3.2.73 \(\int \csc ^4(e+f x) (a+b \sin (e+f x))^3 \, dx\) [173]

Optimal. Leaf size=109 \[ -\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac {a \left (2 a^2+9 b^2\right ) \cot (e+f x)}{3 f}-\frac {7 a^2 b \cot (e+f x) \csc (e+f x)}{6 f}-\frac {a^2 \cot (e+f x) \csc ^2(e+f x) (a+b \sin (e+f x))}{3 f} \]

[Out]

-1/2*b*(3*a^2+2*b^2)*arctanh(cos(f*x+e))/f-1/3*a*(2*a^2+9*b^2)*cot(f*x+e)/f-7/6*a^2*b*cot(f*x+e)*csc(f*x+e)/f-
1/3*a^2*cot(f*x+e)*csc(f*x+e)^2*(a+b*sin(f*x+e))/f

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Rubi [A]
time = 0.14, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2871, 3100, 2827, 3852, 8, 3855} \begin {gather*} -\frac {a \left (2 a^2+9 b^2\right ) \cot (e+f x)}{3 f}-\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac {7 a^2 b \cot (e+f x) \csc (e+f x)}{6 f}-\frac {a^2 \cot (e+f x) \csc ^2(e+f x) (a+b \sin (e+f x))}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4*(a + b*Sin[e + f*x])^3,x]

[Out]

-1/2*(b*(3*a^2 + 2*b^2)*ArcTanh[Cos[e + f*x]])/f - (a*(2*a^2 + 9*b^2)*Cot[e + f*x])/(3*f) - (7*a^2*b*Cot[e + f
*x]*Csc[e + f*x])/(6*f) - (a^2*Cot[e + f*x]*Csc[e + f*x]^2*(a + b*Sin[e + f*x]))/(3*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2871

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/
(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e
 + f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 +
c*d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 +
d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], 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[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3100

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \csc ^4(e+f x) (a+b \sin (e+f x))^3 \, dx &=-\frac {a^2 \cot (e+f x) \csc ^2(e+f x) (a+b \sin (e+f x))}{3 f}+\frac {1}{3} \int \csc ^3(e+f x) \left (7 a^2 b+a \left (2 a^2+9 b^2\right ) \sin (e+f x)+b \left (a^2+3 b^2\right ) \sin ^2(e+f x)\right ) \, dx\\ &=-\frac {7 a^2 b \cot (e+f x) \csc (e+f x)}{6 f}-\frac {a^2 \cot (e+f x) \csc ^2(e+f x) (a+b \sin (e+f x))}{3 f}+\frac {1}{6} \int \csc ^2(e+f x) \left (2 a \left (2 a^2+9 b^2\right )+3 b \left (3 a^2+2 b^2\right ) \sin (e+f x)\right ) \, dx\\ &=-\frac {7 a^2 b \cot (e+f x) \csc (e+f x)}{6 f}-\frac {a^2 \cot (e+f x) \csc ^2(e+f x) (a+b \sin (e+f x))}{3 f}+\frac {1}{2} \left (b \left (3 a^2+2 b^2\right )\right ) \int \csc (e+f x) \, dx+\frac {1}{3} \left (a \left (2 a^2+9 b^2\right )\right ) \int \csc ^2(e+f x) \, dx\\ &=-\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac {7 a^2 b \cot (e+f x) \csc (e+f x)}{6 f}-\frac {a^2 \cot (e+f x) \csc ^2(e+f x) (a+b \sin (e+f x))}{3 f}-\frac {\left (a \left (2 a^2+9 b^2\right )\right ) \text {Subst}(\int 1 \, dx,x,\cot (e+f x))}{3 f}\\ &=-\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac {a \left (2 a^2+9 b^2\right ) \cot (e+f x)}{3 f}-\frac {7 a^2 b \cot (e+f x) \csc (e+f x)}{6 f}-\frac {a^2 \cot (e+f x) \csc ^2(e+f x) (a+b \sin (e+f x))}{3 f}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(525\) vs. \(2(109)=218\).
time = 6.14, size = 525, normalized size = 4.82 \begin {gather*} \frac {\left (-2 a^3 \cos \left (\frac {1}{2} (e+f x)\right )-9 a b^2 \cos \left (\frac {1}{2} (e+f x)\right )\right ) \csc \left (\frac {1}{2} (e+f x)\right ) (b+a \csc (e+f x))^3 \sin ^3(e+f x)}{6 f (a+b \sin (e+f x))^3}-\frac {3 a^2 b \csc ^2\left (\frac {1}{2} (e+f x)\right ) (b+a \csc (e+f x))^3 \sin ^3(e+f x)}{8 f (a+b \sin (e+f x))^3}-\frac {a^3 \cot \left (\frac {1}{2} (e+f x)\right ) \csc ^2\left (\frac {1}{2} (e+f x)\right ) (b+a \csc (e+f x))^3 \sin ^3(e+f x)}{24 f (a+b \sin (e+f x))^3}+\frac {\left (-3 a^2 b-2 b^3\right ) (b+a \csc (e+f x))^3 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right ) \sin ^3(e+f x)}{2 f (a+b \sin (e+f x))^3}+\frac {\left (3 a^2 b+2 b^3\right ) (b+a \csc (e+f x))^3 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin ^3(e+f x)}{2 f (a+b \sin (e+f x))^3}+\frac {3 a^2 b (b+a \csc (e+f x))^3 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin ^3(e+f x)}{8 f (a+b \sin (e+f x))^3}+\frac {(b+a \csc (e+f x))^3 \sec \left (\frac {1}{2} (e+f x)\right ) \left (2 a^3 \sin \left (\frac {1}{2} (e+f x)\right )+9 a b^2 \sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin ^3(e+f x)}{6 f (a+b \sin (e+f x))^3}+\frac {a^3 (b+a \csc (e+f x))^3 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin ^3(e+f x) \tan \left (\frac {1}{2} (e+f x)\right )}{24 f (a+b \sin (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4*(a + b*Sin[e + f*x])^3,x]

[Out]

((-2*a^3*Cos[(e + f*x)/2] - 9*a*b^2*Cos[(e + f*x)/2])*Csc[(e + f*x)/2]*(b + a*Csc[e + f*x])^3*Sin[e + f*x]^3)/
(6*f*(a + b*Sin[e + f*x])^3) - (3*a^2*b*Csc[(e + f*x)/2]^2*(b + a*Csc[e + f*x])^3*Sin[e + f*x]^3)/(8*f*(a + b*
Sin[e + f*x])^3) - (a^3*Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^2*(b + a*Csc[e + f*x])^3*Sin[e + f*x]^3)/(24*f*(a +
b*Sin[e + f*x])^3) + ((-3*a^2*b - 2*b^3)*(b + a*Csc[e + f*x])^3*Log[Cos[(e + f*x)/2]]*Sin[e + f*x]^3)/(2*f*(a
+ b*Sin[e + f*x])^3) + ((3*a^2*b + 2*b^3)*(b + a*Csc[e + f*x])^3*Log[Sin[(e + f*x)/2]]*Sin[e + f*x]^3)/(2*f*(a
 + b*Sin[e + f*x])^3) + (3*a^2*b*(b + a*Csc[e + f*x])^3*Sec[(e + f*x)/2]^2*Sin[e + f*x]^3)/(8*f*(a + b*Sin[e +
 f*x])^3) + ((b + a*Csc[e + f*x])^3*Sec[(e + f*x)/2]*(2*a^3*Sin[(e + f*x)/2] + 9*a*b^2*Sin[(e + f*x)/2])*Sin[e
 + f*x]^3)/(6*f*(a + b*Sin[e + f*x])^3) + (a^3*(b + a*Csc[e + f*x])^3*Sec[(e + f*x)/2]^2*Sin[e + f*x]^3*Tan[(e
 + f*x)/2])/(24*f*(a + b*Sin[e + f*x])^3)

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Maple [A]
time = 0.46, size = 99, normalized size = 0.91

method result size
derivativedivides \(\frac {a^{3} \left (-\frac {2}{3}-\frac {\left (\csc ^{2}\left (f x +e \right )\right )}{3}\right ) \cot \left (f x +e \right )+3 a^{2} b \left (-\frac {\csc \left (f x +e \right ) \cot \left (f x +e \right )}{2}+\frac {\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{2}\right )-3 \cot \left (f x +e \right ) a \,b^{2}+b^{3} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{f}\) \(99\)
default \(\frac {a^{3} \left (-\frac {2}{3}-\frac {\left (\csc ^{2}\left (f x +e \right )\right )}{3}\right ) \cot \left (f x +e \right )+3 a^{2} b \left (-\frac {\csc \left (f x +e \right ) \cot \left (f x +e \right )}{2}+\frac {\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{2}\right )-3 \cot \left (f x +e \right ) a \,b^{2}+b^{3} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{f}\) \(99\)
risch \(\frac {a \left (-18 i b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+9 a b \,{\mathrm e}^{5 i \left (f x +e \right )}+12 i a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+36 i b^{2} {\mathrm e}^{2 i \left (f x +e \right )}-4 i a^{2}-18 i b^{2}-9 a b \,{\mathrm e}^{i \left (f x +e \right )}\right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{3}}-\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{2 f}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{f}+\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{2 f}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{f}\) \(186\)
norman \(\frac {-\frac {a^{3}}{24 f}+\frac {a^{3} \left (\tan ^{12}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{24 f}-\frac {21 a^{2} b \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8 f}-\frac {9 a^{2} b \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}-\frac {33 a^{2} b \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8 f}-\frac {a \left (a^{2}+3 b^{2}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}+\frac {a \left (a^{2}+3 b^{2}\right ) \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}-\frac {a \left (7 a^{2}+24 b^{2}\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8 f}+\frac {a \left (7 a^{2}+24 b^{2}\right ) \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8 f}-\frac {3 a^{2} b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{8 f}+\frac {3 a^{2} b \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8 f}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}+\frac {b \left (3 a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}\) \(290\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^4*(a+b*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

1/f*(a^3*(-2/3-1/3*csc(f*x+e)^2)*cot(f*x+e)+3*a^2*b*(-1/2*csc(f*x+e)*cot(f*x+e)+1/2*ln(csc(f*x+e)-cot(f*x+e)))
-3*cot(f*x+e)*a*b^2+b^3*ln(csc(f*x+e)-cot(f*x+e)))

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Maxima [A]
time = 0.28, size = 127, normalized size = 1.17 \begin {gather*} \frac {9 \, a^{2} b {\left (\frac {2 \, \cos \left (f x + e\right )}{\cos \left (f x + e\right )^{2} - 1} - \log \left (\cos \left (f x + e\right ) + 1\right ) + \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - 6 \, b^{3} {\left (\log \left (\cos \left (f x + e\right ) + 1\right ) - \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - \frac {36 \, a b^{2}}{\tan \left (f x + e\right )} - \frac {4 \, {\left (3 \, \tan \left (f x + e\right )^{2} + 1\right )} a^{3}}{\tan \left (f x + e\right )^{3}}}{12 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

1/12*(9*a^2*b*(2*cos(f*x + e)/(cos(f*x + e)^2 - 1) - log(cos(f*x + e) + 1) + log(cos(f*x + e) - 1)) - 6*b^3*(l
og(cos(f*x + e) + 1) - log(cos(f*x + e) - 1)) - 36*a*b^2/tan(f*x + e) - 4*(3*tan(f*x + e)^2 + 1)*a^3/tan(f*x +
 e)^3)/f

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Fricas [A]
time = 0.40, size = 203, normalized size = 1.86 \begin {gather*} \frac {18 \, a^{2} b \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 4 \, {\left (2 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (3 \, a^{2} b + 2 \, b^{3} - {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) \sin \left (f x + e\right ) - 3 \, {\left (3 \, a^{2} b + 2 \, b^{3} - {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) \sin \left (f x + e\right ) + 12 \, {\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (f x + e\right )}{12 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )} \sin \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

1/12*(18*a^2*b*cos(f*x + e)*sin(f*x + e) - 4*(2*a^3 + 9*a*b^2)*cos(f*x + e)^3 + 3*(3*a^2*b + 2*b^3 - (3*a^2*b
+ 2*b^3)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2)*sin(f*x + e) - 3*(3*a^2*b + 2*b^3 - (3*a^2*b + 2*b^3)*cos
(f*x + e)^2)*log(-1/2*cos(f*x + e) + 1/2)*sin(f*x + e) + 12*(a^3 + 3*a*b^2)*cos(f*x + e))/((f*cos(f*x + e)^2 -
 f)*sin(f*x + e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (e + f x \right )}\right )^{3} \csc ^{4}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**4*(a+b*sin(f*x+e))**3,x)

[Out]

Integral((a + b*sin(e + f*x))**3*csc(e + f*x)**4, x)

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Giac [A]
time = 0.45, size = 201, normalized size = 1.84 \begin {gather*} \frac {a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 9 \, a^{2} b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 9 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 36 \, a b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 12 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) - \frac {66 \, a^{2} b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 44 \, b^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 9 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 36 \, a b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 9 \, a^{2} b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a^{3}}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}}}{24 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e))^3,x, algorithm="giac")

[Out]

1/24*(a^3*tan(1/2*f*x + 1/2*e)^3 + 9*a^2*b*tan(1/2*f*x + 1/2*e)^2 + 9*a^3*tan(1/2*f*x + 1/2*e) + 36*a*b^2*tan(
1/2*f*x + 1/2*e) + 12*(3*a^2*b + 2*b^3)*log(abs(tan(1/2*f*x + 1/2*e))) - (66*a^2*b*tan(1/2*f*x + 1/2*e)^3 + 44
*b^3*tan(1/2*f*x + 1/2*e)^3 + 9*a^3*tan(1/2*f*x + 1/2*e)^2 + 36*a*b^2*tan(1/2*f*x + 1/2*e)^2 + 9*a^2*b*tan(1/2
*f*x + 1/2*e) + a^3)/tan(1/2*f*x + 1/2*e)^3)/f

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Mupad [B]
time = 6.78, size = 150, normalized size = 1.38 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (\frac {3\,a^2\,b}{2}+b^3\right )}{f}+\frac {a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{24\,f}-\frac {{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (3\,a^3+12\,a\,b^2\right )+\frac {a^3}{3}+3\,a^2\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{8\,f}+\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {3\,a^3}{8}+\frac {3\,a\,b^2}{2}\right )}{f}+\frac {3\,a^2\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{8\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x))^3/sin(e + f*x)^4,x)

[Out]

(log(tan(e/2 + (f*x)/2))*((3*a^2*b)/2 + b^3))/f + (a^3*tan(e/2 + (f*x)/2)^3)/(24*f) - (cot(e/2 + (f*x)/2)^3*(t
an(e/2 + (f*x)/2)^2*(12*a*b^2 + 3*a^3) + a^3/3 + 3*a^2*b*tan(e/2 + (f*x)/2)))/(8*f) + (tan(e/2 + (f*x)/2)*((3*
a*b^2)/2 + (3*a^3)/8))/f + (3*a^2*b*tan(e/2 + (f*x)/2)^2)/(8*f)

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