\(\int \sin ^m(c+d x) (a+b \tan (c+d x)) \, dx\) [81]

Optimal result

Integrand size = 19, antiderivative size = 109 \[ \int \sin ^m(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sin ^{1+m}(c+d x)}{d (1+m) \sqrt {\cos ^2(c+d x)}}+\frac {b \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},\sin ^2(c+d x)\right ) \sin ^{2+m}(c+d x)}{d (2+m)} \]

[Out]

b*hypergeom([1, 1+1/2*m],[2+1/2*m],sin(d*x+c)^2)*sin(d*x+c)^(2+m)/d/(2+m)+a*cos(d*x+c)*hypergeom([1/2, 1/2+1/2
*m],[3/2+1/2*m],sin(d*x+c)^2)*sin(d*x+c)^(1+m)/d/(1+m)/(cos(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {4486, 2722, 2644, 371} \[ \int \sin ^m(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \cos (c+d x) \sin ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d (m+1) \sqrt {\cos ^2(c+d x)}}+\frac {b \sin ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},\sin ^2(c+d x)\right )}{d (m+2)} \]

[In]

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

[Out]

(a*Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(1 + m))/(d*(1 + m)*
Sqrt[Cos[c + d*x]^2]) + (b*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(2 + m))/(d
*(2 + m))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 4486

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps \begin{align*} \text {integral}& = \int \left (a \sin ^m(c+d x)+b \sec (c+d x) \sin ^{1+m}(c+d x)\right ) \, dx \\ & = a \int \sin ^m(c+d x) \, dx+b \int \sec (c+d x) \sin ^{1+m}(c+d x) \, dx \\ & = \frac {a \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sin ^{1+m}(c+d x)}{d (1+m) \sqrt {\cos ^2(c+d x)}}+\frac {b \text {Subst}\left (\int \frac {x^{1+m}}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {a \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sin ^{1+m}(c+d x)}{d (1+m) \sqrt {\cos ^2(c+d x)}}+\frac {b \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},\sin ^2(c+d x)\right ) \sin ^{2+m}(c+d x)}{d (2+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00 \[ \int \sin ^m(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) \sin ^{1+m}(c+d x)}{d (1+m)}+\frac {b \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},\sin ^2(c+d x)\right ) \sin ^{2+m}(c+d x)}{d (2+m)} \]

[In]

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

[Out]

(a*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*Sec[c + d*x]*Sin[c + d*x]
^(1 + m))/(d*(1 + m)) + (b*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(2 + m))/(d
*(2 + m))

Maple [F]

\[\int \left (\sin ^{m}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )d x\]

[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)

Fricas [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x)) \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )^{m} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x)) \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right ) \sin ^{m}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x)) \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )^{m} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x)) \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )^{m} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sin ^m(c+d x) (a+b \tan (c+d x)) \, dx=\int {\sin \left (c+d\,x\right )}^m\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right ) \,d x \]

[In]

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

[Out]

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