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

Optimal. Leaf size=124 \[ \frac {a \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {b \cos ^2(c+d x)^{\frac {1+m}{2}} \, _2F_1\left (\frac {1+m}{2},\frac {1}{2} (1+m+n);\frac {1}{2} (3+m+n);\sin ^2(c+d x)\right ) \sin ^n(c+d x) \tan ^{1+m}(c+d x)}{d (1+m+n)} \]

[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)

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Rubi [A]
time = 0.10, antiderivative size = 124, normalized size of antiderivative = 1.00, 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 = {3313, 3557, 371, 2682, 2657} \begin {gather*} \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)} \end {gather*}

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 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*} \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 \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 \, _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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 12.46, size = 1065, normalized size = 8.59 \begin {gather*} \frac {2 \left (a+b \sin ^n(c+d x)\right ) \left (a (1+m+n) F_1\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) F_1\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)}{d \left (\sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (a (1+m+n) F_1\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) F_1\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 )-16 m \cos \left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x) \sec (c+d x) \sin ^5\left (\frac {1}{2} (c+d x)\right ) \left (a (1+m+n) F_1\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) F_1\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 )+2 m \csc (c+d x) \sec (c+d x) \left (a (1+m+n) F_1\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) F_1\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 )+2 (1+m) \tan \left (\frac {1}{2} (c+d x)\right ) \left (b n F_1\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)+\frac {a (1+m+n) \left (-F_1\left (\frac {3+m}{2};m,2;\frac {5+m}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+m F_1\left (\frac {3+m}{2};1+m,1;\frac {5+m}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{3+m}+b n F_1\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 )+\frac {b (1+m+n) \left (-\left ((1+n) F_1\left (\frac {1}{2} (3+m+n);m,2+n;\frac {1}{2} (5+m+n);\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )+m F_1\left (\frac {1}{2} (3+m+n);1+m,1+n;\frac {1}{2} (5+m+n);\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{1+n} \sin ^n(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{3+m+n}\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(a + b*Sin[c + d*x]^n)*(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)/(d*(Sec[(c + d*x)/2]^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)*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)
- 16*m*Cos[(c + d*x)/2]*Csc[c + d*x]^3*Sec[c + d*x]*Sin[(c + d*x)/2]^5*(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) + 2*m*Csc[c + d*x]*Sec[c
+ d*x]*(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] + 2*(1 + m)*Tan[(c + d*x)/2]*(b*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)
+ (a*(1 + m + n)*(-AppellF1[(3 + m)/2, m, 2, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + m*AppellF1[
(3 + m)/2, 1 + m, 1, (5 + 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*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] + (b*(1 + m + n)*(-((1 + n)*AppellF1[(3 + m + n)/2, m, 2 + n
, (5 + m + n)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]) + m*AppellF1[(3 + m + n)/2, 1 + m, 1 + n, (5 + m +
n)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(Sec[(c + d*x)/2]^2)^(1 + n)*Sin[c + d*x]^n*Tan[(c + d*x)/2])/
(3 + m + n))))

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Maple [F]
time = 0.73, size = 0, normalized size = 0.00 \[\int \left (a +b \left (\sin ^{n}\left (d x +c \right )\right )\right ) \left (\tan ^{m}\left (d x +c \right )\right )\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.43, size = 23, normalized size = 0.19 \begin {gather*} {\rm integral}\left ({\left (b \sin \left (d x + c\right )^{n} + a\right )} \tan \left (d x + c\right )^{m}, x\right ) \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin ^{n}{\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}\, dx \end {gather*}

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (a+b\,{\sin \left (c+d\,x\right )}^n\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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