3.1184 \(\int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=411 \[ \frac {2 \left (11 a^2-3 b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{3 a^2 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {2 \left (a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{3 a b^2 d (a+b \sin (c+d x))^{3/2}}-\frac {16 a \left (32 a^2-15 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{21 b^6 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {8 \left (32 a^2-11 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{21 b^5 d}+\frac {8 \left (24 a^2-7 b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{21 a b^4 d}-\frac {2 \left (80 a^2-21 b^2\right ) \sin ^2(c+d x) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{21 a^2 b^3 d}+\frac {8 \left (64 a^4-46 a^2 b^2+3 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{21 b^6 d \sqrt {a+b \sin (c+d x)}} \]

[Out]

-2/3*(a^2-b^2)*cos(d*x+c)*sin(d*x+c)^3/a/b^2/d/(a+b*sin(d*x+c))^(3/2)+2/3*(11*a^2-3*b^2)*cos(d*x+c)*sin(d*x+c)
^3/a^2/b^2/d/(a+b*sin(d*x+c))^(1/2)-8/21*(32*a^2-11*b^2)*cos(d*x+c)*(a+b*sin(d*x+c))^(1/2)/b^5/d+8/21*(24*a^2-
7*b^2)*cos(d*x+c)*sin(d*x+c)*(a+b*sin(d*x+c))^(1/2)/a/b^4/d-2/21*(80*a^2-21*b^2)*cos(d*x+c)*sin(d*x+c)^2*(a+b*
sin(d*x+c))^(1/2)/a^2/b^3/d+16/21*a*(32*a^2-15*b^2)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d
*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*(a+b*sin(d*x+c))^(1/2)/b^6/d/((a+b*sin(d*x+c)
)/(a+b))^(1/2)-8/21*(64*a^4-46*a^2*b^2+3*b^4)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*El
lipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/b^6/d/(a+b*sin(d*x+c
))^(1/2)

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Rubi [A]  time = 0.92, antiderivative size = 411, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {2891, 3049, 3023, 2752, 2663, 2661, 2655, 2653} \[ \frac {2 \left (11 a^2-3 b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{3 a^2 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {2 \left (a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{3 a b^2 d (a+b \sin (c+d x))^{3/2}}-\frac {2 \left (80 a^2-21 b^2\right ) \sin ^2(c+d x) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{21 a^2 b^3 d}+\frac {8 \left (24 a^2-7 b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{21 a b^4 d}-\frac {8 \left (32 a^2-11 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{21 b^5 d}+\frac {8 \left (-46 a^2 b^2+64 a^4+3 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{21 b^6 d \sqrt {a+b \sin (c+d x)}}-\frac {16 a \left (32 a^2-15 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{21 b^6 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(-2*(a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x]^3)/(3*a*b^2*d*(a + b*Sin[c + d*x])^(3/2)) + (2*(11*a^2 - 3*b^2)*Cos[
c + d*x]*Sin[c + d*x]^3)/(3*a^2*b^2*d*Sqrt[a + b*Sin[c + d*x]]) - (8*(32*a^2 - 11*b^2)*Cos[c + d*x]*Sqrt[a + b
*Sin[c + d*x]])/(21*b^5*d) + (8*(24*a^2 - 7*b^2)*Cos[c + d*x]*Sin[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(21*a*b^4
*d) - (2*(80*a^2 - 21*b^2)*Cos[c + d*x]*Sin[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]])/(21*a^2*b^3*d) - (16*a*(32*a^
2 - 15*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(21*b^6*d*Sqrt[(a + b*Sin[c
+ d*x])/(a + b)]) + (8*(64*a^4 - 46*a^2*b^2 + 3*b^4)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b
*Sin[c + d*x])/(a + b)])/(21*b^6*d*Sqrt[a + b*Sin[c + d*x]])

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]

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 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 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 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 2891

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[((a^2 - b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*b^2*d*
f*(m + 1)), x] + (-Dist[1/(a^2*b^2*(m + 1)*(m + 2)), Int[(a + b*Sin[e + f*x])^(m + 2)*(d*Sin[e + f*x])^n*Simp[
a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m + n + 3) + a*b*(m + 2)*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*(m +
n + 2)*(m + n + 4))*Sin[e + f*x]^2, x], x], x] + Simp[((a^2*(n - m + 1) - b^2*(m + n + 2))*Cos[e + f*x]*(a +
b*Sin[e + f*x])^(m + 2)*(d*Sin[e + f*x])^(n + 1))/(a^2*b^2*d*f*(m + 1)*(m + 2)), x]) /; FreeQ[{a, b, d, e, f,
n}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*m, 2*n] && LtQ[m, -1] &&  !LtQ[n, -1] && (LtQ[m, -2] || EqQ[m + n +
4, 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 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])))

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx &=-\frac {2 \left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{3 a b^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 \left (11 a^2-3 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{3 a^2 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {4 \int \frac {\sin ^2(c+d x) \left (\frac {15}{4} \left (4 a^2-b^2\right )-\frac {1}{2} a b \sin (c+d x)-\frac {1}{4} \left (80 a^2-21 b^2\right ) \sin ^2(c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx}{3 a^2 b^2}\\ &=-\frac {2 \left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{3 a b^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 \left (11 a^2-3 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{3 a^2 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {2 \left (80 a^2-21 b^2\right ) \cos (c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{21 a^2 b^3 d}-\frac {8 \int \frac {\sin (c+d x) \left (-\frac {1}{2} a \left (80 a^2-21 b^2\right )+\frac {5}{2} a^2 b \sin (c+d x)+\frac {5}{2} a \left (24 a^2-7 b^2\right ) \sin ^2(c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx}{21 a^2 b^3}\\ &=-\frac {2 \left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{3 a b^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 \left (11 a^2-3 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{3 a^2 b^2 d \sqrt {a+b \sin (c+d x)}}+\frac {8 \left (24 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x) \sqrt {a+b \sin (c+d x)}}{21 a b^4 d}-\frac {2 \left (80 a^2-21 b^2\right ) \cos (c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{21 a^2 b^3 d}-\frac {16 \int \frac {\frac {5}{2} a^2 \left (24 a^2-7 b^2\right )-10 a^3 b \sin (c+d x)-\frac {15}{4} a^2 \left (32 a^2-11 b^2\right ) \sin ^2(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{105 a^2 b^4}\\ &=-\frac {2 \left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{3 a b^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 \left (11 a^2-3 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{3 a^2 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {8 \left (32 a^2-11 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{21 b^5 d}+\frac {8 \left (24 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x) \sqrt {a+b \sin (c+d x)}}{21 a b^4 d}-\frac {2 \left (80 a^2-21 b^2\right ) \cos (c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{21 a^2 b^3 d}-\frac {32 \int \frac {\frac {15}{8} a^2 b \left (16 a^2-3 b^2\right )+\frac {15}{4} a^3 \left (32 a^2-15 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{315 a^2 b^5}\\ &=-\frac {2 \left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{3 a b^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 \left (11 a^2-3 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{3 a^2 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {8 \left (32 a^2-11 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{21 b^5 d}+\frac {8 \left (24 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x) \sqrt {a+b \sin (c+d x)}}{21 a b^4 d}-\frac {2 \left (80 a^2-21 b^2\right ) \cos (c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{21 a^2 b^3 d}-\frac {\left (8 a \left (32 a^2-15 b^2\right )\right ) \int \sqrt {a+b \sin (c+d x)} \, dx}{21 b^6}+\frac {\left (4 \left (64 a^4-46 a^2 b^2+3 b^4\right )\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{21 b^6}\\ &=-\frac {2 \left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{3 a b^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 \left (11 a^2-3 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{3 a^2 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {8 \left (32 a^2-11 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{21 b^5 d}+\frac {8 \left (24 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x) \sqrt {a+b \sin (c+d x)}}{21 a b^4 d}-\frac {2 \left (80 a^2-21 b^2\right ) \cos (c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{21 a^2 b^3 d}-\frac {\left (8 a \left (32 a^2-15 b^2\right ) \sqrt {a+b \sin (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}} \, dx}{21 b^6 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (4 \left (64 a^4-46 a^2 b^2+3 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}} \, dx}{21 b^6 \sqrt {a+b \sin (c+d x)}}\\ &=-\frac {2 \left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{3 a b^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 \left (11 a^2-3 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{3 a^2 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {8 \left (32 a^2-11 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{21 b^5 d}+\frac {8 \left (24 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x) \sqrt {a+b \sin (c+d x)}}{21 a b^4 d}-\frac {2 \left (80 a^2-21 b^2\right ) \cos (c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{21 a^2 b^3 d}-\frac {16 a \left (32 a^2-15 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{21 b^6 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {8 \left (64 a^4-46 a^2 b^2+3 b^4\right ) F\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{21 b^6 d \sqrt {a+b \sin (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 7.96, size = 257, normalized size = 0.63 \[ \frac {32 a \left (32 a^2-15 b^2\right ) (a+b)^2 \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2} E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )-16 \left (64 a^4-46 a^2 b^2+3 b^4\right ) (a+b) \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )-\frac {1}{2} b \cos (c+d x) \left (1024 a^4+1280 a^3 b \sin (c+d x)-288 a^2 b^2-8 \left (8 a^2 b^2-3 b^4\right ) \cos (2 (c+d x))-516 a b^3 \sin (c+d x)+12 a b^3 \sin (3 (c+d x))+3 b^4 \cos (4 (c+d x))-27 b^4\right )}{42 b^6 d (a+b \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(32*a*(a + b)^2*(32*a^2 - 15*b^2)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*((a + b*Sin[c + d*x])/(a + b
))^(3/2) - 16*(a + b)*(64*a^4 - 46*a^2*b^2 + 3*b^4)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*((a + b*Si
n[c + d*x])/(a + b))^(3/2) - (b*Cos[c + d*x]*(1024*a^4 - 288*a^2*b^2 - 27*b^4 - 8*(8*a^2*b^2 - 3*b^4)*Cos[2*(c
+ d*x)] + 3*b^4*Cos[4*(c + d*x)] + 1280*a^3*b*Sin[c + d*x] - 516*a*b^3*Sin[c + d*x] + 12*a*b^3*Sin[3*(c + d*x
)]))/2)/(42*b^6*d*(a + b*Sin[c + d*x])^(3/2))

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fricas [F]  time = 1.20, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (\cos \left (d x + c\right )^{6} - \cos \left (d x + c\right )^{4}\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((cos(d*x + c)^6 - cos(d*x + c)^4)*sqrt(b*sin(d*x + c) + a)/(3*a*b^2*cos(d*x + c)^2 - a^3 - 3*a*b^2 +
(b^3*cos(d*x + c)^2 - 3*a^2*b - b^3)*sin(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^4*sin(d*x + c)^2/(b*sin(d*x + c) + a)^(5/2), x)

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maple [B]  time = 1.90, size = 1642, normalized size = 4.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^2/(a+b*sin(d*x+c))^(5/2),x)

[Out]

-2/21*(6*a*b^5*sin(d*x+c)*cos(d*x+c)^4+(160*a^3*b^3-66*a*b^5)*cos(d*x+c)^2*sin(d*x+c)+4*(b/(a-b)*sin(d*x+c)+a/
(a-b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*b*(64*EllipticF((b/(a-b)*
sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*b-48*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a
+b))^(1/2))*a^3*b^2-46*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^3+27*EllipticF(
(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^4+3*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),(
(a-b)/(a+b))^(1/2))*b^5-64*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5+94*EllipticE(
(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^2-30*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2
),((a-b)/(a+b))^(1/2))*a*b^4)*sin(d*x+c)+3*b^6*cos(d*x+c)^6+(-16*a^2*b^4+3*b^6)*cos(d*x+c)^4+(128*a^4*b^2-28*a
^2*b^4-6*b^6)*cos(d*x+c)^2+256*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b
)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5*b-192*(b/(a-
b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticF
((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*b^2-184*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a
+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)
,((a-b)/(a+b))^(1/2))*a^3*b^3+108*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(
a-b)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^4+12*(b
/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*Ellip
ticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^5-256*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/
(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/
2),((a-b)/(a+b))^(1/2))*a^6+376*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-
b)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*b^2-120*(b/
(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*Ellipt
icE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^4)/(a+b*sin(d*x+c))^(3/2)/b^7/cos(d*x+c)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^4*sin(d*x + c)^2/(b*sin(d*x + c) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*sin(c + d*x)^2)/(a + b*sin(c + d*x))^(5/2),x)

[Out]

int((cos(c + d*x)^4*sin(c + d*x)^2)/(a + b*sin(c + d*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**2/(a+b*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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