3.1493 \(\int (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx\)

Optimal. Leaf size=162 \[ -\frac{\left (4 a^2+15 a b+12 b^2\right ) \log (1-\sin (c+d x))}{8 d}+\frac{\left (15 a b-4 \left (a^2+3 b^2\right )\right ) \log (\sin (c+d x)+1)}{8 d}-\frac{2 a b \sin (c+d x)}{d}+\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) (4 a+5 b \sin (c+d x))}{4 d}-\frac{b^2 \sin ^2(c+d x)}{2 d} \]

[Out]

-((4*a^2 + 15*a*b + 12*b^2)*Log[1 - Sin[c + d*x]])/(8*d) + ((15*a*b - 4*(a^2 + 3*b^2))*Log[1 + Sin[c + d*x]])/
(8*d) - (2*a*b*Sin[c + d*x])/d - (b^2*Sin[c + d*x]^2)/(2*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])*(4*a + 5*b*Sin[c + d*x]))/(4*d)

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Rubi [A]  time = 0.27027, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2721, 1645, 1810, 633, 31} \[ -\frac{\left (4 a^2+15 a b+12 b^2\right ) \log (1-\sin (c+d x))}{8 d}+\frac{\left (15 a b-4 \left (a^2+3 b^2\right )\right ) \log (\sin (c+d x)+1)}{8 d}-\frac{2 a b \sin (c+d x)}{d}+\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) (4 a+5 b \sin (c+d x))}{4 d}-\frac{b^2 \sin ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^2*Tan[c + d*x]^5,x]

[Out]

-((4*a^2 + 15*a*b + 12*b^2)*Log[1 - Sin[c + d*x]])/(8*d) + ((15*a*b - 4*(a^2 + 3*b^2))*Log[1 + Sin[c + d*x]])/
(8*d) - (2*a*b*Sin[c + d*x])/d - (b^2*Sin[c + d*x]^2)/(2*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])*(4*a + 5*b*Sin[c + d*x]))/(4*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 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 (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5 (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^6-4 a b^4 x-4 b^4 x^2-4 a b^2 x^3-4 b^2 x^4\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 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)) (4 a+5 b \sin (c+d x))}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{14 a b^6+8 b^4 \left (a^2+2 b^2\right ) x+16 a b^4 x^2+8 b^4 x^3}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^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)) (4 a+5 b \sin (c+d x))}{4 d}+\frac{\operatorname{Subst}\left (\int \left (-16 a b^4-8 b^4 x+\frac{2 \left (15 a b^6+4 b^4 \left (a^2+3 b^2\right ) x\right )}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^4 d}\\ &=-\frac{2 a b \sin (c+d x)}{d}-\frac{b^2 \sin ^2(c+d x)}{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)) (4 a+5 b \sin (c+d x))}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{15 a b^6+4 b^4 \left (a^2+3 b^2\right ) x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{4 b^4 d}\\ &=-\frac{2 a b \sin (c+d x)}{d}-\frac{b^2 \sin ^2(c+d x)}{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)) (4 a+5 b \sin (c+d x))}{4 d}+\frac{\left (4 a^2-15 a b+12 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{8 d}+\frac{\left (4 a^2+15 a b+12 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{b-x} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=-\frac{\left (4 a^2+15 a b+12 b^2\right ) \log (1-\sin (c+d x))}{8 d}-\frac{\left (4 a^2-15 a b+12 b^2\right ) \log (1+\sin (c+d x))}{8 d}-\frac{2 a b \sin (c+d x)}{d}-\frac{b^2 \sin ^2(c+d x)}{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)) (4 a+5 b \sin (c+d x))}{4 d}\\ \end{align*}

Mathematica [A]  time = 2.15948, size = 164, normalized size = 1.01 \[ \frac{-2 \left (4 a^2+15 a b+12 b^2\right ) \log (1-\sin (c+d x))-2 \left (4 a^2-15 a b+12 b^2\right ) \log (\sin (c+d x)+1)+\frac{(a-b)^2}{(\sin (c+d x)+1)^2}-\frac{(7 a-11 b) (a-b)}{\sin (c+d x)+1}-32 a b \sin (c+d x)+\frac{(a+b) (7 a+11 b)}{\sin (c+d x)-1}+\frac{(a+b)^2}{(\sin (c+d x)-1)^2}-8 b^2 \sin ^2(c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^2*Tan[c + d*x]^5,x]

[Out]

(-2*(4*a^2 + 15*a*b + 12*b^2)*Log[1 - Sin[c + d*x]] - 2*(4*a^2 - 15*a*b + 12*b^2)*Log[1 + Sin[c + d*x]] + (a +
b)^2/(-1 + Sin[c + d*x])^2 + ((a + b)*(7*a + 11*b))/(-1 + Sin[c + d*x]) - 32*a*b*Sin[c + d*x] - 8*b^2*Sin[c +
d*x]^2 + (a - b)^2/(1 + Sin[c + d*x])^2 - ((7*a - 11*b)*(a - b))/(1 + Sin[c + d*x]))/(16*d)

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Maple [A]  time = 0.076, size = 270, normalized size = 1.7 \begin{align*}{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\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 ) ^{7}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{3\,ab \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,ab \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d}}-{\frac{5\,ab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{15\,ab\sin \left ( dx+c \right ) }{4\,d}}+{\frac{15\,ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{2\,d}}-{\frac{3\,{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{2}}{2\,d}}-3\,{\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)^5*(a+b*sin(d*x+c))^2,x)

[Out]

1/4/d*a^2*tan(d*x+c)^4-1/2/d*a^2*tan(d*x+c)^2-1/d*a^2*ln(cos(d*x+c))+1/2/d*a*b*sin(d*x+c)^7/cos(d*x+c)^4-3/4/d
*a*b*sin(d*x+c)^7/cos(d*x+c)^2-3/4*a*b*sin(d*x+c)^5/d-5/4*a*b*sin(d*x+c)^3/d-15/4*a*b*sin(d*x+c)/d+15/4/d*a*b*
ln(sec(d*x+c)+tan(d*x+c))+1/4/d*b^2*sin(d*x+c)^8/cos(d*x+c)^4-1/2/d*b^2*sin(d*x+c)^8/cos(d*x+c)^2-1/2*b^2*sin(
d*x+c)^6/d-3/4*b^2*sin(d*x+c)^4/d-3/2*b^2*sin(d*x+c)^2/d-3/d*b^2*ln(cos(d*x+c))

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Maxima [A]  time = 0.993926, size = 212, normalized size = 1.31 \begin{align*} -\frac{4 \, b^{2} \sin \left (d x + c\right )^{2} + 16 \, a b \sin \left (d x + c\right ) +{\left (4 \, a^{2} - 15 \, a b + 12 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (4 \, a^{2} + 15 \, a b + 12 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (9 \, a b \sin \left (d x + c\right )^{3} - 7 \, a b \sin \left (d x + c\right ) + 2 \,{\left (2 \, a^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )^{2} - 3 \, a^{2} - 5 \, 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)^5*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.44838, size = 436, normalized size = 2.69 \begin{align*} \frac{4 \, b^{2} \cos \left (d x + c\right )^{6} - 2 \, b^{2} \cos \left (d x + c\right )^{4} -{\left (4 \, a^{2} - 15 \, a b + 12 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (4 \, a^{2} + 15 \, a b + 12 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 4 \,{\left (2 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, b^{2} - 2 \,{\left (8 \, a b \cos \left (d x + c\right )^{4} + 9 \, 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)^5*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/8*(4*b^2*cos(d*x + c)^6 - 2*b^2*cos(d*x + c)^4 - (4*a^2 - 15*a*b + 12*b^2)*cos(d*x + c)^4*log(sin(d*x + c) +
1) - (4*a^2 + 15*a*b + 12*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 4*(2*a^2 + 3*b^2)*cos(d*x + c)^2 + 2*a
^2 + 2*b^2 - 2*(8*a*b*cos(d*x + c)^4 + 9*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)**5*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.25318, size = 236, normalized size = 1.46 \begin{align*} -\frac{4 \, b^{2} \sin \left (d x + c\right )^{2} + 16 \, a b \sin \left (d x + c\right ) +{\left (4 \, a^{2} - 15 \, a b + 12 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) +{\left (4 \, a^{2} + 15 \, a b + 12 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (3 \, a^{2} \sin \left (d x + c\right )^{4} + 9 \, b^{2} \sin \left (d x + c\right )^{4} + 9 \, a b \sin \left (d x + c\right )^{3} - 2 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, b^{2} \sin \left (d x + c\right )^{2} - 7 \, a b \sin \left (d x + c\right ) + 4 \, b^{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)^5*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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