\(\int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx\) [80]

Optimal result

Integrand size = 21, antiderivative size = 179 \[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a^2 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sin ^{1+m}(c+d x)}{d (1+m) \sqrt {\cos ^2(c+d x)}}+\frac {2 a b \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},\sin ^2(c+d x)\right ) \sin ^{2+m}(c+d x)}{d (2+m)}+\frac {b^2 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) \sin ^{3+m}(c+d x)}{d (3+m)} \]

[Out]

2*a*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^2*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)+b^2*hypergeom([3/2, 3/2+1/2*
m],[5/2+1/2*m],sin(d*x+c)^2)*sec(d*x+c)*sin(d*x+c)^(3+m)*(cos(d*x+c)^2)^(1/2)/d/(3+m)

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4486, 2722, 2644, 371, 2657} \[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a^2 \cos (c+d x) \sin ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d (m+1) \sqrt {\cos ^2(c+d x)}}+\frac {2 a b \sin ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},\sin ^2(c+d x)\right )}{d (m+2)}+\frac {b^2 \sqrt {\cos ^2(c+d x)} \sec (c+d x) \sin ^{m+3}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+3}{2},\frac {m+5}{2},\sin ^2(c+d x)\right )}{d (m+3)} \]

[In]

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

[Out]

(a^2*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]) + (2*a*b*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(2 +
m))/(d*(2 + m)) + (b^2*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[3/2, (3 + m)/2, (5 + m)/2, Sin[c + d*x]^2]*Sec[c
+ d*x]*Sin[c + d*x]^(3 + m))/(d*(3 + 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 2657

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b^(2*IntPart[
(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b
, e, f, m, n}, x]

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^2 \sin ^m(c+d x)+2 a b \sec (c+d x) \sin ^{1+m}(c+d x)+b^2 \sec ^2(c+d x) \sin ^{2+m}(c+d x)\right ) \, dx \\ & = a^2 \int \sin ^m(c+d x) \, dx+(2 a b) \int \sec (c+d x) \sin ^{1+m}(c+d x) \, dx+b^2 \int \sec ^2(c+d x) \sin ^{2+m}(c+d x) \, dx \\ & = \frac {a^2 \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^2 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) \sin ^{3+m}(c+d x)}{d (3+m)}+\frac {(2 a b) \text {Subst}\left (\int \frac {x^{1+m}}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {a^2 \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 {2 a 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)}+\frac {b^2 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) \sin ^{3+m}(c+d x)}{d (3+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.27 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.93 \[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\sin ^{1+m}(c+d x) \left (\frac {a^2 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x)}{1+m}+\frac {b \sin (c+d x) \left (2 a (3+m) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},\sin ^2(c+d x)\right )+b (2+m) \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},\sin ^2(c+d x)\right ) \tan (c+d x)\right )}{(2+m) (3+m)}\right )}{d} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (\sin ^{m}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}d x\]

[In]

int(sin(d*x+c)^m*(a+b*tan(d*x+c))^2,x)

[Out]

int(sin(d*x+c)^m*(a+b*tan(d*x+c))^2,x)

Fricas [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(sin(d*x+c)^m*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

integral((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)*sin(d*x + c)^m, x)

Sympy [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \sin ^{m}{\left (c + d x \right )}\, dx \]

[In]

integrate(sin(d*x+c)**m*(a+b*tan(d*x+c))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*sin(c + d*x)**m, x)

Maxima [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(sin(d*x+c)^m*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^2*sin(d*x + c)^m, x)

Giac [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(sin(d*x+c)^m*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c) + a)^2*sin(d*x + c)^m, x)

Mupad [F(-1)]

Timed out. \[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\int {\sin \left (c+d\,x\right )}^m\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^2 \,d x \]

[In]

int(sin(c + d*x)^m*(a + b*tan(c + d*x))^2,x)

[Out]

int(sin(c + d*x)^m*(a + b*tan(c + d*x))^2, x)