3.1346 \(\int \frac {\tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=126 \[ \frac {\sec ^2(c+d x) (a-b \sin (c+d x))}{2 d \left (a^2-b^2\right )}-\frac {a^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}+\frac {(2 a+b) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac {(2 a-b) \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]

[Out]

1/4*(2*a+b)*ln(1-sin(d*x+c))/(a+b)^2/d+1/4*(2*a-b)*ln(1+sin(d*x+c))/(a-b)^2/d-a^3*ln(a+b*sin(d*x+c))/(a^2-b^2)
^2/d+1/2*sec(d*x+c)^2*(a-b*sin(d*x+c))/(a^2-b^2)/d

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Rubi [A]  time = 0.20, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2721, 1647, 801} \[ -\frac {a^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}+\frac {\sec ^2(c+d x) (a-b \sin (c+d x))}{2 d \left (a^2-b^2\right )}+\frac {(2 a+b) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac {(2 a-b) \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3/(a + b*Sin[c + d*x]),x]

[Out]

((2*a + b)*Log[1 - Sin[c + d*x]])/(4*(a + b)^2*d) + ((2*a - b)*Log[1 + Sin[c + d*x]])/(4*(a - b)^2*d) - (a^3*L
og[a + b*Sin[c + d*x]])/((a^2 - b^2)^2*d) + (Sec[c + d*x]^2*(a - b*Sin[c + d*x]))/(2*(a^2 - b^2)*d)

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.55, size = 117, normalized size = 0.93 \[ \frac {-\frac {4 a^3 \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}-\frac {1}{(a+b) (\sin (c+d x)-1)}+\frac {1}{(a-b) (\sin (c+d x)+1)}+\frac {(2 a+b) \log (1-\sin (c+d x))}{(a+b)^2}+\frac {(2 a-b) \log (\sin (c+d x)+1)}{(a-b)^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3/(a + b*Sin[c + d*x]),x]

[Out]

(((2*a + b)*Log[1 - Sin[c + d*x]])/(a + b)^2 + ((2*a - b)*Log[1 + Sin[c + d*x]])/(a - b)^2 - (4*a^3*Log[a + b*
Sin[c + d*x]])/((a - b)^2*(a + b)^2) - 1/((a + b)*(-1 + Sin[c + d*x])) + 1/((a - b)*(1 + Sin[c + d*x])))/(4*d)

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fricas [A]  time = 0.89, size = 157, normalized size = 1.25 \[ -\frac {4 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{3} - 3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a^{3} + 2 \, a b^{2} + 2 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(4*a^3*cos(d*x + c)^2*log(b*sin(d*x + c) + a) - (2*a^3 + 3*a^2*b - b^3)*cos(d*x + c)^2*log(sin(d*x + c) +
1) - (2*a^3 - 3*a^2*b + b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*a^3 + 2*a*b^2 + 2*(a^2*b - b^3)*sin(d*
x + c))/((a^4 - 2*a^2*b^2 + b^4)*d*cos(d*x + c)^2)

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giac [A]  time = 0.24, size = 177, normalized size = 1.40 \[ -\frac {\frac {4 \, a^{3} b \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac {{\left (2 \, a - b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (2 \, a + b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (a^{3} \sin \left (d x + c\right )^{2} - a^{2} b \sin \left (d x + c\right ) + b^{3} \sin \left (d x + c\right ) - a b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/4*(4*a^3*b*log(abs(b*sin(d*x + c) + a))/(a^4*b - 2*a^2*b^3 + b^5) - (2*a - b)*log(abs(sin(d*x + c) + 1))/(a
^2 - 2*a*b + b^2) - (2*a + b)*log(abs(sin(d*x + c) - 1))/(a^2 + 2*a*b + b^2) + 2*(a^3*sin(d*x + c)^2 - a^2*b*s
in(d*x + c) + b^3*sin(d*x + c) - a*b^2)/((a^4 - 2*a^2*b^2 + b^4)*(sin(d*x + c)^2 - 1)))/d

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maple [A]  time = 0.41, size = 164, normalized size = 1.30 \[ -\frac {1}{d \left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right ) a}{2 d \left (a +b \right )^{2}}+\frac {\ln \left (\sin \left (d x +c \right )-1\right ) b}{4 d \left (a +b \right )^{2}}-\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {1}{d \left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {a \ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{2} d}-\frac {b \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*sin(d*x+c)^3/(a+b*sin(d*x+c)),x)

[Out]

-1/d/(4*a+4*b)/(sin(d*x+c)-1)+1/2/d/(a+b)^2*ln(sin(d*x+c)-1)*a+1/4/d/(a+b)^2*ln(sin(d*x+c)-1)*b-1/d*a^3/(a+b)^
2/(a-b)^2*ln(a+b*sin(d*x+c))+1/d/(4*a-4*b)/(1+sin(d*x+c))+1/2*a*ln(1+sin(d*x+c))/(a-b)^2/d-1/4*b*ln(1+sin(d*x+
c))/(a-b)^2/d

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maxima [A]  time = 0.32, size = 142, normalized size = 1.13 \[ -\frac {\frac {4 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (2 \, a - b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (2 \, a + b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (b \sin \left (d x + c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/4*(4*a^3*log(b*sin(d*x + c) + a)/(a^4 - 2*a^2*b^2 + b^4) - (2*a - b)*log(sin(d*x + c) + 1)/(a^2 - 2*a*b + b
^2) - (2*a + b)*log(sin(d*x + c) - 1)/(a^2 + 2*a*b + b^2) - 2*(b*sin(d*x + c) - a)/((a^2 - b^2)*sin(d*x + c)^2
- a^2 + b^2))/d

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mupad [B]  time = 12.29, size = 217, normalized size = 1.72 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (2\,a-b\right )}{2\,d\,{\left (a-b\right )}^2}-\frac {\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2-b^2}-\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^2-b^2}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a^2-b^2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a^3\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (2\,a+b\right )}{2\,d\,{\left (a+b\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^3/(cos(c + d*x)^3*(a + b*sin(c + d*x))),x)

[Out]

(log(tan(c/2 + (d*x)/2) + 1)*(2*a - b))/(2*d*(a - b)^2) - ((b*tan(c/2 + (d*x)/2))/(a^2 - b^2) - (2*a*tan(c/2 +
(d*x)/2)^2)/(a^2 - b^2) + (b*tan(c/2 + (d*x)/2)^3)/(a^2 - b^2))/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/
2)^2 + 1)) - (a^3*log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2))/(d*(a^4 + b^4 - 2*a^2*b^2)) + (log
(tan(c/2 + (d*x)/2) - 1)*(2*a + b))/(2*d*(a + b)^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*sin(d*x+c)**3/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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