3.576 \(\int (a+b \sin ^n(c+d x)) \tan ^m(c+d x) \, dx\)
Optimal. Leaf size=124 \[ \frac{a \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\tan ^2(c+d x)\right )}{d (m+1)}+\frac{b \cos ^2(c+d x)^{\frac{m+1}{2}} \tan ^{m+1}(c+d x) \sin ^n(c+d x) \, _2F_1\left (\frac{m+1}{2},\frac{1}{2} (m+n+1);\frac{1}{2} (m+n+3);\sin ^2(c+d x)\right )}{d (m+n+1)} \]
[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))
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Rubi [A] time = 0.161395, antiderivative size = 124, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used =
{3234, 3476, 364, 2602, 2577} \[ \frac{a \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\tan ^2(c+d x)\right )}{d (m+1)}+\frac{b \cos ^2(c+d x)^{\frac{m+1}{2}} \tan ^{m+1}(c+d x) \sin ^n(c+d x) \, _2F_1\left (\frac{m+1}{2},\frac{1}{2} (m+n+1);\frac{1}{2} (m+n+3);\sin ^2(c+d x)\right )}{d (m+n+1)} \]
Antiderivative was successfully verified.
[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 3234
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 3476
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]
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 2602
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 2577
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)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]
Rubi steps
\begin{align*} \int \left (a+b \sin ^n(c+d x)\right ) \tan ^m(c+d x) \, dx &=\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 \operatorname{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 \, _2F_1\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}} \, _2F_1\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] time = 13.5209, size = 1395, normalized size = 11.25 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[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))))
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Maple [F] time = 180., size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ( \sin \left ( dx+c \right ) \right ) ^{n} \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right )^{n} + a\right )} \tan \left (d x + c\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sin \left (d x + c\right )^{n} + a\right )} \tan \left (d x + c\right )^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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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((a+b*sin(d*x+c)**n)*tan(d*x+c)**m,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right )^{n} + a\right )} \tan \left (d x + c\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)