right ) \sin (2 (c+d x))+b \left (\left (452 a^2+513 b^2\right ) \sin (3 (c+d x))+7 b (46 a \sin (4 (c+d x))+9 b \sin (5 (c+d x)))\right )\right )\right ) (a+b \cos (c+d x))}{5544 b^3 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(16*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b*(2*a^4*b + 663*a^2*b^3 + 135*b^5)*EllipticF[(c + d*x)/2, (2*b)/(a +
b)] + a*(8*a^4 + 51*a^2*b^2 + 741*b^4)*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/
2, (2*b)/(a + b)])) - b*(a + b*Cos[c + d*x])*((64*a^4 - 3732*a^2*b^2 - 2610*b^4)*Sin[c + d*x] - b*(4*(6*a^3 +
619*a*b^2)*Sin[2*(c + d*x)] + b*((452*a^2 + 513*b^2)*Sin[3*(c + d*x)] + 7*b*(46*a*Sin[4*(c + d*x)] + 9*b*Sin[5
*(c + d*x)])))))/(5544*b^3*d*Sqrt[a + b*Cos[c + d*x]])

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Maple [B]  time = 3.718, size = 1140, normalized size = 3.1 \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)^3*(a+b*cos(d*x+c))^(5/2),x)

[Out]

-2/693*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4032*b^6*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+(-7168*a*b^5-10080*b^6)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(4384*a^2*b^4+14336*a*b^5+11376*b^6)*
sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-928*a^3*b^3-6576*a^2*b^4-13232*a*b^5-6984*b^6)*sin(1/2*d*x+1/2*c)^6*
cos(1/2*d*x+1/2*c)+(-4*a^4*b^2+928*a^3*b^3+5024*a^2*b^4+6064*a*b^5+2772*b^6)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+
1/2*c)+(8*a^5*b+2*a^4*b^2-642*a^3*b^3-1416*a^2*b^4-1338*a*b^5-558*b^6)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)
-8*(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^6-49*(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^4*b^2-78*(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^2*b^4+135*(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
))*b^6+8*(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^6-8*(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*b+51*(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^2-51*(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^3*b^3+741*(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(co
s(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^4-741*(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^5)/b^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*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} \cos \left (d x + c\right )^{5} + 2 \, a b \cos \left (d x + c\right )^{4} + a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt{b \cos \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((b^2*cos(d*x + c)^5 + 2*a*b*cos(d*x + c)^4 + a^2*cos(d*x + c)^3)*sqrt(b*cos(d*x + c) + a), x)

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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)**3*(a+b*cos(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 \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^3, x)