3.202 \(\int \frac {1}{(a+b \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=182 \[ \frac {a \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac {b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {5 a b \cos (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {b \cos (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^3} \]

[Out]

a*(2*a^2+3*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(7/2)/d+1/3*b*cos(d*x+c)/(a^2-b^2)/
d/(a+b*sin(d*x+c))^3+5/6*a*b*cos(d*x+c)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))^2+1/6*b*(11*a^2+4*b^2)*cos(d*x+c)/(a^2-
b^2)^3/d/(a+b*sin(d*x+c))

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Rubi [A]  time = 0.22, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2664, 2754, 12, 2660, 618, 204} \[ \frac {a \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac {b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {5 a b \cos (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {b \cos (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^(-4),x]

[Out]

(a*(2*a^2 + 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) + (b*Cos[c + d*x])/
(3*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^3) + (5*a*b*Cos[c + d*x])/(6*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x])^2) + (
b*(11*a^2 + 4*b^2)*Cos[c + d*x])/(6*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 2754

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

Rubi steps

\begin {align*} \int \frac {1}{(a+b \sin (c+d x))^4} \, dx &=\frac {b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^3}-\frac {\int \frac {-3 a+2 b \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac {b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^3}+\frac {5 a b \cos (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac {\int \frac {2 \left (3 a^2+2 b^2\right )-5 a b \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{6 \left (a^2-b^2\right )^2}\\ &=\frac {b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^3}+\frac {5 a b \cos (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac {b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {\int -\frac {3 a \left (2 a^2+3 b^2\right )}{a+b \sin (c+d x)} \, dx}{6 \left (a^2-b^2\right )^3}\\ &=\frac {b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^3}+\frac {5 a b \cos (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac {b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {\left (a \left (2 a^2+3 b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=\frac {b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^3}+\frac {5 a b \cos (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac {b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {\left (a \left (2 a^2+3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d}\\ &=\frac {b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^3}+\frac {5 a b \cos (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac {b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {\left (2 a \left (2 a^2+3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d}\\ &=\frac {a \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}+\frac {b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^3}+\frac {5 a b \cos (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac {b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.02, size = 157, normalized size = 0.86 \[ \frac {\frac {6 a \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac {b \cos (c+d x) \left (18 a^4+b^2 \left (11 a^2+4 b^2\right ) \sin ^2(c+d x)+3 a b \left (9 a^2+b^2\right ) \sin (c+d x)-5 a^2 b^2+2 b^4\right )}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))^3}}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^(-4),x]

[Out]

((6*a*(2*a^2 + 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(7/2) + (b*Cos[c + d*x]*(1
8*a^4 - 5*a^2*b^2 + 2*b^4 + 3*a*b*(9*a^2 + b^2)*Sin[c + d*x] + b^2*(11*a^2 + 4*b^2)*Sin[c + d*x]^2))/((a - b)^
3*(a + b)^3*(a + b*Sin[c + d*x])^3))/(6*d)

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fricas [B]  time = 0.59, size = 965, normalized size = 5.30 \[ \left [\frac {2 \, {\left (11 \, a^{4} b^{3} - 7 \, a^{2} b^{5} - 4 \, b^{7}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (9 \, a^{5} b^{2} - 8 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, {\left (2 \, a^{6} + 9 \, a^{4} b^{2} + 9 \, a^{2} b^{4} - 3 \, {\left (2 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (6 \, a^{5} b + 11 \, a^{3} b^{3} + 3 \, a b^{5} - {\left (2 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 12 \, {\left (3 \, a^{6} b - 2 \, a^{4} b^{3} - b^{7}\right )} \cos \left (d x + c\right )}{12 \, {\left (3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{11} - a^{9} b^{2} - 6 \, a^{7} b^{4} + 14 \, a^{5} b^{6} - 11 \, a^{3} b^{8} + 3 \, a b^{10}\right )} d + {\left ({\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{2} - {\left (3 \, a^{10} b - 11 \, a^{8} b^{3} + 14 \, a^{6} b^{5} - 6 \, a^{4} b^{7} - a^{2} b^{9} + b^{11}\right )} d\right )} \sin \left (d x + c\right )\right )}}, \frac {{\left (11 \, a^{4} b^{3} - 7 \, a^{2} b^{5} - 4 \, b^{7}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (9 \, a^{5} b^{2} - 8 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, {\left (2 \, a^{6} + 9 \, a^{4} b^{2} + 9 \, a^{2} b^{4} - 3 \, {\left (2 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (6 \, a^{5} b + 11 \, a^{3} b^{3} + 3 \, a b^{5} - {\left (2 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 6 \, {\left (3 \, a^{6} b - 2 \, a^{4} b^{3} - b^{7}\right )} \cos \left (d x + c\right )}{6 \, {\left (3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{11} - a^{9} b^{2} - 6 \, a^{7} b^{4} + 14 \, a^{5} b^{6} - 11 \, a^{3} b^{8} + 3 \, a b^{10}\right )} d + {\left ({\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{2} - {\left (3 \, a^{10} b - 11 \, a^{8} b^{3} + 14 \, a^{6} b^{5} - 6 \, a^{4} b^{7} - a^{2} b^{9} + b^{11}\right )} d\right )} \sin \left (d x + c\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

[1/12*(2*(11*a^4*b^3 - 7*a^2*b^5 - 4*b^7)*cos(d*x + c)^3 - 6*(9*a^5*b^2 - 8*a^3*b^4 - a*b^6)*cos(d*x + c)*sin(
d*x + c) - 3*(2*a^6 + 9*a^4*b^2 + 9*a^2*b^4 - 3*(2*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)^2 + (6*a^5*b + 11*a^3*b^3
+ 3*a*b^5 - (2*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x
+ c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(
b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 12*(3*a^6*b - 2*a^4*b^3 - b^7)*cos(d*x + c))/(3*(a^9*b
^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c)^2 - (a^11 - a^9*b^2 - 6*a^7*b^4 + 14*a^5*b^6 -
11*a^3*b^8 + 3*a*b^10)*d + ((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*d*cos(d*x + c)^2 - (3*a^10*b
- 11*a^8*b^3 + 14*a^6*b^5 - 6*a^4*b^7 - a^2*b^9 + b^11)*d)*sin(d*x + c)), 1/6*((11*a^4*b^3 - 7*a^2*b^5 - 4*b^
7)*cos(d*x + c)^3 - 3*(9*a^5*b^2 - 8*a^3*b^4 - a*b^6)*cos(d*x + c)*sin(d*x + c) + 3*(2*a^6 + 9*a^4*b^2 + 9*a^2
*b^4 - 3*(2*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)^2 + (6*a^5*b + 11*a^3*b^3 + 3*a*b^5 - (2*a^3*b^3 + 3*a*b^5)*cos(
d*x + c)^2)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - 6*(3*
a^6*b - 2*a^4*b^3 - b^7)*cos(d*x + c))/(3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c
)^2 - (a^11 - a^9*b^2 - 6*a^7*b^4 + 14*a^5*b^6 - 11*a^3*b^8 + 3*a*b^10)*d + ((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7
- 4*a^2*b^9 + b^11)*d*cos(d*x + c)^2 - (3*a^10*b - 11*a^8*b^3 + 14*a^6*b^5 - 6*a^4*b^7 - a^2*b^9 + b^11)*d)*si
n(d*x + c))]

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giac [B]  time = 0.36, size = 510, normalized size = 2.80 \[ \frac {\frac {3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {27 \, a^{6} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18 \, a^{4} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, a^{7} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 81 \, a^{5} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 36 \, a^{3} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 12 \, a b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 108 \, a^{6} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 42 \, a^{4} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, a^{2} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, a^{7} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, a^{5} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 18 \, a^{3} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 81 \, a^{6} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, a^{4} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{2} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, a^{7} b - 5 \, a^{5} b^{3} + 2 \, a^{3} b^{5}}{{\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/3*(3*(2*a^3 + 3*a*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a
^2 - b^2)))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + (27*a^6*b^2*tan(1/2*d*x + 1/2*c)^5 - 18*a^
4*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*a^2*b^6*tan(1/2*d*x + 1/2*c)^5 + 18*a^7*b*tan(1/2*d*x + 1/2*c)^4 + 81*a^5*b^3
*tan(1/2*d*x + 1/2*c)^4 - 36*a^3*b^5*tan(1/2*d*x + 1/2*c)^4 + 12*a*b^7*tan(1/2*d*x + 1/2*c)^4 + 108*a^6*b^2*ta
n(1/2*d*x + 1/2*c)^3 + 42*a^4*b^4*tan(1/2*d*x + 1/2*c)^3 - 8*a^2*b^6*tan(1/2*d*x + 1/2*c)^3 + 8*b^8*tan(1/2*d*
x + 1/2*c)^3 + 36*a^7*b*tan(1/2*d*x + 1/2*c)^2 + 120*a^5*b^3*tan(1/2*d*x + 1/2*c)^2 - 18*a^3*b^5*tan(1/2*d*x +
1/2*c)^2 + 12*a*b^7*tan(1/2*d*x + 1/2*c)^2 + 81*a^6*b^2*tan(1/2*d*x + 1/2*c) - 12*a^4*b^4*tan(1/2*d*x + 1/2*c
) + 6*a^2*b^6*tan(1/2*d*x + 1/2*c) + 18*a^7*b - 5*a^5*b^3 + 2*a^3*b^5)/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6
)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^3))/d

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maple [B]  time = 0.21, size = 1733, normalized size = 9.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(d*x+c))^4,x)

[Out]

9/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b^2*a^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*
c)^5-6/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b^4*a/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1
/2*c)^5+2/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b^6/a/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*
x+1/2*c)^5+6/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b*a^4/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2
*d*x+1/2*c)^4+27/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b^3*a^2/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*t
an(1/2*d*x+1/2*c)^4-12/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b^5/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)
*tan(1/2*d*x+1/2*c)^4+4/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b^7/a^2/(a^6-3*a^4*b^2+3*a^2*b^4
-b^6)*tan(1/2*d*x+1/2*c)^4+36/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*a^3*b^2/(a^6-3*a^4*b^2+3*a
^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^3+14/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*a*b^4/(a^6-3*a^4*b^2
+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^3-8/3/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3/a*b^6/(a^6-3*a^
4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^3+8/3/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3/a^3*b^8/(a
^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^3+12/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*a^4*
b/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^2+40/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*
a^2*b^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^2-6/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+
a)^3*b^5/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^2+4/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b
+a)^3/a^2*b^7/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^2+27/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/
2*c)*b+a)^3*b^2*a^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)-4/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x
+1/2*c)*b+a)^3*b^4*a/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)+2/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*
x+1/2*c)*b+a)^3*b^6/a/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)+6/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d
*x+1/2*c)*b+a)^3*b/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*a^4-5/3/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3
*b^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*a^2+2/3/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b^5/(a^6-3*a^
4*b^2+3*a^2*b^4-b^6)+2/d*a^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+
2*b)/(a^2-b^2)^(1/2))+3/d*a/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2
*b)/(a^2-b^2)^(1/2))*b^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 10.38, size = 708, normalized size = 3.89 \[ \frac {\frac {18\,a^4\,b-5\,a^2\,b^3+2\,b^5}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (6\,a^6\,b+20\,a^4\,b^3-3\,a^2\,b^5+2\,b^7\right )}{a^2\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (6\,a^6\,b+27\,a^4\,b^3-12\,a^2\,b^5+4\,b^7\right )}{a^2\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (27\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{a\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (9\,a^4\,b-6\,a^2\,b^3+2\,b^5\right )}{a\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^2+2\,b^2\right )\,\left (18\,a^4\,b-5\,a^2\,b^3+2\,b^5\right )}{3\,a^3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (3\,a^3+12\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a^3+12\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (12\,a^2\,b+8\,b^3\right )+a^3+6\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {a\,\mathrm {atan}\left (\frac {\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+3\,b^2\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}+\frac {a\,\left (2\,a^2+3\,b^2\right )\,\left (2\,a^6\,b-6\,a^4\,b^3+6\,a^2\,b^5-2\,b^7\right )}{2\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{2\,a^3+3\,a\,b^2}\right )\,\left (2\,a^2+3\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sin(c + d*x))^4,x)

[Out]

((18*a^4*b + 2*b^5 - 5*a^2*b^3)/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)^2*(6*a^6*b + 2
*b^7 - 3*a^2*b^5 + 20*a^4*b^3))/(a^2*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (tan(c/2 + (d*x)/2)^4*(6*a^6*b + 4
*b^7 - 12*a^2*b^5 + 27*a^4*b^3))/(a^2*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (b*tan(c/2 + (d*x)/2)*(27*a^4*b +
2*b^5 - 4*a^2*b^3))/(a*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (b*tan(c/2 + (d*x)/2)^5*(9*a^4*b + 2*b^5 - 6*a^
2*b^3))/(a*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (2*b*tan(c/2 + (d*x)/2)^3*(3*a^2 + 2*b^2)*(18*a^4*b + 2*b^5
- 5*a^2*b^3))/(3*a^3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)))/(d*(a^3*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^2
*(12*a*b^2 + 3*a^3) + tan(c/2 + (d*x)/2)^4*(12*a*b^2 + 3*a^3) + tan(c/2 + (d*x)/2)^3*(12*a^2*b + 8*b^3) + a^3
+ 6*a^2*b*tan(c/2 + (d*x)/2) + 6*a^2*b*tan(c/2 + (d*x)/2)^5)) + (a*atan((((a^2*tan(c/2 + (d*x)/2)*(2*a^2 + 3*b
^2))/((a + b)^(7/2)*(a - b)^(7/2)) + (a*(2*a^2 + 3*b^2)*(2*a^6*b - 2*b^7 + 6*a^2*b^5 - 6*a^4*b^3))/(2*(a + b)^
(7/2)*(a - b)^(7/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)))*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/(3*a*b^2 + 2*a^
3))*(2*a^2 + 3*b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2))

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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(1/(a+b*sin(d*x+c))**4,x)

[Out]

Timed out

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