3.1494 \(\int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=150 \[ -\frac{\left (3 a^2+16 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac{\left (3 a^2-16 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac{\sec ^2(c+d x) (5 a \sin (c+d x)+7 b) (a+b \sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac{b^2 \sin (c+d x)}{d} \]

[Out]

-((3*a^2 + 16*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/(16*d) + ((3*a^2 - 16*a*b + 15*b^2)*Log[1 + Sin[c + d*x]])/
(16*d) - (b^2*Sin[c + d*x])/d - (Sec[c + d*x]^2*(7*b + 5*a*Sin[c + d*x])*(a + b*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 = 0.284945, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2837, 12, 1645, 1810, 633, 31} \[ -\frac{\left (3 a^2+16 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac{\left (3 a^2-16 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac{\sec ^2(c+d x) (5 a \sin (c+d x)+7 b) (a+b \sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac{b^2 \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + b*Sin[c + d*x])^2*Tan[c + d*x]^4,x]

[Out]

-((3*a^2 + 16*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/(16*d) + ((3*a^2 - 16*a*b + 15*b^2)*Log[1 + Sin[c + d*x]])/
(16*d) - (b^2*Sin[c + d*x])/d - (Sec[c + d*x]^2*(7*b + 5*a*Sin[c + d*x])*(a + b*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 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, 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 \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{x^4 (a+x)^2}{b^4 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b \operatorname{Subst}\left (\int \frac{x^4 (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{\operatorname{Subst}\left (\int \frac{(a+x) \left (-a b^4-3 b^4 x-4 a b^2 x^2-4 b^2 x^3\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b d}\\ &=-\frac{\sec ^2(c+d x) (7 b+5 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{b^4 \left (3 a^2+7 b^2\right )+16 a b^4 x+8 b^4 x^2}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^3 d}\\ &=-\frac{\sec ^2(c+d x) (7 b+5 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac{\operatorname{Subst}\left (\int \left (-8 b^4+\frac{3 b^4 \left (a^2+5 b^2\right )+16 a b^4 x}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^3 d}\\ &=-\frac{b^2 \sin (c+d x)}{d}-\frac{\sec ^2(c+d x) (7 b+5 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{3 b^4 \left (a^2+5 b^2\right )+16 a b^4 x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^3 d}\\ &=-\frac{b^2 \sin (c+d x)}{d}-\frac{\sec ^2(c+d x) (7 b+5 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac{\left (3 a^2+16 a b+15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac{\left (8 a b^4-\frac{3}{2} b^3 \left (a^2+5 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{8 b^3 d}\\ &=-\frac{\left (3 a^2+16 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d}-\frac{\left (16 a b-3 \left (a^2+5 b^2\right )\right ) \log (1+\sin (c+d x))}{16 d}-\frac{b^2 \sin (c+d x)}{d}-\frac{\sec ^2(c+d x) (7 b+5 a \sin (c+d x)) (a+b \sin (c+d x))}{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 = 1.03108, size = 151, normalized size = 1.01 \[ \frac{-\left (3 a^2+16 a b+15 b^2\right ) \log (1-\sin (c+d x))+\left (3 a^2-16 a b+15 b^2\right ) \log (\sin (c+d x)+1)-\frac{(a-b)^2}{(\sin (c+d x)+1)^2}+\frac{(5 a-9 b) (a-b)}{\sin (c+d x)+1}+\frac{(a+b) (5 a+9 b)}{\sin (c+d x)-1}+\frac{(a+b)^2}{(\sin (c+d x)-1)^2}-16 b^2 \sin (c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + b*Sin[c + d*x])^2*Tan[c + d*x]^4,x]

[Out]

(-((3*a^2 + 16*a*b + 15*b^2)*Log[1 - Sin[c + d*x]]) + (3*a^2 - 16*a*b + 15*b^2)*Log[1 + Sin[c + d*x]] + (a + b
)^2/(-1 + Sin[c + d*x])^2 + ((a + b)*(5*a + 9*b))/(-1 + Sin[c + d*x]) - 16*b^2*Sin[c + d*x] - (a - b)^2/(1 + S
in[c + d*x])^2 + ((5*a - 9*b)*(a - b))/(1 + Sin[c + d*x]))/(16*d)

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Maple [A]  time = 0.073, size = 262, normalized size = 1.8 \begin{align*}{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}{a}^{2}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}{a}^{2}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{3\,{a}^{2}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{ab \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{2\,d}}-{\frac{ab \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}-2\,{\frac{ab\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{3\,{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d}}-{\frac{5\,{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{15\,{b}^{2}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{15\,{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)^4*(a+b*sin(d*x+c))^2,x)

[Out]

1/4/d*a^2*sin(d*x+c)^5/cos(d*x+c)^4-1/8/d*a^2*sin(d*x+c)^5/cos(d*x+c)^2-1/8*a^2*sin(d*x+c)^3/d-3/8*a^2*sin(d*x
+c)/d+3/8/d*a^2*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*a*b*tan(d*x+c)^4-1/d*a*b*tan(d*x+c)^2-2/d*a*b*ln(cos(d*x+c))+1
/4/d*b^2*sin(d*x+c)^7/cos(d*x+c)^4-3/8/d*b^2*sin(d*x+c)^7/cos(d*x+c)^2-3/8*b^2*sin(d*x+c)^5/d-5/8*b^2*sin(d*x+
c)^3/d-15/8*b^2*sin(d*x+c)/d+15/8/d*b^2*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.969534, size = 200, normalized size = 1.33 \begin{align*} -\frac{16 \, b^{2} \sin \left (d x + c\right ) -{\left (3 \, a^{2} - 16 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (3 \, a^{2} + 16 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (16 \, a b \sin \left (d x + c\right )^{2} +{\left (5 \, a^{2} + 9 \, b^{2}\right )} \sin \left (d x + c\right )^{3} - 12 \, a b -{\left (3 \, a^{2} + 7 \, 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)^4*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/16*(16*b^2*sin(d*x + c) - (3*a^2 - 16*a*b + 15*b^2)*log(sin(d*x + c) + 1) + (3*a^2 + 16*a*b + 15*b^2)*log(s
in(d*x + c) - 1) - 2*(16*a*b*sin(d*x + c)^2 + (5*a^2 + 9*b^2)*sin(d*x + c)^3 - 12*a*b - (3*a^2 + 7*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 = 2.22936, size = 374, normalized size = 2.49 \begin{align*} \frac{{\left (3 \, a^{2} - 16 \, a b + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, a^{2} + 16 \, a b + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 32 \, a b \cos \left (d x + c\right )^{2} + 8 \, a b - 2 \,{\left (8 \, b^{2} \cos \left (d x + c\right )^{4} +{\left (5 \, a^{2} + 9 \, 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)^4*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*((3*a^2 - 16*a*b + 15*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (3*a^2 + 16*a*b + 15*b^2)*cos(d*x + c)^
4*log(-sin(d*x + c) + 1) - 32*a*b*cos(d*x + c)^2 + 8*a*b - 2*(8*b^2*cos(d*x + c)^4 + (5*a^2 + 9*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)**4*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.23788, size = 212, normalized size = 1.41 \begin{align*} -\frac{16 \, b^{2} \sin \left (d x + c\right ) -{\left (3 \, a^{2} - 16 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) +{\left (3 \, a^{2} + 16 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (12 \, a b \sin \left (d x + c\right )^{4} + 5 \, a^{2} \sin \left (d x + c\right )^{3} + 9 \, b^{2} \sin \left (d x + c\right )^{3} - 8 \, a b \sin \left (d x + c\right )^{2} - 3 \, a^{2} \sin \left (d x + c\right ) - 7 \, b^{2} \sin \left (d x + c\right )\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)^4*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*(16*b^2*sin(d*x + c) - (3*a^2 - 16*a*b + 15*b^2)*log(abs(sin(d*x + c) + 1)) + (3*a^2 + 16*a*b + 15*b^2)*
log(abs(sin(d*x + c) - 1)) - 2*(12*a*b*sin(d*x + c)^4 + 5*a^2*sin(d*x + c)^3 + 9*b^2*sin(d*x + c)^3 - 8*a*b*si
n(d*x + c)^2 - 3*a^2*sin(d*x + c) - 7*b^2*sin(d*x + c))/(sin(d*x + c)^2 - 1)^2)/d