3.1495 \(\int \sec ^2(c+d x) (a+b \sin (c+d x))^2 \tan ^3(c+d x) \, dx\)

Optimal. Leaf size=116 \[ -\frac{b (3 a+4 b) \log (1-\sin (c+d x))}{8 d}+\frac{b (3 a-4 b) \log (\sin (c+d x)+1)}{8 d}+\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) (2 a+3 b \sin (c+d x))}{4 d} \]

[Out]

-(b*(3*a + 4*b)*Log[1 - Sin[c + d*x]])/(8*d) + ((3*a - 4*b)*b*Log[1 + Sin[c + d*x]])/(8*d) + (Sec[c + d*x]^4*(
a + b*Sin[c + d*x])^2)/(4*d) - (Sec[c + d*x]^2*(a + b*Sin[c + d*x])*(2*a + 3*b*Sin[c + d*x]))/(4*d)

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Rubi [A]  time = 0.223091, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2837, 12, 1645, 633, 31} \[ -\frac{b (3 a+4 b) \log (1-\sin (c+d x))}{8 d}+\frac{b (3 a-4 b) \log (\sin (c+d x)+1)}{8 d}+\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) (2 a+3 b \sin (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(3*a + 4*b)*Log[1 - Sin[c + d*x]])/(8*d) + ((3*a - 4*b)*b*Log[1 + Sin[c + d*x]])/(8*d) + (Sec[c + d*x]^4*(
a + b*Sin[c + d*x])^2)/(4*d) - (Sec[c + d*x]^2*(a + b*Sin[c + d*x])*(2*a + 3*b*Sin[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 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 \sec ^2(c+d x) (a+b \sin (c+d x))^2 \tan ^3(c+d x) \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{x^3 (a+x)^2}{b^3 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^2 \operatorname{Subst}\left (\int \frac{x^3 (a+x)^2}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{(a+x) \left (-2 b^4-4 a b^2 x-4 b^2 x^2\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) (2 a+3 b \sin (c+d x))}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{6 a b^4+8 b^4 x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^2 d}\\ &=\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) (2 a+3 b \sin (c+d x))}{4 d}-\frac{((3 a-4 b) b) \operatorname{Subst}\left (\int \frac{1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{8 d}+\frac{(b (3 a+4 b)) \operatorname{Subst}\left (\int \frac{1}{b-x} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=-\frac{b (3 a+4 b) \log (1-\sin (c+d x))}{8 d}+\frac{(3 a-4 b) b \log (1+\sin (c+d x))}{8 d}+\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) (2 a+3 b \sin (c+d x))}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.364141, size = 129, normalized size = 1.11 \[ \frac{a^2 \tan ^4(c+d x)}{4 d}+\frac{2 a b \tan ^3(c+d x) \sec (c+d x)}{d}-\frac{a b \left (6 \tan (c+d x) \sec ^3(c+d x)-3 \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )\right )}{4 d}-\frac{b^2 \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*Sec[c + d*x]*Tan[c + d*x]^3)/d + (a^2*Tan[c + d*x]^4)/(4*d) - (b^2*(4*Log[Cos[c + d*x]] + 2*Tan[c + d*x
]^2 - Tan[c + d*x]^4))/(4*d) - (a*b*(6*Sec[c + d*x]^3*Tan[c + d*x] - 3*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*T
an[c + d*x])))/(4*d)

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Maple [A]  time = 0.067, size = 168, normalized size = 1.5 \begin{align*}{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}{a}^{2}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{3\,ab\sin \left ( dx+c \right ) }{4\,d}}+{\frac{3\,ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\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(sec(d*x+c)^5*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.00683, size = 166, normalized size = 1.43 \begin{align*} \frac{{\left (3 \, a b - 4 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, a b + 4 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac{2 \,{\left (5 \, a b \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right ) + 2 \,{\left (a^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} - 3 \, b^{2}\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.1148, size = 313, normalized size = 2.7 \begin{align*} \frac{{\left (3 \, a b - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 4 \,{\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, b^{2} - 2 \,{\left (5 \, a b \cos \left (d x + c\right )^{2} - 2 \, a b\right )} \sin \left (d x + c\right )}{8 \, 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)^3*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.23428, size = 176, normalized size = 1.52 \begin{align*} \frac{{\left (3 \, a b - 4 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (3 \, a b + 4 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (3 \, b^{2} \sin \left (d x + c\right )^{4} + 5 \, a b \sin \left (d x + c\right )^{3} + 2 \, a^{2} \sin \left (d x + c\right )^{2} - 2 \, b^{2} \sin \left (d x + c\right )^{2} - 3 \, a b \sin \left (d x + c\right ) - a^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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