\(\int (a+b \sin ^n(c+d x)) \tan ^m(c+d x) \, dx\) [576]

Optimal result

Integrand size = 21, antiderivative size = 124 \[ \int \left (a+b \sin ^n(c+d x)\right ) \tan ^m(c+d x) \, dx=\frac {a \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {b \cos ^2(c+d x)^{\frac {1+m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {1}{2} (1+m+n),\frac {1}{2} (3+m+n),\sin ^2(c+d x)\right ) \sin ^n(c+d x) \tan ^{1+m}(c+d x)}{d (1+m+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 124, 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 = {3313, 3557, 371, 2682, 2657} \[ \int \left (a+b \sin ^n(c+d x)\right ) \tan ^m(c+d x) \, dx=\frac {a \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1)}+\frac {b \cos ^2(c+d x)^{\frac {m+1}{2}} \tan ^{m+1}(c+d x) \sin ^n(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},\frac {1}{2} (m+n+1),\frac {1}{2} (m+n+3),\sin ^2(c+d x)\right )}{d (m+n+1)} \]

[In]

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

[Out]

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

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 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 2682

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[a*Cos[e + f*
x]^(n + 1)*((b*Tan[e + f*x])^(n + 1)/(b*(a*Sin[e + f*x])^(n + 1))), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^
n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[n]

Rule 3313

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol]
:> Int[ExpandTrig[(d*tan[e + f*x])^m*(a + b*(c*sin[e + f*x])^n)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n},
x] && IGtQ[p, 0]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 14.19 (sec) , antiderivative size = 1395, normalized size of antiderivative = 11.25 \[ \int \left (a+b \sin ^n(c+d x)\right ) \tan ^m(c+d x) \, dx=\frac {2 \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^m \left (a (1+m+n) \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+b (1+m) \operatorname {AppellF1}\left (\frac {1}{2} (1+m+n),m,1+n,\frac {1}{2} (3+m+n),\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^n \sin ^n(c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \tan ^m(c+d x) \left (a \tan ^m(c+d x)+b \sin ^n(c+d x) \tan ^m(c+d x)\right )}{d (1+m) (1+m+n) \left (\frac {2 m \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^m \sec ^2(c+d x) \left (a (1+m+n) \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+b (1+m) \operatorname {AppellF1}\left (\frac {1}{2} (1+m+n),m,1+n,\frac {1}{2} (3+m+n),\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^n \sin ^n(c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \tan ^{-1+m}(c+d x)}{(1+m) (1+m+n)}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^m \left (a (1+m+n) \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+b (1+m) \operatorname {AppellF1}\left (\frac {1}{2} (1+m+n),m,1+n,\frac {1}{2} (3+m+n),\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^n \sin ^n(c+d x)\right ) \tan ^m(c+d x)}{(1+m) (1+m+n)}+\frac {2 m \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{-1+m} \left (a (1+m+n) \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+b (1+m) \operatorname {AppellF1}\left (\frac {1}{2} (1+m+n),m,1+n,\frac {1}{2} (3+m+n),\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^n \sin ^n(c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \left (-\sec ^2\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)+\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right ) \tan ^m(c+d x)}{(1+m) (1+m+n)}+\frac {2 \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^m \tan \left (\frac {1}{2} (c+d x)\right ) \left (b (1+m) n \operatorname {AppellF1}\left (\frac {1}{2} (1+m+n),m,1+n,\frac {1}{2} (3+m+n),\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )^n \sin ^{-1+n}(c+d x)+b (1+m) n \operatorname {AppellF1}\left (\frac {1}{2} (1+m+n),m,1+n,\frac {1}{2} (3+m+n),\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^n \sin ^n(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )+a (1+m+n) \left (-\frac {(1+m) \operatorname {AppellF1}\left (1+\frac {1+m}{2},m,2,1+\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{3+m}+\frac {m (1+m) \operatorname {AppellF1}\left (1+\frac {1+m}{2},1+m,1,1+\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{3+m}\right )+b (1+m) \sec ^2\left (\frac {1}{2} (c+d x)\right )^n \sin ^n(c+d x) \left (-\frac {(1+n) (1+m+n) \operatorname {AppellF1}\left (1+\frac {1}{2} (1+m+n),m,2+n,1+\frac {1}{2} (3+m+n),\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{3+m+n}+\frac {m (1+m+n) \operatorname {AppellF1}\left (1+\frac {1}{2} (1+m+n),1+m,1+n,1+\frac {1}{2} (3+m+n),\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{3+m+n}\right )\right ) \tan ^m(c+d x)}{(1+m) (1+m+n)}\right )} \]

[In]

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

[Out]

(2*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^m*(a*(1 + m + n)*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, Tan[(c + d*x)/2]^2,
-Tan[(c + d*x)/2]^2] + b*(1 + m)*AppellF1[(1 + m + n)/2, m, 1 + n, (3 + m + n)/2, Tan[(c + d*x)/2]^2, -Tan[(c
+ d*x)/2]^2]*(Sec[(c + d*x)/2]^2)^n*Sin[c + d*x]^n)*Tan[(c + d*x)/2]*Tan[c + d*x]^m*(a*Tan[c + d*x]^m + b*Sin
[c + d*x]^n*Tan[c + d*x]^m))/(d*(1 + m)*(1 + m + n)*((2*m*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^m*Sec[c + d*x]^2*(
a*(1 + m + n)*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + b*(1 + m)*Appell
F1[(1 + m + n)/2, m, 1 + n, (3 + m + n)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(Sec[(c + d*x)/2]^2)^n*Sin
[c + d*x]^n)*Tan[(c + d*x)/2]*Tan[c + d*x]^(-1 + m))/((1 + m)*(1 + m + n)) + (Sec[(c + d*x)/2]^2*(Cos[c + d*x]
*Sec[(c + d*x)/2]^2)^m*(a*(1 + m + n)*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/
2]^2] + b*(1 + m)*AppellF1[(1 + m + n)/2, m, 1 + n, (3 + m + n)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(S
ec[(c + d*x)/2]^2)^n*Sin[c + d*x]^n)*Tan[c + d*x]^m)/((1 + m)*(1 + m + n)) + (2*m*(Cos[c + d*x]*Sec[(c + d*x)/
2]^2)^(-1 + m)*(a*(1 + m + n)*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] +
b*(1 + m)*AppellF1[(1 + m + n)/2, m, 1 + n, (3 + m + n)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(Sec[(c +
d*x)/2]^2)^n*Sin[c + d*x]^n)*Tan[(c + d*x)/2]*(-(Sec[(c + d*x)/2]^2*Sin[c + d*x]) + Cos[c + d*x]*Sec[(c + d*x)
/2]^2*Tan[(c + d*x)/2])*Tan[c + d*x]^m)/((1 + m)*(1 + m + n)) + (2*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^m*Tan[(c
+ d*x)/2]*(b*(1 + m)*n*AppellF1[(1 + m + n)/2, m, 1 + n, (3 + m + n)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^
2]*Cos[c + d*x]*(Sec[(c + d*x)/2]^2)^n*Sin[c + d*x]^(-1 + n) + b*(1 + m)*n*AppellF1[(1 + m + n)/2, m, 1 + n, (
3 + m + n)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(Sec[(c + d*x)/2]^2)^n*Sin[c + d*x]^n*Tan[(c + d*x)/2]
+ a*(1 + m + n)*(-(((1 + m)*AppellF1[1 + (1 + m)/2, m, 2, 1 + (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]
^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(3 + m)) + (m*(1 + m)*AppellF1[1 + (1 + m)/2, 1 + m, 1, 1 + (3 + m)/2
, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(3 + m)) + b*(1 + m)*(Sec[(c +
d*x)/2]^2)^n*Sin[c + d*x]^n*(-(((1 + n)*(1 + m + n)*AppellF1[1 + (1 + m + n)/2, m, 2 + n, 1 + (3 + m + n)/2,
Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(3 + m + n)) + (m*(1 + m + n)*Ap
pellF1[1 + (1 + m + n)/2, 1 + m, 1 + n, 1 + (3 + m + n)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d
*x)/2]^2*Tan[(c + d*x)/2])/(3 + m + n)))*Tan[c + d*x]^m)/((1 + m)*(1 + m + n))))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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