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

Optimal. Leaf size=150 \[ \frac{b \left (6 a^2+5 b^2\right ) \sin (c+d x)}{2 d}+\frac{3 a b^2 \sin ^2(c+d x)}{2 d}+\frac{(a+b)^2 (2 a+5 b) \log (1-\sin (c+d x))}{4 d}+\frac{(2 a-5 b) (a-b)^2 \log (\sin (c+d x)+1)}{4 d}+\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac{b^3 \sin ^3(c+d x)}{3 d} \]

[Out]

((a + b)^2*(2*a + 5*b)*Log[1 - Sin[c + d*x]])/(4*d) + ((2*a - 5*b)*(a - b)^2*Log[1 + Sin[c + d*x]])/(4*d) + (b
*(6*a^2 + 5*b^2)*Sin[c + d*x])/(2*d) + (3*a*b^2*Sin[c + d*x]^2)/(2*d) + (b^3*Sin[c + d*x]^3)/(3*d) + (Sec[c +
d*x]^2*(a + b*Sin[c + d*x])^3)/(2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.241513, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2721, 1645, 1629, 633, 31} \[ \frac{b \left (6 a^2+5 b^2\right ) \sin (c+d x)}{2 d}+\frac{3 a b^2 \sin ^2(c+d x)}{2 d}+\frac{(a+b)^2 (2 a+5 b) \log (1-\sin (c+d x))}{4 d}+\frac{(2 a-5 b) (a-b)^2 \log (\sin (c+d x)+1)}{4 d}+\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac{b^3 \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)^2*(2*a + 5*b)*Log[1 - Sin[c + d*x]])/(4*d) + ((2*a - 5*b)*(a - b)^2*Log[1 + Sin[c + d*x]])/(4*d) + (b
*(6*a^2 + 5*b^2)*Sin[c + d*x])/(2*d) + (3*a*b^2*Sin[c + d*x]^2)/(2*d) + (b^3*Sin[c + d*x]^3)/(3*d) + (Sec[c +
d*x]^2*(a + b*Sin[c + d*x])^3)/(2*d)

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]

Rule 1645

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

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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 31

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

Rubi steps

\begin{align*} \int (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^3 (a+x)^3}{\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))^3}{2 d}+\frac{\operatorname{Subst}\left (\int \frac{(a+x)^2 \left (-3 b^4-2 a b^2 x-2 b^2 x^2\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac{\operatorname{Subst}\left (\int \left (6 a^2 b^2+5 b^4+6 a b^2 x+2 b^2 x^2-\frac{9 a^2 b^4+5 b^6+2 a b^2 \left (a^2+6 b^2\right ) x}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac{b \left (6 a^2+5 b^2\right ) \sin (c+d x)}{2 d}+\frac{3 a b^2 \sin ^2(c+d x)}{2 d}+\frac{b^3 \sin ^3(c+d x)}{3 d}+\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^3}{2 d}-\frac{\operatorname{Subst}\left (\int \frac{9 a^2 b^4+5 b^6+2 a b^2 \left (a^2+6 b^2\right ) x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac{b \left (6 a^2+5 b^2\right ) \sin (c+d x)}{2 d}+\frac{3 a b^2 \sin ^2(c+d x)}{2 d}+\frac{b^3 \sin ^3(c+d x)}{3 d}+\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^3}{2 d}-\frac{\left ((2 a-5 b) (a-b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{4 d}-\frac{\left ((a+b)^2 (2 a+5 b)\right ) \operatorname{Subst}\left (\int \frac{1}{b-x} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=\frac{(a+b)^2 (2 a+5 b) \log (1-\sin (c+d x))}{4 d}+\frac{(2 a-5 b) (a-b)^2 \log (1+\sin (c+d x))}{4 d}+\frac{b \left (6 a^2+5 b^2\right ) \sin (c+d x)}{2 d}+\frac{3 a b^2 \sin ^2(c+d x)}{2 d}+\frac{b^3 \sin ^3(c+d x)}{3 d}+\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^3}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.263131, size = 141, normalized size = 0.94 \[ \frac{12 b \left (3 a^2+2 b^2\right ) \sin (c+d x)+18 a b^2 \sin ^2(c+d x)+\frac{3 (a-b)^3}{\sin (c+d x)+1}-\frac{3 (a+b)^3}{\sin (c+d x)-1}+3 (2 a-5 b) (a-b)^2 \log (\sin (c+d x)+1)+3 (a+b)^2 (2 a+5 b) \log (1-\sin (c+d x))+4 b^3 \sin ^3(c+d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + b)^2*(2*a + 5*b)*Log[1 - Sin[c + d*x]] + 3*(2*a - 5*b)*(a - b)^2*Log[1 + Sin[c + d*x]] - (3*(a + b)^3)
/(-1 + Sin[c + d*x]) + 12*b*(3*a^2 + 2*b^2)*Sin[c + d*x] + 18*a*b^2*Sin[c + d*x]^2 + 4*b^3*Sin[c + d*x]^3 + (3
*(a - b)^3)/(1 + Sin[c + d*x]))/(12*d)

________________________________________________________________________________________

Maple [B]  time = 0.056, size = 279, normalized size = 1.9 \begin{align*}{\frac{{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{3\,{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{9\,{a}^{2}b\sin \left ( dx+c \right ) }{2\,d}}-{\frac{9\,{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{2\,d}}+3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}a{b}^{2}}{d}}+6\,{\frac{a{b}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{2\,d}}+{\frac{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}{b}^{3}}{6\,d}}+{\frac{5\,{b}^{3}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{5\,{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*a^3*tan(d*x+c)^2+1/d*a^3*ln(cos(d*x+c))+3/2/d*a^2*b*sin(d*x+c)^5/cos(d*x+c)^2+3/2/d*a^2*b*sin(d*x+c)^3+9
/2*a^2*b*sin(d*x+c)/d-9/2/d*a^2*b*ln(sec(d*x+c)+tan(d*x+c))+3/2/d*a*b^2*sin(d*x+c)^6/cos(d*x+c)^2+3/2/d*a*b^2*
sin(d*x+c)^4+3*a*b^2*sin(d*x+c)^2/d+6/d*a*b^2*ln(cos(d*x+c))+1/2/d*b^3*sin(d*x+c)^7/cos(d*x+c)^2+1/2/d*b^3*sin
(d*x+c)^5+5/6*b^3*sin(d*x+c)^3/d+5/2/d*b^3*sin(d*x+c)-5/2/d*b^3*ln(sec(d*x+c)+tan(d*x+c))

________________________________________________________________________________________

Maxima [A]  time = 1.54646, size = 219, normalized size = 1.46 \begin{align*} \frac{4 \, b^{3} \sin \left (d x + c\right )^{3} + 18 \, a b^{2} \sin \left (d x + c\right )^{2} + 3 \,{\left (2 \, a^{3} - 9 \, a^{2} b + 12 \, a b^{2} - 5 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (2 \, a^{3} + 9 \, a^{2} b + 12 \, a b^{2} + 5 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 12 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right ) - \frac{6 \,{\left (a^{3} + 3 \, a b^{2} +{\left (3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*b^3*sin(d*x + c)^3 + 18*a*b^2*sin(d*x + c)^2 + 3*(2*a^3 - 9*a^2*b + 12*a*b^2 - 5*b^3)*log(sin(d*x + c)
+ 1) + 3*(2*a^3 + 9*a^2*b + 12*a*b^2 + 5*b^3)*log(sin(d*x + c) - 1) + 12*(3*a^2*b + 2*b^3)*sin(d*x + c) - 6*(
a^3 + 3*a*b^2 + (3*a^2*b + b^3)*sin(d*x + c))/(sin(d*x + c)^2 - 1))/d

________________________________________________________________________________________

Fricas [A]  time = 1.72138, size = 470, normalized size = 3.13 \begin{align*} -\frac{18 \, a b^{2} \cos \left (d x + c\right )^{4} - 9 \, a b^{2} \cos \left (d x + c\right )^{2} - 3 \,{\left (2 \, a^{3} - 9 \, a^{2} b + 12 \, a b^{2} - 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, a^{3} + 9 \, a^{2} b + 12 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 6 \, a^{3} - 18 \, a b^{2} + 2 \,{\left (2 \, b^{3} \cos \left (d x + c\right )^{4} - 9 \, a^{2} b - 3 \, b^{3} - 2 \,{\left (9 \, a^{2} b + 7 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(18*a*b^2*cos(d*x + c)^4 - 9*a*b^2*cos(d*x + c)^2 - 3*(2*a^3 - 9*a^2*b + 12*a*b^2 - 5*b^3)*cos(d*x + c)^
2*log(sin(d*x + c) + 1) - 3*(2*a^3 + 9*a^2*b + 12*a*b^2 + 5*b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 6*a^3
- 18*a*b^2 + 2*(2*b^3*cos(d*x + c)^4 - 9*a^2*b - 3*b^3 - 2*(9*a^2*b + 7*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d
*cos(d*x + c)^2)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out