3.161 \(\int (a+b \sin (c+d x))^3 \tan (c+d x) \, dx\)

Optimal. Leaf size=105 \[ -\frac {b \left (3 a^2+b^2\right ) \sin (c+d x)}{d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}-\frac {(a-b)^3 \log (\sin (c+d x)+1)}{2 d}-\frac {(a+b)^3 \log (1-\sin (c+d x))}{2 d}-\frac {b^3 \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2721, 801, 633, 31} \[ -\frac {b \left (3 a^2+b^2\right ) \sin (c+d x)}{d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}-\frac {(a-b)^3 \log (\sin (c+d x)+1)}{2 d}-\frac {(a+b)^3 \log (1-\sin (c+d x))}{2 d}-\frac {b^3 \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

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

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Mathematica [A]  time = 0.20, size = 90, normalized size = 0.86 \[ -\frac {6 b \left (3 a^2+b^2\right ) \sin (c+d x)+9 a b^2 \sin ^2(c+d x)+3 \left ((a-b)^3 \log (\sin (c+d x)+1)+(a+b)^3 \log (1-\sin (c+d x))\right )+2 b^3 \sin ^3(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(3*((a + b)^3*Log[1 - Sin[c + d*x]] + (a - b)^3*Log[1 + Sin[c + d*x]]) + 6*b*(3*a^2 + b^2)*Sin[c + d*x] +
9*a*b^2*Sin[c + d*x]^2 + 2*b^3*Sin[c + d*x]^3)/d

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fricas [A]  time = 0.47, size = 116, normalized size = 1.10 \[ \frac {9 \, a b^{2} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (b^{3} \cos \left (d x + c\right )^{2} - 9 \, a^{2} b - 4 \, b^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(9*a*b^2*cos(d*x + c)^2 - 3*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*log(sin(d*x + c) + 1) - 3*(a^3 + 3*a^2*b + 3*a
*b^2 + b^3)*log(-sin(d*x + c) + 1) + 2*(b^3*cos(d*x + c)^2 - 9*a^2*b - 4*b^3)*sin(d*x + c))/d

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giac [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((a+b*sin(d*x+c))^3*tan(d*x+c),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.14, size = 139, normalized size = 1.32 \[ -\frac {a^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {3 a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {3 a^{2} b \sin \left (d x +c \right )}{d}-\frac {3 a \,b^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}-\frac {3 a \,b^{2} \ln \left (\cos \left (d x +c \right )\right )}{d}-\frac {b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}-\frac {b^{3} \sin \left (d x +c \right )}{d}+\frac {b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.30, size = 113, normalized size = 1.08 \[ -\frac {2 \, b^{3} \sin \left (d x + c\right )^{3} + 9 \, a b^{2} \sin \left (d x + c\right )^{2} + 3 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, {\left (3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*b^3*sin(d*x + c)^3 + 9*a*b^2*sin(d*x + c)^2 + 3*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*log(sin(d*x + c) + 1)
+ 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*log(sin(d*x + c) - 1) + 6*(3*a^2*b + b^3)*sin(d*x + c))/d

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mupad [B]  time = 6.75, size = 226, normalized size = 2.15 \[ \frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (a^3+3\,a\,b^2\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2\,b+2\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (6\,a^2\,b+2\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (12\,a^2\,b+\frac {20\,b^3}{3}\right )+6\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left (a-b\right )}^3}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left (a+b\right )}^3}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(tan(c/2 + (d*x)/2)^2 + 1)*(3*a*b^2 + a^3))/d - (tan(c/2 + (d*x)/2)*(6*a^2*b + 2*b^3) + tan(c/2 + (d*x)/2)
^5*(6*a^2*b + 2*b^3) + tan(c/2 + (d*x)/2)^3*(12*a^2*b + (20*b^3)/3) + 6*a*b^2*tan(c/2 + (d*x)/2)^2 + 6*a*b^2*t
an(c/2 + (d*x)/2)^4)/(d*(3*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) - (log(t
an(c/2 + (d*x)/2) + 1)*(a - b)^3)/d - (log(tan(c/2 + (d*x)/2) - 1)*(a + b)^3)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \tan {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sin(c + d*x))**3*tan(c + d*x), x)

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