3.1145 \(\int \cos ^3(c+d x) \cot (c+d x) \sqrt{a+b \sin (c+d x)} \, dx\)
Optimal. Leaf size=338 \[ -\frac{2 \left (8 a^2-45 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{105 b^2 d}-\frac{2 \left (-53 a^2 b^2+8 a^4-60 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{105 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{2 a \left (8 a^2-51 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{105 b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 b^2 d}-\frac{2 \sin (c+d x) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{7 b d}+\frac{2 a \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}} \]
[Out]
(-2*(8*a^2 - 45*b^2)*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(105*b^2*d) + (8*a*Cos[c + d*x]*(a + b*Sin[c + d*x
])^(3/2))/(35*b^2*d) - (2*Cos[c + d*x]*Sin[c + d*x]*(a + b*Sin[c + d*x])^(3/2))/(7*b*d) + (2*a*(8*a^2 - 51*b^2
)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(105*b^3*d*Sqrt[(a + b*Sin[c + d*x])/
(a + b)]) - (2*(8*a^4 - 53*a^2*b^2 - 60*b^4)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c +
d*x])/(a + b)])/(105*b^3*d*Sqrt[a + b*Sin[c + d*x]]) + (2*a*EllipticPi[2, (c - Pi/2 + d*x)/2, (2*b)/(a + b)]*S
qrt[(a + b*Sin[c + d*x])/(a + b)])/(d*Sqrt[a + b*Sin[c + d*x]])
________________________________________________________________________________________
Rubi [A] time = 0.892257, antiderivative size = 338, normalized size of antiderivative
= 1., number of steps used = 10, number of rules used = 10, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\)
= 0.345, Rules used = {2895, 3049, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805}
\[ -\frac{2 \left (8 a^2-45 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{105 b^2 d}-\frac{2 \left (-53 a^2 b^2+8 a^4-60 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{105 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{2 a \left (8 a^2-51 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{105 b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 b^2 d}-\frac{2 \sin (c+d x) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{7 b d}+\frac{2 a \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^3*Cot[c + d*x]*Sqrt[a + b*Sin[c + d*x]],x]
[Out]
(-2*(8*a^2 - 45*b^2)*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(105*b^2*d) + (8*a*Cos[c + d*x]*(a + b*Sin[c + d*x
])^(3/2))/(35*b^2*d) - (2*Cos[c + d*x]*Sin[c + d*x]*(a + b*Sin[c + d*x])^(3/2))/(7*b*d) + (2*a*(8*a^2 - 51*b^2
)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(105*b^3*d*Sqrt[(a + b*Sin[c + d*x])/
(a + b)]) - (2*(8*a^4 - 53*a^2*b^2 - 60*b^4)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c +
d*x])/(a + b)])/(105*b^3*d*Sqrt[a + b*Sin[c + d*x]]) + (2*a*EllipticPi[2, (c - Pi/2 + d*x)/2, (2*b)/(a + b)]*S
qrt[(a + b*Sin[c + d*x])/(a + b)])/(d*Sqrt[a + b*Sin[c + d*x]])
Rule 2895
Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(a*(n + 3)*Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(b^2*d*f*(m
+ n + 3)*(m + n + 4)), x] + (-Dist[1/(b^2*(m + n + 3)*(m + n + 4)), Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x
])^m*Simp[a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n + 4) + a*b*m*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*
(m + n + 3)*(m + n + 5))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*(a + b*Sin[e
+ f*x])^(m + 1))/(b*d^2*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[
m, 0] || IntegersQ[2*m, 2*n]) && !m < -1 && !LtQ[n, -1] && NeQ[m + n + 3, 0] && NeQ[m + n + 4, 0]
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 3059
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*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] && NeQ[c^2 - d^2, 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]
Rule 3002
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^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 2807
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] && !GtQ[c + d, 0]
Rule 2805
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), 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[c + d, 0]
Rubi steps
\begin{align*} \int \cos ^3(c+d x) \cot (c+d x) \sqrt{a+b \sin (c+d x)} \, dx &=\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2}}{7 b d}-\frac{4 \int \csc (c+d x) \sqrt{a+b \sin (c+d x)} \left (-\frac{35 b^2}{4}+\frac{1}{2} a b \sin (c+d x)-\frac{1}{4} \left (8 a^2-45 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{35 b^2}\\ &=-\frac{2 \left (8 a^2-45 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{105 b^2 d}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2}}{7 b d}-\frac{8 \int \frac{\csc (c+d x) \left (-\frac{105 a b^2}{8}-\frac{1}{4} b \left (a^2+30 b^2\right ) \sin (c+d x)-\frac{1}{8} a \left (8 a^2-51 b^2\right ) \sin ^2(c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{105 b^2}\\ &=-\frac{2 \left (8 a^2-45 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{105 b^2 d}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2}}{7 b d}+\frac{8 \int \frac{\csc (c+d x) \left (\frac{105 a b^3}{8}-\frac{1}{8} \left (8 a^4-53 a^2 b^2-60 b^4\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{105 b^3}+\frac{\left (a \left (8 a^2-51 b^2\right )\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{105 b^3}\\ &=-\frac{2 \left (8 a^2-45 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{105 b^2 d}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2}}{7 b d}+a \int \frac{\csc (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx-\frac{\left (8 a^4-53 a^2 b^2-60 b^4\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{105 b^3}+\frac{\left (a \left (8 a^2-51 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}{105 b^3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}\\ &=-\frac{2 \left (8 a^2-45 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{105 b^2 d}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2}}{7 b d}+\frac{2 a \left (8 a^2-51 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)}}{105 b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (a \sqrt{\frac{a+b \sin (c+d x)}{a+b}}\right ) \int \frac{\csc (c+d x)}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{\sqrt{a+b \sin (c+d x)}}-\frac{\left (\left (8 a^4-53 a^2 b^2-60 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}{105 b^3 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{2 \left (8 a^2-45 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{105 b^2 d}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2}}{7 b d}+\frac{2 a \left (8 a^2-51 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)}}{105 b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 \left (8 a^4-53 a^2 b^2-60 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}}}{105 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{2 a \Pi \left (2;\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}}}{d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}
Mathematica [C] time = 3.31863, size = 435, normalized size = 1.29 \[ \frac{2 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (8 a^2-6 a b \sin (c+d x)+15 b^2 \cos (2 (c+d x))+75 b^2\right )-\frac{8 b \left (a^2+30 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}-\frac{2 a \left (8 a^2+159 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}+\frac{2 i \left (51 b^2-8 a^2\right ) \sec (c+d x) \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}} \sqrt{\frac{b (\sin (c+d x)+1)}{b-a}} \left (b \left (b \Pi \left (\frac{a+b}{a};i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )}{b^2 \sqrt{-\frac{1}{a+b}}}}{210 b^2 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^3*Cot[c + d*x]*Sqrt[a + b*Sin[c + d*x]],x]
[Out]
(((2*I)*(-8*a^2 + 51*b^2)*(-2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Sin[c + d*x]]], (a
+ b)/(a - b)] + b*(-2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Sin[c + d*x]]], (a + b)/(a - b)] +
b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Sin[c + d*x]]], (a + b)/(a - b)]))*Sec[c + d*
x]*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]*Sqrt[(b*(1 + Sin[c + d*x]))/(-a + b)])/(b^2*Sqrt[-(a + b)^(-1)]) -
(8*b*(a^2 + 30*b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/Sqrt[
a + b*Sin[c + d*x]] - (2*a*(8*a^2 + 159*b^2)*EllipticPi[2, (-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*S
in[c + d*x])/(a + b)])/Sqrt[a + b*Sin[c + d*x]] + 2*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(8*a^2 + 75*b^2 + 15
*b^2*Cos[2*(c + d*x)] - 6*a*b*Sin[c + d*x]))/(210*b^2*d)
________________________________________________________________________________________
Maple [B] time = 1.514, size = 1155, normalized size = 3.4 \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*cot(d*x+c)*(a+b*sin(d*x+c))^(1/2),x)
[Out]
2/105*(15*b^5*sin(d*x+c)^5+8*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b
/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*b-6*((a+b*sin(d*x+c))/(a-b))^(
1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),
((a-b)/(a+b))^(1/2))*a^3*b^2-53*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c)
)*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^3+111*((a+b*sin(d*x+c))/(
a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))
^(1/2),((a-b)/(a+b))^(1/2))*a*b^4-60*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d
*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^5-8*((a+b*sin(d*x+c))/(a
-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^
(1/2),((a-b)/(a+b))^(1/2))*a^5+59*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+
c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^2-51*((a+b*sin(d*x+c))/
(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b)
)^(1/2),((a-b)/(a+b))^(1/2))*a*b^4-105*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin
(d*x+c))*b/(a-b))^(1/2)*b^4*EllipticPi(((a+b*sin(d*x+c))/(a-b))^(1/2),(a-b)/a,((a-b)/(a+b))^(1/2))*a+105*((a+b
*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*b^5*EllipticPi(((a+b
*sin(d*x+c))/(a-b))^(1/2),(a-b)/a,((a-b)/(a+b))^(1/2))+18*a*b^4*sin(d*x+c)^4-a^2*b^3*sin(d*x+c)^3-60*b^5*sin(d
*x+c)^3-4*a^3*b^2*sin(d*x+c)^2-63*a*b^4*sin(d*x+c)^2+a^2*b^3*sin(d*x+c)+45*b^5*sin(d*x+c)+4*a^3*b^2+45*a*b^4)/
b^4/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{3} \cot \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*cot(d*x+c)*(a+b*sin(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate(sqrt(b*sin(d*x + c) + a)*cos(d*x + c)^3*cot(d*x + c), x)
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Fricas [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*cot(d*x+c)*(a+b*sin(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
Timed out
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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*cot(d*x+c)*(a+b*sin(d*x+c))**(1/2),x)
[Out]
Timed out
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Giac [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*cot(d*x+c)*(a+b*sin(d*x+c))^(1/2),x, algorithm="giac")
[Out]
Timed out