3.150 \(\int (a+b \sin (c+d x))^2 \tan ^3(c+d x) \, dx\)

Optimal. Leaf size=111 \[ \frac{2 a b \sin (c+d x)}{d}+\frac{(a+b) (a+2 b) \log (1-\sin (c+d x))}{2 d}+\frac{(a-2 b) (a-b) \log (\sin (c+d x)+1)}{2 d}+\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^2}{2 d}+\frac{b^2 \sin ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.172935, antiderivative size = 111, 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} \[ \frac{2 a b \sin (c+d x)}{d}+\frac{(a+b) (a+2 b) \log (1-\sin (c+d x))}{2 d}+\frac{(a-2 b) (a-b) \log (\sin (c+d x)+1)}{2 d}+\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^2}{2 d}+\frac{b^2 \sin ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^2*Tan[c + d*x]^3,x]

[Out]

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

Mathematica [A]  time = 0.431531, size = 108, normalized size = 0.97 \[ \frac{\frac{(a-b)^2}{\sin (c+d x)+1}+8 a b \sin (c+d x)-\frac{(a+b)^2}{\sin (c+d x)-1}+2 (a-2 b) (a-b) \log (\sin (c+d x)+1)+2 (a+b) (a+2 b) \log (1-\sin (c+d x))+2 b^2 \sin ^2(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^2*Tan[c + d*x]^3,x]

[Out]

(2*(a + b)*(a + 2*b)*Log[1 - Sin[c + d*x]] + 2*(a - 2*b)*(a - b)*Log[1 + Sin[c + d*x]] - (a + b)^2/(-1 + Sin[c
+ d*x]) + 8*a*b*Sin[c + d*x] + 2*b^2*Sin[c + d*x]^2 + (a - b)^2/(1 + Sin[c + d*x]))/(4*d)

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Maple [A]  time = 0.049, size = 172, normalized size = 1.6 \begin{align*}{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{a}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d}}+3\,{\frac{ab\sin \left ( dx+c \right ) }{d}}-3\,{\frac{ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{2\,d}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{2}}{d}}+2\,{\frac{{b}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x+c))^2*tan(d*x+c)^3,x)

[Out]

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

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Maxima [A]  time = 1.67617, size = 142, normalized size = 1.28 \begin{align*} \frac{b^{2} \sin \left (d x + c\right )^{2} + 4 \, a b \sin \left (d x + c\right ) +{\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \, a b \sin \left (d x + c\right ) + a^{2} + b^{2}}{\sin \left (d x + c\right )^{2} - 1}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^2*tan(d*x+c)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.54829, size = 348, normalized size = 3.14 \begin{align*} -\frac{2 \, b^{2} \cos \left (d x + c\right )^{4} - b^{2} \cos \left (d x + c\right )^{2} - 2 \,{\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a^{2} - 2 \, b^{2} - 4 \,{\left (2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \sin \left (d x + c\right )}{4 \, 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))^2*tan(d*x+c)^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sin{\left (c + d x \right )}\right )^{2} \tan ^{3}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))**2*tan(d*x+c)**3,x)

[Out]

Integral((a + b*sin(c + d*x))**2*tan(c + d*x)**3, x)

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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))^2*tan(d*x+c)^3,x, algorithm="giac")

[Out]

Timed out