3.82 \(\int \frac{\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=765 \[ \text{result too large to display} \]

[Out]

(2^(1 + m)*Hypergeometric2F1[(1 + m)/2, 1 + m, (3 + m)/2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]*(Tan[(c + d*x)
/2]/(1 + Tan[(c + d*x)/2]^2))^m*(1 + Tan[(c + d*x)/2]^2)^m)/(a*d*(1 + m)) + (2^(1 + m)*b*AppellF1[(2 + m)/2, 1
+ m, 1, (4 + m)/2, -Tan[(c + d*x)/2]^2, (a^2*Tan[(c + d*x)/2]^2)/(b - Sqrt[a^2 + b^2])^2]*Tan[(c + d*x)/2]^2*
(Tan[(c + d*x)/2]/(1 + Tan[(c + d*x)/2]^2))^m*(1 + Tan[(c + d*x)/2]^2)^m)/(Sqrt[a^2 + b^2]*(b - Sqrt[a^2 + b^2
])*d*(2 + m)) - (2^(1 + m)*b*AppellF1[(2 + m)/2, 1 + m, 1, (4 + m)/2, -Tan[(c + d*x)/2]^2, (a^2*Tan[(c + d*x)/
2]^2)/(b + Sqrt[a^2 + b^2])^2]*Tan[(c + d*x)/2]^2*(Tan[(c + d*x)/2]/(1 + Tan[(c + d*x)/2]^2))^m*(1 + Tan[(c +
d*x)/2]^2)^m)/(Sqrt[a^2 + b^2]*(b + Sqrt[a^2 + b^2])*d*(2 + m)) + (2^(1 + m)*a*b*AppellF1[(3 + m)/2, 1 + m, 1,
(5 + m)/2, -Tan[(c + d*x)/2]^2, (a^2*Tan[(c + d*x)/2]^2)/(b - Sqrt[a^2 + b^2])^2]*Tan[(c + d*x)/2]^3*(Tan[(c
+ d*x)/2]/(1 + Tan[(c + d*x)/2]^2))^m*(1 + Tan[(c + d*x)/2]^2)^m)/(Sqrt[a^2 + b^2]*(b - Sqrt[a^2 + b^2])^2*d*(
3 + m)) - (2^(1 + m)*a*b*AppellF1[(3 + m)/2, 1 + m, 1, (5 + m)/2, -Tan[(c + d*x)/2]^2, (a^2*Tan[(c + d*x)/2]^2
)/(b + Sqrt[a^2 + b^2])^2]*Tan[(c + d*x)/2]^3*(Tan[(c + d*x)/2]/(1 + Tan[(c + d*x)/2]^2))^m*(1 + Tan[(c + d*x)
/2]^2)^m)/(Sqrt[a^2 + b^2]*(b + Sqrt[a^2 + b^2])^2*d*(3 + m))

________________________________________________________________________________________

Rubi [A]  time = 4.17316, antiderivative size = 765, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {12, 6719, 6728, 364, 959, 510} \[ \frac{a b 2^{m+1} \tan ^3\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}\right )^m \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )^m F_1\left (\frac{m+3}{2};m+1,1;\frac{m+5}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),\frac{a^2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\left (b-\sqrt{a^2+b^2}\right )^2}\right )}{d (m+3) \sqrt{a^2+b^2} \left (b-\sqrt{a^2+b^2}\right )^2}-\frac{a b 2^{m+1} \tan ^3\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}\right )^m \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )^m F_1\left (\frac{m+3}{2};m+1,1;\frac{m+5}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),\frac{a^2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\left (b+\sqrt{a^2+b^2}\right )^2}\right )}{d (m+3) \sqrt{a^2+b^2} \left (\sqrt{a^2+b^2}+b\right )^2}+\frac{b 2^{m+1} \tan ^2\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}\right )^m \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )^m F_1\left (\frac{m+2}{2};m+1,1;\frac{m+4}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),\frac{a^2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\left (b-\sqrt{a^2+b^2}\right )^2}\right )}{d (m+2) \sqrt{a^2+b^2} \left (b-\sqrt{a^2+b^2}\right )}-\frac{b 2^{m+1} \tan ^2\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}\right )^m \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )^m F_1\left (\frac{m+2}{2};m+1,1;\frac{m+4}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),\frac{a^2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\left (b+\sqrt{a^2+b^2}\right )^2}\right )}{d (m+2) \sqrt{a^2+b^2} \left (\sqrt{a^2+b^2}+b\right )}+\frac{2^{m+1} \tan \left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}\right )^m \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )^m \, _2F_1\left (\frac{m+1}{2},m+1;\frac{m+3}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{a d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(1 + m)*Hypergeometric2F1[(1 + m)/2, 1 + m, (3 + m)/2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]*(Tan[(c + d*x)
/2]/(1 + Tan[(c + d*x)/2]^2))^m*(1 + Tan[(c + d*x)/2]^2)^m)/(a*d*(1 + m)) + (2^(1 + m)*b*AppellF1[(2 + m)/2, 1
+ m, 1, (4 + m)/2, -Tan[(c + d*x)/2]^2, (a^2*Tan[(c + d*x)/2]^2)/(b - Sqrt[a^2 + b^2])^2]*Tan[(c + d*x)/2]^2*
(Tan[(c + d*x)/2]/(1 + Tan[(c + d*x)/2]^2))^m*(1 + Tan[(c + d*x)/2]^2)^m)/(Sqrt[a^2 + b^2]*(b - Sqrt[a^2 + b^2
])*d*(2 + m)) - (2^(1 + m)*b*AppellF1[(2 + m)/2, 1 + m, 1, (4 + m)/2, -Tan[(c + d*x)/2]^2, (a^2*Tan[(c + d*x)/
2]^2)/(b + Sqrt[a^2 + b^2])^2]*Tan[(c + d*x)/2]^2*(Tan[(c + d*x)/2]/(1 + Tan[(c + d*x)/2]^2))^m*(1 + Tan[(c +
d*x)/2]^2)^m)/(Sqrt[a^2 + b^2]*(b + Sqrt[a^2 + b^2])*d*(2 + m)) + (2^(1 + m)*a*b*AppellF1[(3 + m)/2, 1 + m, 1,
(5 + m)/2, -Tan[(c + d*x)/2]^2, (a^2*Tan[(c + d*x)/2]^2)/(b - Sqrt[a^2 + b^2])^2]*Tan[(c + d*x)/2]^3*(Tan[(c
+ d*x)/2]/(1 + Tan[(c + d*x)/2]^2))^m*(1 + Tan[(c + d*x)/2]^2)^m)/(Sqrt[a^2 + b^2]*(b - Sqrt[a^2 + b^2])^2*d*(
3 + m)) - (2^(1 + m)*a*b*AppellF1[(3 + m)/2, 1 + m, 1, (5 + m)/2, -Tan[(c + d*x)/2]^2, (a^2*Tan[(c + d*x)/2]^2
)/(b + Sqrt[a^2 + b^2])^2]*Tan[(c + d*x)/2]^3*(Tan[(c + d*x)/2]/(1 + Tan[(c + d*x)/2]^2))^m*(1 + Tan[(c + d*x)
/2]^2)^m)/(Sqrt[a^2 + b^2]*(b + Sqrt[a^2 + b^2])^2*d*(3 + m))

Rule 12

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

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 959

Int[(((g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[(d*(g*x)^n)/x^n, In
t[(x^n*(a + c*x^2)^p)/(d^2 - e^2*x^2), x], x] - Dist[(e*(g*x)^n)/x^n, Int[(x^(n + 1)*(a + c*x^2)^p)/(d^2 - e^2
*x^2), x], x] /; FreeQ[{a, c, d, e, g, n, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&  !IntegersQ[n, 2
*p]

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx &=\frac{2 \operatorname{Subst}\left (\int \frac{2^m \left (1-x^2\right ) \left (\frac{x}{1+x^2}\right )^{1+m}}{x \left (a+2 b x-a x^2\right )} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{d}\\ &=\frac{2^{1+m} \operatorname{Subst}\left (\int \frac{\left (1-x^2\right ) \left (\frac{x}{1+x^2}\right )^{1+m}}{x \left (a+2 b x-a x^2\right )} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{d}\\ &=\frac{\left (2^{1+m} \tan ^{-m}\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m\right ) \operatorname{Subst}\left (\int \frac{x^m \left (1-x^2\right ) \left (1+x^2\right )^{-1-m}}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{d}\\ &=\frac{\left (2^{1+m} \tan ^{-m}\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m\right ) \operatorname{Subst}\left (\int \left (\frac{x^m \left (1+x^2\right )^{-1-m}}{a}-\frac{2 b x^{1+m} \left (1+x^2\right )^{-1-m}}{a \left (a+2 b x-a x^2\right )}\right ) \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{d}\\ &=\frac{\left (2^{1+m} \tan ^{-m}\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m\right ) \operatorname{Subst}\left (\int x^m \left (1+x^2\right )^{-1-m} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a d}-\frac{\left (2^{2+m} b \tan ^{-m}\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+m} \left (1+x^2\right )^{-1-m}}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a d}\\ &=\frac{2^{1+m} \, _2F_1\left (\frac{1+m}{2},1+m;\frac{3+m}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right ) \tan \left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m}{a d (1+m)}-\frac{\left (2^{2+m} b \tan ^{-m}\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m\right ) \operatorname{Subst}\left (\int \left (-\frac{a x^{1+m} \left (1+x^2\right )^{-1-m}}{\sqrt{a^2+b^2} \left (2 b-2 \sqrt{a^2+b^2}-2 a x\right )}+\frac{a x^{1+m} \left (1+x^2\right )^{-1-m}}{\sqrt{a^2+b^2} \left (2 b+2 \sqrt{a^2+b^2}-2 a x\right )}\right ) \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a d}\\ &=\frac{2^{1+m} \, _2F_1\left (\frac{1+m}{2},1+m;\frac{3+m}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right ) \tan \left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m}{a d (1+m)}+\frac{\left (2^{2+m} b \tan ^{-m}\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+m} \left (1+x^2\right )^{-1-m}}{2 b-2 \sqrt{a^2+b^2}-2 a x} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2+b^2} d}-\frac{\left (2^{2+m} b \tan ^{-m}\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+m} \left (1+x^2\right )^{-1-m}}{2 b+2 \sqrt{a^2+b^2}-2 a x} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2+b^2} d}\\ &=\frac{2^{1+m} \, _2F_1\left (\frac{1+m}{2},1+m;\frac{3+m}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right ) \tan \left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m}{a d (1+m)}+\frac{\left (2^{3+m} a b \tan ^{-m}\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m\right ) \operatorname{Subst}\left (\int \frac{x^{2+m} \left (1+x^2\right )^{-1-m}}{\left (2 b-2 \sqrt{a^2+b^2}\right )^2-4 a^2 x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2+b^2} d}-\frac{\left (2^{3+m} a b \tan ^{-m}\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m\right ) \operatorname{Subst}\left (\int \frac{x^{2+m} \left (1+x^2\right )^{-1-m}}{\left (2 b+2 \sqrt{a^2+b^2}\right )^2-4 a^2 x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2+b^2} d}+\frac{\left (2^{3+m} b \left (b-\sqrt{a^2+b^2}\right ) \tan ^{-m}\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+m} \left (1+x^2\right )^{-1-m}}{\left (2 b-2 \sqrt{a^2+b^2}\right )^2-4 a^2 x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2+b^2} d}-\frac{\left (2^{3+m} b \left (b+\sqrt{a^2+b^2}\right ) \tan ^{-m}\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+m} \left (1+x^2\right )^{-1-m}}{\left (2 b+2 \sqrt{a^2+b^2}\right )^2-4 a^2 x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2+b^2} d}\\ &=\frac{2^{1+m} \, _2F_1\left (\frac{1+m}{2},1+m;\frac{3+m}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right ) \tan \left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m}{a d (1+m)}+\frac{2^{1+m} b F_1\left (\frac{2+m}{2};1+m,1;\frac{4+m}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),\frac{a^2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\left (b-\sqrt{a^2+b^2}\right )^2}\right ) \tan ^2\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m}{\sqrt{a^2+b^2} \left (b-\sqrt{a^2+b^2}\right ) d (2+m)}-\frac{2^{1+m} b F_1\left (\frac{2+m}{2};1+m,1;\frac{4+m}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),\frac{a^2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\left (b+\sqrt{a^2+b^2}\right )^2}\right ) \tan ^2\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m}{\sqrt{a^2+b^2} \left (b+\sqrt{a^2+b^2}\right ) d (2+m)}+\frac{2^{1+m} a b F_1\left (\frac{3+m}{2};1+m,1;\frac{5+m}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),\frac{a^2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\left (b-\sqrt{a^2+b^2}\right )^2}\right ) \tan ^3\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m}{\sqrt{a^2+b^2} \left (b-\sqrt{a^2+b^2}\right )^2 d (3+m)}-\frac{2^{1+m} a b F_1\left (\frac{3+m}{2};1+m,1;\frac{5+m}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),\frac{a^2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\left (b+\sqrt{a^2+b^2}\right )^2}\right ) \tan ^3\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^m}{\sqrt{a^2+b^2} \left (b+\sqrt{a^2+b^2}\right )^2 d (3+m)}\\ \end{align*}

Mathematica [F]  time = 12.7916, size = 0, normalized size = 0. \[ \int \frac{\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.164, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{m}}{a+b\tan \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^m/(a+b*tan(d*x+c)),x)

[Out]

int(sin(d*x+c)^m/(a+b*tan(d*x+c)),x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(d*x + c)^m/(b*tan(d*x + c) + a), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sin \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sin(d*x + c)^m/(b*tan(d*x + c) + a), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin ^{m}{\left (c + d x \right )}}{a + b \tan{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**m/(a+b*tan(d*x+c)),x)

[Out]

Integral(sin(c + d*x)**m/(a + b*tan(c + d*x)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(d*x + c)^m/(b*tan(d*x + c) + a), x)