3.1496 \(\int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=93 \[ -\frac{\left (a^2-3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{\sec ^2(c+d x) \left (\left (a^2+3 b^2\right ) \sin (c+d x)+4 a b\right )}{8 d}+\frac{\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^2}{4 d} \]

[Out]

-((a^2 - 3*b^2)*ArcTanh[Sin[c + d*x]])/(8*d) - (Sec[c + d*x]^2*(4*a*b + (a^2 + 3*b^2)*Sin[c + d*x]))/(8*d) + (
Sec[c + d*x]^3*(a + b*Sin[c + d*x])^2*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.194086, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2837, 12, 1645, 778, 206} \[ -\frac{\left (a^2-3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{\sec ^2(c+d x) \left (\left (a^2+3 b^2\right ) \sin (c+d x)+4 a b\right )}{8 d}+\frac{\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^2}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2 - 3*b^2)*ArcTanh[Sin[c + d*x]])/(8*d) - (Sec[c + d*x]^2*(4*a*b + (a^2 + 3*b^2)*Sin[c + d*x]))/(8*d) + (
Sec[c + d*x]^3*(a + b*Sin[c + d*x])^2*Tan[c + d*x])/(4*d)

Rule 2837

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 778

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*(e*f + d*g) -
(c*d*f - a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{x^2 (a+x)^2}{b^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^3 \operatorname{Subst}\left (\int \frac{x^2 (a+x)^2}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac{b \operatorname{Subst}\left (\int \frac{(a+x) \left (-a b^2-3 b^2 x\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=-\frac{\sec ^2(c+d x) \left (4 a b+\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}-\frac{\left (b \left (a^2-3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=-\frac{\left (a^2-3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{\sec ^2(c+d x) \left (4 a b+\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.746953, size = 85, normalized size = 0.91 \[ -\frac{\left (a^2-3 b^2\right ) \tanh ^{-1}(\sin (c+d x))+\frac{1}{4} \sec ^4(c+d x) \left (2 \sin (c+d x) \left (\left (a^2+5 b^2\right ) \cos (2 (c+d x))-3 a^2+b^2\right )+16 a b \cos (2 (c+d x))\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2 - 3*b^2)*ArcTanh[Sin[c + d*x]] + (Sec[c + d*x]^4*(16*a*b*Cos[2*(c + d*x)] + 2*(-3*a^2 + b^2 + (a^2 + 5*
b^2)*Cos[2*(c + d*x)])*Sin[c + d*x]))/4)/(8*d)

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Maple [B]  time = 0.062, size = 209, normalized size = 2.3 \begin{align*}{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{8\,d}}-{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{3\,{b}^{2}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^2*(a+b*sin(d*x+c))^2,x)

[Out]

1/4/d*a^2*sin(d*x+c)^3/cos(d*x+c)^4+1/8/d*a^2*sin(d*x+c)^3/cos(d*x+c)^2+1/8*a^2*sin(d*x+c)/d-1/8/d*a^2*ln(sec(
d*x+c)+tan(d*x+c))+1/2/d*a*b*sin(d*x+c)^4/cos(d*x+c)^4+1/4/d*b^2*sin(d*x+c)^5/cos(d*x+c)^4-1/8/d*b^2*sin(d*x+c
)^5/cos(d*x+c)^2-1/8*b^2*sin(d*x+c)^3/d-3/8*b^2*sin(d*x+c)/d+3/8/d*b^2*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.976745, size = 162, normalized size = 1.74 \begin{align*} -\frac{{\left (a^{2} - 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (a^{2} - 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (8 \, a b \sin \left (d x + c\right )^{2} +{\left (a^{2} + 5 \, b^{2}\right )} \sin \left (d x + c\right )^{3} - 4 \, a b +{\left (a^{2} - 3 \, b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/16*((a^2 - 3*b^2)*log(sin(d*x + c) + 1) - (a^2 - 3*b^2)*log(sin(d*x + c) - 1) - 2*(8*a*b*sin(d*x + c)^2 + (
a^2 + 5*b^2)*sin(d*x + c)^3 - 4*a*b + (a^2 - 3*b^2)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

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Fricas [A]  time = 2.02974, size = 309, normalized size = 3.32 \begin{align*} -\frac{{\left (a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 16 \, a b \cos \left (d x + c\right )^{2} - 8 \, a b + 2 \,{\left ({\left (a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/16*((a^2 - 3*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (a^2 - 3*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1
) + 16*a*b*cos(d*x + c)^2 - 8*a*b + 2*((a^2 + 5*b^2)*cos(d*x + c)^2 - 2*a^2 - 2*b^2)*sin(d*x + c))/(d*cos(d*x
+ c)^4)

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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(sec(d*x+c)**5*sin(d*x+c)**2*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.27209, size = 167, normalized size = 1.8 \begin{align*} -\frac{{\left (a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (a^{2} \sin \left (d x + c\right )^{3} + 5 \, b^{2} \sin \left (d x + c\right )^{3} + 8 \, a b \sin \left (d x + c\right )^{2} + a^{2} \sin \left (d x + c\right ) - 3 \, b^{2} \sin \left (d x + c\right ) - 4 \, a b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*((a^2 - 3*b^2)*log(abs(sin(d*x + c) + 1)) - (a^2 - 3*b^2)*log(abs(sin(d*x + c) - 1)) - 2*(a^2*sin(d*x +
c)^3 + 5*b^2*sin(d*x + c)^3 + 8*a*b*sin(d*x + c)^2 + a^2*sin(d*x + c) - 3*b^2*sin(d*x + c) - 4*a*b)/(sin(d*x +
c)^2 - 1)^2)/d