3.483 \(\int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=185 \[ \frac {b (A b-a B) \tan ^{m+1}(c+d x) \, _2F_1\left (1,m+1;m+2;-\frac {b \tan (c+d x)}{a}\right )}{a d (m+1) \left (a^2+b^2\right )}+\frac {(a A+b B) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{d (m+1) \left (a^2+b^2\right )}-\frac {(A b-a B) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\tan ^2(c+d x)\right )}{d (m+2) \left (a^2+b^2\right )} \]

[Out]

(A*a+B*b)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-tan(d*x+c)^2)*tan(d*x+c)^(1+m)/(a^2+b^2)/d/(1+m)+b*(A*b-B*a)*h
ypergeom([1, 1+m],[2+m],-b*tan(d*x+c)/a)*tan(d*x+c)^(1+m)/a/(a^2+b^2)/d/(1+m)-(A*b-B*a)*hypergeom([1, 1+1/2*m]
,[2+1/2*m],-tan(d*x+c)^2)*tan(d*x+c)^(2+m)/(a^2+b^2)/d/(2+m)

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Rubi [A]  time = 0.31, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3613, 3538, 3476, 364, 3634, 64} \[ \frac {(a A+b B) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{d (m+1) \left (a^2+b^2\right )}+\frac {b (A b-a B) \tan ^{m+1}(c+d x) \, _2F_1\left (1,m+1;m+2;-\frac {b \tan (c+d x)}{a}\right )}{a d (m+1) \left (a^2+b^2\right )}-\frac {(A b-a B) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\tan ^2(c+d x)\right )}{d (m+2) \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*A + b*B)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 + m))/((a^2 + b^2)*d*
(1 + m)) + (b*(A*b - a*B)*Hypergeometric2F1[1, 1 + m, 2 + m, -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(1 + m))/(a*(
a^2 + b^2)*d*(1 + m)) - ((A*b - a*B)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^
(2 + m))/((a^2 + b^2)*d*(2 + m))

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

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 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 3538

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*T
an[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ
[c^2 + d^2, 0] &&  !IntegerQ[2*m]

Rule 3613

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_))/((a_.) + (b_.)*tan[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*x])^n*Simp[a*A + b*B - (A*b - a*B)
*Tan[e + f*x], x], x], x] + Dist[(b*(A*b - a*B))/(a^2 + b^2), Int[((c + d*Tan[e + f*x])^n*(1 + Tan[e + f*x]^2)
)/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rubi steps

\begin {align*} \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx &=\frac {\int \tan ^m(c+d x) (a A+b B-(A b-a B) \tan (c+d x)) \, dx}{a^2+b^2}+\frac {(b (A b-a B)) \int \frac {\tan ^m(c+d x) \left (1+\tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{a^2+b^2}\\ &=-\frac {(A b-a B) \int \tan ^{1+m}(c+d x) \, dx}{a^2+b^2}+\frac {(a A+b B) \int \tan ^m(c+d x) \, dx}{a^2+b^2}+\frac {(b (A b-a B)) \operatorname {Subst}\left (\int \frac {x^m}{a+b x} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=\frac {b (A b-a B) \, _2F_1\left (1,1+m;2+m;-\frac {b \tan (c+d x)}{a}\right ) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (1+m)}-\frac {(A b-a B) \operatorname {Subst}\left (\int \frac {x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d}+\frac {(a A+b B) \operatorname {Subst}\left (\int \frac {x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=\frac {(a A+b B) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{\left (a^2+b^2\right ) d (1+m)}+\frac {b (A b-a B) \, _2F_1\left (1,1+m;2+m;-\frac {b \tan (c+d x)}{a}\right ) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (1+m)}-\frac {(A b-a B) \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{\left (a^2+b^2\right ) d (2+m)}\\ \end {align*}

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Mathematica [A]  time = 0.95, size = 144, normalized size = 0.78 \[ \frac {\tan ^{m+1}(c+d x) \left ((a A+b B) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )+\frac {(A b-a B) \left (b (m+2) \, _2F_1\left (1,m+1;m+2;-\frac {b \tan (c+d x)}{a}\right )-a (m+1) \tan (c+d x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\tan ^2(c+d x)\right )\right )}{a (m+2)}\right )}{d (m+1) \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Tan[c + d*x]^(1 + m)*((a*A + b*B)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2] + ((A*b - a*B)*
(b*(2 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((b*Tan[c + d*x])/a)] - a*(1 + m)*Hypergeometric2F1[1, (2 + m)/
2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]))/(a*(2 + m))))/((a^2 + b^2)*d*(1 + m))

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*tan(d*x + c) + A)*tan(d*x + c)^m/(b*tan(d*x + c) + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*tan(d*x + c)^m/(b*tan(d*x + c) + a), x)

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maple [F]  time = 3.80, size = 0, normalized size = 0.00 \[ \int \frac {\left (\tan ^{m}\left (d x +c \right )\right ) \left (A +B \tan \left (d x +c \right )\right )}{a +b \tan \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*tan(d*x + c)^m/(b*tan(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(c + d*x)^m*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x)),x)

[Out]

int((tan(c + d*x)^m*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**m*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x)

[Out]

Integral((A + B*tan(c + d*x))*tan(c + d*x)**m/(a + b*tan(c + d*x)), x)

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