3.1338 \(\int \frac{\sin ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=268 \[ -\frac{a^3 \cos (c+d x)}{b^2 d \left (a^2-b^2\right )}+\frac{a \cos (c+d x)}{d \left (a^2-b^2\right )}+\frac{2 a^5 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 d \left (a^2-b^2\right )^{3/2}}-\frac{3 b \tan (c+d x)}{2 d \left (a^2-b^2\right )}+\frac{a \sec (c+d x)}{d \left (a^2-b^2\right )}+\frac{b \sin ^2(c+d x) \tan (c+d x)}{2 d \left (a^2-b^2\right )}+\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 b d \left (a^2-b^2\right )}-\frac{a^2 x \left (2 a^2+b^2\right )}{2 b^3 \left (a^2-b^2\right )}+\frac{3 b x}{2 \left (a^2-b^2\right )} \]

[Out]

(3*b*x)/(2*(a^2 - b^2)) - (a^2*(2*a^2 + b^2)*x)/(2*b^3*(a^2 - b^2)) + (2*a^5*ArcTan[(b + a*Tan[(c + d*x)/2])/S
qrt[a^2 - b^2]])/(b^3*(a^2 - b^2)^(3/2)*d) + (a*Cos[c + d*x])/((a^2 - b^2)*d) - (a^3*Cos[c + d*x])/(b^2*(a^2 -
b^2)*d) + (a*Sec[c + d*x])/((a^2 - b^2)*d) + (a^2*Cos[c + d*x]*Sin[c + d*x])/(2*b*(a^2 - b^2)*d) - (3*b*Tan[c
+ d*x])/(2*(a^2 - b^2)*d) + (b*Sin[c + d*x]^2*Tan[c + d*x])/(2*(a^2 - b^2)*d)

________________________________________________________________________________________

Rubi [A]  time = 0.395946, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 13, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.448, Rules used = {2902, 2590, 14, 2591, 288, 321, 203, 2793, 3023, 2735, 2660, 618, 204} \[ -\frac{a^3 \cos (c+d x)}{b^2 d \left (a^2-b^2\right )}+\frac{a \cos (c+d x)}{d \left (a^2-b^2\right )}+\frac{2 a^5 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 d \left (a^2-b^2\right )^{3/2}}-\frac{3 b \tan (c+d x)}{2 d \left (a^2-b^2\right )}+\frac{a \sec (c+d x)}{d \left (a^2-b^2\right )}+\frac{b \sin ^2(c+d x) \tan (c+d x)}{2 d \left (a^2-b^2\right )}+\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 b d \left (a^2-b^2\right )}-\frac{a^2 x \left (2 a^2+b^2\right )}{2 b^3 \left (a^2-b^2\right )}+\frac{3 b x}{2 \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]^3*Tan[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

(3*b*x)/(2*(a^2 - b^2)) - (a^2*(2*a^2 + b^2)*x)/(2*b^3*(a^2 - b^2)) + (2*a^5*ArcTan[(b + a*Tan[(c + d*x)/2])/S
qrt[a^2 - b^2]])/(b^3*(a^2 - b^2)^(3/2)*d) + (a*Cos[c + d*x])/((a^2 - b^2)*d) - (a^3*Cos[c + d*x])/(b^2*(a^2 -
b^2)*d) + (a*Sec[c + d*x])/((a^2 - b^2)*d) + (a^2*Cos[c + d*x]*Sin[c + d*x])/(2*b*(a^2 - b^2)*d) - (3*b*Tan[c
+ d*x])/(2*(a^2 - b^2)*d) + (b*Sin[c + d*x]^2*Tan[c + d*x])/(2*(a^2 - b^2)*d)

Rule 2902

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[(a*d^2)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[(b*d)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a^2*d^2)/(g^2*(a^2 - b^
2)), Int[((g*Cos[e + f*x])^(p + 2)*(d*Sin[e + f*x])^(n - 2))/(a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 203

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

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 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 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 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 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 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])

Rubi steps

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

Mathematica [A]  time = 1.52499, size = 221, normalized size = 0.82 \[ \frac{\frac{2 a^2 b^2 (c+d x)+4 a^4 (c+d x)-4 a b^3-6 b^4 (c+d x)}{b^5-a^2 b^3}+\frac{8 a^5 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{3/2}}-\frac{4 a \cos (c+d x)}{b^2}+\frac{4 \sin \left (\frac{1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{4 \sin \left (\frac{1}{2} (c+d x)\right )}{(a-b) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{\sin (2 (c+d x))}{b}}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]^3*Tan[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.108, size = 283, normalized size = 1.1 \begin{align*} -64\,{\frac{1}{d \left ( 64\,a+64\,b \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }}-{\frac{1}{bd} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a}{d{b}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{1}{bd}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{a}{d{b}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){a}^{2}}{d{b}^{3}}}-3\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{bd}}+2\,{\frac{{a}^{5}}{d \left ( a-b \right ) \left ( a+b \right ){b}^{3}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+64\,{\frac{1}{d \left ( 64\,a-64\,b \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-64/d/(64*a+64*b)/(tan(1/2*d*x+1/2*c)-1)-1/d/b/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-2/d/b^2/(1+tan(
1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^2*a+1/d/b/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)-2/d/b^2/(1+tan(
1/2*d*x+1/2*c)^2)^2*a-2/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a^2-3/d/b*arctan(tan(1/2*d*x+1/2*c))+2/d/(a-b)/(a+b)*
a^5/b^3/(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))+64/d/(64*a-64*b)/(tan(1/2*d*x
+1/2*c)+1)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.8168, size = 1138, normalized size = 4.25 \begin{align*} \left [\frac{\sqrt{-a^{2} + b^{2}} a^{5} \cos \left (d x + c\right ) \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 ) + 2 \, a^{3} b^{3} - 2 \, a b^{5} -{\left (2 \, a^{6} - a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} d x \cos \left (d x + c\right ) - 2 \,{\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{2} -{\left (2 \, a^{2} b^{4} - 2 \, b^{6} -{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d \cos \left (d x + c\right )}, -\frac{2 \, \sqrt{a^{2} - b^{2}} a^{5} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - 2 \, a^{3} b^{3} + 2 \, a b^{5} +{\left (2 \, a^{6} - a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} d x \cos \left (d x + c\right ) + 2 \,{\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{2} +{\left (2 \, a^{2} b^{4} - 2 \, b^{6} -{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d \cos \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(-a^2 + b^2)*a^5*cos(d*x + c)*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)) + 2*a^3*b^3 - 2*a*b^5 - (2*a^6 - a^4*b^2 - 4*a^2*b^4 + 3*b^6)*d*x*cos(d*x + c) - 2*(a^5*b - 2*a^3*
b^3 + a*b^5)*cos(d*x + c)^2 - (2*a^2*b^4 - 2*b^6 - (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2)*sin(d*x + c))/(
(a^4*b^3 - 2*a^2*b^5 + b^7)*d*cos(d*x + c)), -1/2*(2*sqrt(a^2 - b^2)*a^5*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^
2 - b^2)*cos(d*x + c)))*cos(d*x + c) - 2*a^3*b^3 + 2*a*b^5 + (2*a^6 - a^4*b^2 - 4*a^2*b^4 + 3*b^6)*d*x*cos(d*x
+ c) + 2*(a^5*b - 2*a^3*b^3 + a*b^5)*cos(d*x + c)^2 + (2*a^2*b^4 - 2*b^6 - (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*
x + c)^2)*sin(d*x + c))/((a^4*b^3 - 2*a^2*b^5 + b^7)*d*cos(d*x + c))]

________________________________________________________________________________________

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(sec(d*x+c)**2*sin(d*x+c)**5/(a+b*sin(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.2004, size = 281, normalized size = 1.05 \begin{align*} \frac{\frac{4 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )} a^{5}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt{a^{2} - b^{2}}} + \frac{4 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}} - \frac{{\left (2 \, a^{2} + 3 \, b^{2}\right )}{\left (d x + c\right )}}{b^{3}} - \frac{2 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*(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^5/((
a^2*b^3 - b^5)*sqrt(a^2 - b^2)) + 4*(b*tan(1/2*d*x + 1/2*c) - a)/((a^2 - b^2)*(tan(1/2*d*x + 1/2*c)^2 - 1)) -
(2*a^2 + 3*b^2)*(d*x + c)/b^3 - 2*(b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2
*c) + 2*a)/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*b^2))/d