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

Optimal. Leaf size=390 \[ -\frac{2 \left (a^2-b^2-c^2\right ) \left (15 a^2 e+56 a d+25 e \left (b^2+c^2\right )\right ) \sqrt{\frac{a+b \cos (x)+c \sin (x)}{a+\sqrt{b^2+c^2}}} \text{EllipticF}\left (\frac{1}{2} \left (x-\tan ^{-1}(b,c)\right ),\frac{2 \sqrt{b^2+c^2}}{a+\sqrt{b^2+c^2}}\right )}{105 \sqrt{a+b \cos (x)+c \sin (x)}}+\frac{2 \left (161 a^2 d+15 a^3 e+145 a e \left (b^2+c^2\right )+63 d \left (b^2+c^2\right )\right ) \sqrt{a+b \cos (x)+c \sin (x)} E\left (\frac{1}{2} \left (x-\tan ^{-1}(b,c)\right )|\frac{2 \sqrt{b^2+c^2}}{a+\sqrt{b^2+c^2}}\right )}{105 \sqrt{\frac{a+b \cos (x)+c \sin (x)}{a+\sqrt{b^2+c^2}}}}-\frac{2}{105} \sqrt{a+b \cos (x)+c \sin (x)} \left (c \cos (x) \left (15 a^2 e+56 a d+25 e \left (b^2+c^2\right )\right )-b \sin (x) \left (15 a^2 e+56 a d+25 e \left (b^2+c^2\right )\right )\right )-\frac{2}{35} (a+b \cos (x)+c \sin (x))^{3/2} (c \cos (x) (5 a e+7 d)-b \sin (x) (5 a e+7 d))-\frac{2}{7} (a+b \cos (x)+c \sin (x))^{5/2} (c e \cos (x)-b e \sin (x)) \]

[Out]

(2*(161*a^2*d + 63*(b^2 + c^2)*d + 15*a^3*e + 145*a*(b^2 + c^2)*e)*EllipticE[(x - ArcTan[b, c])/2, (2*Sqrt[b^2
 + c^2])/(a + Sqrt[b^2 + c^2])]*Sqrt[a + b*Cos[x] + c*Sin[x]])/(105*Sqrt[(a + b*Cos[x] + c*Sin[x])/(a + Sqrt[b
^2 + c^2])]) - (2*(a^2 - b^2 - c^2)*(56*a*d + 15*a^2*e + 25*(b^2 + c^2)*e)*EllipticF[(x - ArcTan[b, c])/2, (2*
Sqrt[b^2 + c^2])/(a + Sqrt[b^2 + c^2])]*Sqrt[(a + b*Cos[x] + c*Sin[x])/(a + Sqrt[b^2 + c^2])])/(105*Sqrt[a + b
*Cos[x] + c*Sin[x]]) - (2*(a + b*Cos[x] + c*Sin[x])^(5/2)*(c*e*Cos[x] - b*e*Sin[x]))/7 - (2*(a + b*Cos[x] + c*
Sin[x])^(3/2)*(c*(7*d + 5*a*e)*Cos[x] - b*(7*d + 5*a*e)*Sin[x]))/35 - (2*Sqrt[a + b*Cos[x] + c*Sin[x]]*(c*(56*
a*d + 15*a^2*e + 25*(b^2 + c^2)*e)*Cos[x] - b*(56*a*d + 15*a^2*e + 25*(b^2 + c^2)*e)*Sin[x]))/105

________________________________________________________________________________________

Rubi [A]  time = 0.887684, antiderivative size = 390, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3146, 3149, 3119, 2653, 3127, 2661} \[ -\frac{2 \left (a^2-b^2-c^2\right ) \left (15 a^2 e+56 a d+25 e \left (b^2+c^2\right )\right ) \sqrt{\frac{a+b \cos (x)+c \sin (x)}{a+\sqrt{b^2+c^2}}} F\left (\frac{1}{2} \left (x-\tan ^{-1}(b,c)\right )|\frac{2 \sqrt{b^2+c^2}}{a+\sqrt{b^2+c^2}}\right )}{105 \sqrt{a+b \cos (x)+c \sin (x)}}+\frac{2 \left (161 a^2 d+15 a^3 e+145 a e \left (b^2+c^2\right )+63 d \left (b^2+c^2\right )\right ) \sqrt{a+b \cos (x)+c \sin (x)} E\left (\frac{1}{2} \left (x-\tan ^{-1}(b,c)\right )|\frac{2 \sqrt{b^2+c^2}}{a+\sqrt{b^2+c^2}}\right )}{105 \sqrt{\frac{a+b \cos (x)+c \sin (x)}{a+\sqrt{b^2+c^2}}}}-\frac{2}{105} \sqrt{a+b \cos (x)+c \sin (x)} \left (c \cos (x) \left (15 a^2 e+56 a d+25 e \left (b^2+c^2\right )\right )-b \sin (x) \left (15 a^2 e+56 a d+25 e \left (b^2+c^2\right )\right )\right )-\frac{2}{35} (a+b \cos (x)+c \sin (x))^{3/2} (c \cos (x) (5 a e+7 d)-b \sin (x) (5 a e+7 d))-\frac{2}{7} (a+b \cos (x)+c \sin (x))^{5/2} (c e \cos (x)-b e \sin (x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(161*a^2*d + 63*(b^2 + c^2)*d + 15*a^3*e + 145*a*(b^2 + c^2)*e)*EllipticE[(x - ArcTan[b, c])/2, (2*Sqrt[b^2
 + c^2])/(a + Sqrt[b^2 + c^2])]*Sqrt[a + b*Cos[x] + c*Sin[x]])/(105*Sqrt[(a + b*Cos[x] + c*Sin[x])/(a + Sqrt[b
^2 + c^2])]) - (2*(a^2 - b^2 - c^2)*(56*a*d + 15*a^2*e + 25*(b^2 + c^2)*e)*EllipticF[(x - ArcTan[b, c])/2, (2*
Sqrt[b^2 + c^2])/(a + Sqrt[b^2 + c^2])]*Sqrt[(a + b*Cos[x] + c*Sin[x])/(a + Sqrt[b^2 + c^2])])/(105*Sqrt[a + b
*Cos[x] + c*Sin[x]]) - (2*(a + b*Cos[x] + c*Sin[x])^(5/2)*(c*e*Cos[x] - b*e*Sin[x]))/7 - (2*(a + b*Cos[x] + c*
Sin[x])^(3/2)*(c*(7*d + 5*a*e)*Cos[x] - b*(7*d + 5*a*e)*Sin[x]))/35 - (2*Sqrt[a + b*Cos[x] + c*Sin[x]]*(c*(56*
a*d + 15*a^2*e + 25*(b^2 + c^2)*e)*Cos[x] - b*(56*a*d + 15*a^2*e + 25*(b^2 + c^2)*e)*Sin[x]))/105

Rule 3146

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

Rule 3149

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

Rule 3119

Int[Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*C
os[d + e*x] + c*Sin[d + e*x]]/Sqrt[(a + b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])], Int[Sqrt[a/(a
 + Sqrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]])/(a + Sqrt[b^2 + c^2])], x], x] /; FreeQ[{a
, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[b^2 + c^2, 0] &&  !GtQ[a + Sqrt[b^2 + c^2], 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 3127

Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a +
b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])]/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]], Int[1/Sqrt[
a/(a + Sqrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]])/(a + Sqrt[b^2 + c^2])], x], x] /; Free
Q[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[b^2 + c^2, 0] &&  !GtQ[a + Sqrt[b^2 + c^2], 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]

Rubi steps

\begin{align*} \int (a+b \cos (x)+c \sin (x))^{5/2} (d+b e \cos (x)+c e \sin (x)) \, dx &=-\frac{2}{7} (a+b \cos (x)+c \sin (x))^{5/2} (c e \cos (x)-b e \sin (x))+\frac{2 \int (a+b \cos (x)+c \sin (x))^{3/2} \left (\frac{1}{2} a \left (7 a d+5 \left (b^2+c^2\right ) e\right )+\frac{1}{2} a b (7 d+5 a e) \cos (x)+\frac{1}{2} a c (7 d+5 a e) \sin (x)\right ) \, dx}{7 a}\\ &=-\frac{2}{7} (a+b \cos (x)+c \sin (x))^{5/2} (c e \cos (x)-b e \sin (x))-\frac{2}{35} (a+b \cos (x)+c \sin (x))^{3/2} (c (7 d+5 a e) \cos (x)-b (7 d+5 a e) \sin (x))+\frac{4 \int \sqrt{a+b \cos (x)+c \sin (x)} \left (\frac{1}{4} a^2 \left (35 a^2 d+21 \left (b^2+c^2\right ) d+40 a \left (b^2+c^2\right ) e\right )+\frac{1}{4} a^2 b \left (56 a d+15 a^2 e+25 \left (b^2+c^2\right ) e\right ) \cos (x)+\frac{1}{4} a^2 c \left (56 a d+15 a^2 e+25 \left (b^2+c^2\right ) e\right ) \sin (x)\right ) \, dx}{35 a^2}\\ &=-\frac{2}{7} (a+b \cos (x)+c \sin (x))^{5/2} (c e \cos (x)-b e \sin (x))-\frac{2}{35} (a+b \cos (x)+c \sin (x))^{3/2} (c (7 d+5 a e) \cos (x)-b (7 d+5 a e) \sin (x))-\frac{2}{105} \sqrt{a+b \cos (x)+c \sin (x)} \left (c \left (56 a d+15 a^2 e+25 \left (b^2+c^2\right ) e\right ) \cos (x)-b \left (56 a d+15 a^2 e+25 \left (b^2+c^2\right ) e\right ) \sin (x)\right )+\frac{8 \int \frac{\frac{1}{8} a^3 \left (105 a^3 d+119 a \left (b^2+c^2\right ) d+135 a^2 \left (b^2+c^2\right ) e+25 \left (b^2+c^2\right )^2 e\right )+\frac{1}{8} a^3 b \left (161 a^2 d+63 \left (b^2+c^2\right ) d+15 a^3 e+145 a \left (b^2+c^2\right ) e\right ) \cos (x)+\frac{1}{8} a^3 c \left (161 a^2 d+63 \left (b^2+c^2\right ) d+15 a^3 e+145 a \left (b^2+c^2\right ) e\right ) \sin (x)}{\sqrt{a+b \cos (x)+c \sin (x)}} \, dx}{105 a^3}\\ &=-\frac{2}{7} (a+b \cos (x)+c \sin (x))^{5/2} (c e \cos (x)-b e \sin (x))-\frac{2}{35} (a+b \cos (x)+c \sin (x))^{3/2} (c (7 d+5 a e) \cos (x)-b (7 d+5 a e) \sin (x))-\frac{2}{105} \sqrt{a+b \cos (x)+c \sin (x)} \left (c \left (56 a d+15 a^2 e+25 \left (b^2+c^2\right ) e\right ) \cos (x)-b \left (56 a d+15 a^2 e+25 \left (b^2+c^2\right ) e\right ) \sin (x)\right )-\frac{1}{105} \left (\left (a^2-b^2-c^2\right ) \left (56 a d+15 a^2 e+25 \left (b^2+c^2\right ) e\right )\right ) \int \frac{1}{\sqrt{a+b \cos (x)+c \sin (x)}} \, dx+\frac{1}{105} \left (161 a^2 d+63 \left (b^2+c^2\right ) d+15 a^3 e+145 a \left (b^2+c^2\right ) e\right ) \int \sqrt{a+b \cos (x)+c \sin (x)} \, dx\\ &=-\frac{2}{7} (a+b \cos (x)+c \sin (x))^{5/2} (c e \cos (x)-b e \sin (x))-\frac{2}{35} (a+b \cos (x)+c \sin (x))^{3/2} (c (7 d+5 a e) \cos (x)-b (7 d+5 a e) \sin (x))-\frac{2}{105} \sqrt{a+b \cos (x)+c \sin (x)} \left (c \left (56 a d+15 a^2 e+25 \left (b^2+c^2\right ) e\right ) \cos (x)-b \left (56 a d+15 a^2 e+25 \left (b^2+c^2\right ) e\right ) \sin (x)\right )+\frac{\left (\left (161 a^2 d+63 \left (b^2+c^2\right ) d+15 a^3 e+145 a \left (b^2+c^2\right ) e\right ) \sqrt{a+b \cos (x)+c \sin (x)}\right ) \int \sqrt{\frac{a}{a+\sqrt{b^2+c^2}}+\frac{\sqrt{b^2+c^2} \cos \left (x-\tan ^{-1}(b,c)\right )}{a+\sqrt{b^2+c^2}}} \, dx}{105 \sqrt{\frac{a+b \cos (x)+c \sin (x)}{a+\sqrt{b^2+c^2}}}}-\frac{\left (\left (a^2-b^2-c^2\right ) \left (56 a d+15 a^2 e+25 \left (b^2+c^2\right ) e\right ) \sqrt{\frac{a+b \cos (x)+c \sin (x)}{a+\sqrt{b^2+c^2}}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+\sqrt{b^2+c^2}}+\frac{\sqrt{b^2+c^2} \cos \left (x-\tan ^{-1}(b,c)\right )}{a+\sqrt{b^2+c^2}}}} \, dx}{105 \sqrt{a+b \cos (x)+c \sin (x)}}\\ &=\frac{2 \left (161 a^2 d+63 \left (b^2+c^2\right ) d+15 a^3 e+145 a \left (b^2+c^2\right ) e\right ) E\left (\frac{1}{2} \left (x-\tan ^{-1}(b,c)\right )|\frac{2 \sqrt{b^2+c^2}}{a+\sqrt{b^2+c^2}}\right ) \sqrt{a+b \cos (x)+c \sin (x)}}{105 \sqrt{\frac{a+b \cos (x)+c \sin (x)}{a+\sqrt{b^2+c^2}}}}-\frac{2 \left (a^2-b^2-c^2\right ) \left (56 a d+15 a^2 e+25 \left (b^2+c^2\right ) e\right ) F\left (\frac{1}{2} \left (x-\tan ^{-1}(b,c)\right )|\frac{2 \sqrt{b^2+c^2}}{a+\sqrt{b^2+c^2}}\right ) \sqrt{\frac{a+b \cos (x)+c \sin (x)}{a+\sqrt{b^2+c^2}}}}{105 \sqrt{a+b \cos (x)+c \sin (x)}}-\frac{2}{7} (a+b \cos (x)+c \sin (x))^{5/2} (c e \cos (x)-b e \sin (x))-\frac{2}{35} (a+b \cos (x)+c \sin (x))^{3/2} (c (7 d+5 a e) \cos (x)-b (7 d+5 a e) \sin (x))-\frac{2}{105} \sqrt{a+b \cos (x)+c \sin (x)} \left (c \left (56 a d+15 a^2 e+25 \left (b^2+c^2\right ) e\right ) \cos (x)-b \left (56 a d+15 a^2 e+25 \left (b^2+c^2\right ) e\right ) \sin (x)\right )\\ \end{align*}

Mathematica [C]  time = 6.90226, size = 7823, normalized size = 20.06 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B]  time = 16.03, size = 3502, normalized size = 9. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-b^2*sin(x-arctan(-b,c))-c^2*sin(x-arctan(-b,c))-a*(b^2+c^2)^(1/2))*cos(x-arctan(-b,c))^2/(b^2+c^2)^(1/2))^
(1/2)/(b^2+c^2)*((b^6*e+3*b^4*c^2*e+3*b^2*c^4*e+c^6*e)*(-2/7/(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))^2*(cos(x-arct
an(-b,c))^2*((b^2+c^2)^(1/2)*sin(x-arctan(-b,c))+a))^(1/2)+12/35/(b^2+c^2)*a*sin(x-arctan(-b,c))*(cos(x-arctan
(-b,c))^2*((b^2+c^2)^(1/2)*sin(x-arctan(-b,c))+a))^(1/2)-2/3*(5/7+24/35/(b^2+c^2)*a^2)/(b^2+c^2)^(1/2)*(cos(x-
arctan(-b,c))^2*((b^2+c^2)^(1/2)*sin(x-arctan(-b,c))+a))^(1/2)+2*(-4/35/(b^2+c^2)*a^2+5/21)*(1/(b^2+c^2)^(1/2)
*a-1)*((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2)*((-sin(x-arctan(-b,c))+1)*(b^2+c^2
)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+sin(x-arctan(-b,c)))*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2)/(cos(x
-arctan(-b,c))^2*((b^2+c^2)^(1/2)*sin(x-arctan(-b,c))+a))^(1/2)*EllipticF(((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c
))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2))+2*(-48*a^3-44*a*b^2-44*a*c^
2)/(105*(b^2+c^2)^(1/2)*b^2+105*(b^2+c^2)^(1/2)*c^2)*(1/(b^2+c^2)^(1/2)*a-1)*((-(b^2+c^2)^(1/2)*sin(x-arctan(-
b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2)*((-sin(x-arctan(-b,c))+1)*(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+
sin(x-arctan(-b,c)))*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2)/(cos(x-arctan(-b,c))^2*((b^2+c^2)^(1/2)*sin(x
-arctan(-b,c))+a))^(1/2)*((-1/(b^2+c^2)^(1/2)*a-1)*EllipticE(((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))-a)/(-a+(b^
2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2))+EllipticF(((-(b^2+c^2)^(1/2)*sin(x-arcta
n(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2))))+(3*(b^2+c^2)^(1/2)*
a*b^4*e+6*(b^2+c^2)^(1/2)*a*b^2*c^2*e+3*(b^2+c^2)^(1/2)*a*c^4*e+(b^2+c^2)^(1/2)*b^4*d+2*(b^2+c^2)^(1/2)*b^2*c^
2*d+(b^2+c^2)^(1/2)*c^4*d)*(-2/5/(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))*(cos(x-arctan(-b,c))^2*((b^2+c^2)^(1/2)*s
in(x-arctan(-b,c))+a))^(1/2)+8/15/(b^2+c^2)*a*(cos(x-arctan(-b,c))^2*((b^2+c^2)^(1/2)*sin(x-arctan(-b,c))+a))^
(1/2)+4/15/(b^2+c^2)^(1/2)*a*(1/(b^2+c^2)^(1/2)*a-1)*((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))-a)/(-a+(b^2+c^2)^(
1/2)))^(1/2)*((-sin(x-arctan(-b,c))+1)*(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+sin(x-arctan(-b,c)))*(b^
2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2)/(cos(x-arctan(-b,c))^2*((b^2+c^2)^(1/2)*sin(x-arctan(-b,c))+a))^(1/2)
*EllipticF(((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+
c^2)^(1/2)))^(1/2))+2*(3/5+8/15/(b^2+c^2)*a^2)*(1/(b^2+c^2)^(1/2)*a-1)*((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))-
a)/(-a+(b^2+c^2)^(1/2)))^(1/2)*((-sin(x-arctan(-b,c))+1)*(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+sin(x-
arctan(-b,c)))*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2)/(cos(x-arctan(-b,c))^2*((b^2+c^2)^(1/2)*sin(x-arcta
n(-b,c))+a))^(1/2)*((-1/(b^2+c^2)^(1/2)*a-1)*EllipticE(((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))-a)/(-a+(b^2+c^2)
^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2))+EllipticF(((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c
))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2))))+(3*a^2*b^4*e+6*a^2*b^2*c^
2*e+3*a^2*c^4*e+3*a*b^4*d+6*a*b^2*c^2*d+3*a*c^4*d)*(-2/3/(b^2+c^2)^(1/2)*(cos(x-arctan(-b,c))^2*((b^2+c^2)^(1/
2)*sin(x-arctan(-b,c))+a))^(1/2)+2/3*(1/(b^2+c^2)^(1/2)*a-1)*((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))-a)/(-a+(b^
2+c^2)^(1/2)))^(1/2)*((-sin(x-arctan(-b,c))+1)*(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+sin(x-arctan(-b,
c)))*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2)/(cos(x-arctan(-b,c))^2*((b^2+c^2)^(1/2)*sin(x-arctan(-b,c))+a
))^(1/2)*EllipticF(((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/
(a+(b^2+c^2)^(1/2)))^(1/2))-4/3/(b^2+c^2)^(1/2)*a*(1/(b^2+c^2)^(1/2)*a-1)*((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c
))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2)*((-sin(x-arctan(-b,c))+1)*(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+sin
(x-arctan(-b,c)))*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2)/(cos(x-arctan(-b,c))^2*((b^2+c^2)^(1/2)*sin(x-ar
ctan(-b,c))+a))^(1/2)*((-1/(b^2+c^2)^(1/2)*a-1)*EllipticE(((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))-a)/(-a+(b^2+c
^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2))+EllipticF(((-(b^2+c^2)^(1/2)*sin(x-arctan(-
b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2))))+2*((b^2+c^2)^(1/2)*a^3
*b^2*e+(b^2+c^2)^(1/2)*a^3*c^2*e+(b^2+c^2)^(3/2)*a^2*d+2*a^2*b^2*d*(b^2+c^2)^(1/2)+2*a^2*c^2*d*(b^2+c^2)^(1/2)
)*(1/(b^2+c^2)^(1/2)*a-1)*((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2)*((-sin(x-arcta
n(-b,c))+1)*(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+sin(x-arctan(-b,c)))*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^
(1/2)))^(1/2)/(cos(x-arctan(-b,c))^2*((b^2+c^2)^(1/2)*sin(x-arctan(-b,c))+a))^(1/2)*((-1/(b^2+c^2)^(1/2)*a-1)*
EllipticE(((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c
^2)^(1/2)))^(1/2))+EllipticF(((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^
2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2)))+2*a^3*b^2*d*(1/(b^2+c^2)^(1/2)*a-1)*((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,
c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2)*((-sin(x-arctan(-b,c))+1)*(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+si
n(x-arctan(-b,c)))*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2)/(-(-b^2*sin(x-arctan(-b,c))-c^2*sin(x-arctan(-b
,c))-a*(b^2+c^2)^(1/2))*cos(x-arctan(-b,c))^2/(b^2+c^2)^(1/2))^(1/2)*EllipticF(((-(b^2+c^2)^(1/2)*sin(x-arctan
(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2))+2*a^3*c^2*d*(1/(b^2+c^
2)^(1/2)*a-1)*((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2)*((-sin(x-arctan(-b,c))+1)*
(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+sin(x-arctan(-b,c)))*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2
)/(-(-b^2*sin(x-arctan(-b,c))-c^2*sin(x-arctan(-b,c))-a*(b^2+c^2)^(1/2))*cos(x-arctan(-b,c))^2/(b^2+c^2)^(1/2)
)^(1/2)*EllipticF(((-(b^2+c^2)^(1/2)*sin(x-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(
a+(b^2+c^2)^(1/2)))^(1/2)))/cos(x-arctan(-b,c))/((b^2*sin(x-arctan(-b,c))+c^2*sin(x-arctan(-b,c))+a*(b^2+c^2)^
(1/2))/(b^2+c^2)^(1/2))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(((b^3 - 3*b*c^2)*e*cos(x)^3 + 2*a*c^2*e + ((b^2 - c^2)*d + 2*(a*b^2 - a*c^2)*e)*cos(x)^2 + (a^2 + c^2
)*d + (2*a*b*d + (a^2*b + 3*b*c^2)*e)*cos(x) + ((3*b^2*c - c^3)*e*cos(x)^2 + 2*a*c*d + (a^2*c + c^3)*e + 2*(2*
a*b*c*e + b*c*d)*cos(x))*sin(x))*sqrt(b*cos(x) + c*sin(x) + 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((a+b*cos(x)+c*sin(x))**(5/2)*(d+b*e*cos(x)+c*e*sin(x)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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