3.236 \(\int \sin (a+b x) \tan (c+d x) \, dx\)

Optimal. Leaf size=143 \[ -\frac {i e^{-i (a+b x)} \, _2F_1\left (1,-\frac {b}{2 d};1-\frac {b}{2 d};-e^{2 i (c+d x)}\right )}{b}-\frac {i e^{i (a+b x)} \, _2F_1\left (1,\frac {b}{2 d};\frac {b}{2 d}+1;-e^{2 i (c+d x)}\right )}{b}+\frac {i e^{-i (a+b x)}}{2 b}+\frac {i e^{i (a+b x)}}{2 b} \]

[Out]

1/2*I/b/exp(I*(b*x+a))+1/2*I*exp(I*(b*x+a))/b-I*hypergeom([1, -1/2*b/d],[1-1/2*b/d],-exp(2*I*(d*x+c)))/b/exp(I
*(b*x+a))-I*exp(I*(b*x+a))*hypergeom([1, 1/2*b/d],[1+1/2*b/d],-exp(2*I*(d*x+c)))/b

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Rubi [A]  time = 0.11, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {4557, 2194, 2251} \[ -\frac {i e^{-i (a+b x)} \, _2F_1\left (1,-\frac {b}{2 d};1-\frac {b}{2 d};-e^{2 i (c+d x)}\right )}{b}-\frac {i e^{i (a+b x)} \, _2F_1\left (1,\frac {b}{2 d};\frac {b}{2 d}+1;-e^{2 i (c+d x)}\right )}{b}+\frac {i e^{-i (a+b x)}}{2 b}+\frac {i e^{i (a+b x)}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]*Tan[c + d*x],x]

[Out]

(I/2)/(b*E^(I*(a + b*x))) + ((I/2)*E^(I*(a + b*x)))/b - (I*Hypergeometric2F1[1, -b/(2*d), 1 - b/(2*d), -E^((2*
I)*(c + d*x))])/(b*E^(I*(a + b*x))) - (I*E^(I*(a + b*x))*Hypergeometric2F1[1, b/(2*d), 1 + b/(2*d), -E^((2*I)*
(c + d*x))])/b

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2251

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[(a^p*G^(h*(f + g*x))*Hypergeometric2F1[-p, (g*h*Log[G])/(d*e*Log[F]), (g*h*Log[G])/(d*e*Log[F]) + 1, Simplify
[-((b*F^(e*(c + d*x)))/a)]])/(g*h*Log[G]), x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] ||
GtQ[a, 0])

Rule 4557

Int[Sin[(a_.) + (b_.)*(x_)]*Tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[1/(E^(I*(a + b*x))*2) - E^(I*(a + b*x))/
2 - 1/(E^(I*(a + b*x))*(1 + E^(2*I*(c + d*x)))) + E^(I*(a + b*x))/(1 + E^(2*I*(c + d*x))), x] /; FreeQ[{a, b,
c, d}, x] && NeQ[b^2 - d^2, 0]

Rubi steps

\begin {align*} \int \sin (a+b x) \tan (c+d x) \, dx &=\int \left (\frac {1}{2} e^{-i (a+b x)}-\frac {1}{2} e^{i (a+b x)}-\frac {e^{-i (a+b x)}}{1+e^{2 i (c+d x)}}+\frac {e^{i (a+b x)}}{1+e^{2 i (c+d x)}}\right ) \, dx\\ &=\frac {1}{2} \int e^{-i (a+b x)} \, dx-\frac {1}{2} \int e^{i (a+b x)} \, dx-\int \frac {e^{-i (a+b x)}}{1+e^{2 i (c+d x)}} \, dx+\int \frac {e^{i (a+b x)}}{1+e^{2 i (c+d x)}} \, dx\\ &=\frac {i e^{-i (a+b x)}}{2 b}+\frac {i e^{i (a+b x)}}{2 b}-\frac {i e^{-i (a+b x)} \, _2F_1\left (1,-\frac {b}{2 d};1-\frac {b}{2 d};-e^{2 i (c+d x)}\right )}{b}-\frac {i e^{i (a+b x)} \, _2F_1\left (1,\frac {b}{2 d};1+\frac {b}{2 d};-e^{2 i (c+d x)}\right )}{b}\\ \end {align*}

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Mathematica [A]  time = 1.81, size = 116, normalized size = 0.81 \[ -\frac {i e^{-i (a+b x)} \left (2 e^{2 i (a+b x)} \, _2F_1\left (1,\frac {b}{2 d};\frac {b}{2 d}+1;-e^{2 i (c+d x)}\right )-e^{2 i (a+b x)}+2 \, _2F_1\left (1,-\frac {b}{2 d};1-\frac {b}{2 d};-e^{2 i (c+d x)}\right )-1\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]*Tan[c + d*x],x]

[Out]

((-1/2*I)*(-1 - E^((2*I)*(a + b*x)) + 2*Hypergeometric2F1[1, -1/2*b/d, 1 - b/(2*d), -E^((2*I)*(c + d*x))] + 2*
E^((2*I)*(a + b*x))*Hypergeometric2F1[1, b/(2*d), 1 + b/(2*d), -E^((2*I)*(c + d*x))]))/(b*E^(I*(a + b*x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.46, size = 0, normalized size = 0.00 \[ \int \sin \left (b x +a \right ) \tan \left (d x +c \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)*tan(c + d*x),x)

[Out]

int(sin(a + b*x)*tan(c + d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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