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]

-1/16*(3*a^2+16*a*b+15*b^2)*ln(1-sin(d*x+c))/d+1/16*(3*a^2-16*a*b+15*b^2)*ln(1+sin(d*x+c))/d-b^2*sin(d*x+c)/d-
1/8*sec(d*x+c)^2*(7*b+5*a*sin(d*x+c))*(a+b*sin(d*x+c))/d+1/4*sec(d*x+c)^3*(a+b*sin(d*x+c))^2*tan(d*x+c)/d

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Rubi [A]  time = 0.28, antiderivative size = 150, normalized size of antiderivative = 1.00, 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 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 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 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]

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*}

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Mathematica [A]  time = 1.02, 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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fricas [A]  time = 0.47, size = 151, normalized size = 1.01 \[ \frac {{\left (3 \, a^{2} - 16 \, a b + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a^{2} + 16 \, a b + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 32 \, a b \cos \left (d x + c\right )^{2} + 8 \, a b - 2 \, {\left (8 \, b^{2} \cos \left (d x + c\right )^{4} + {\left (5 \, a^{2} + 9 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]

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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giac [A]  time = 0.29, size = 157, normalized size = 1.05 \[ -\frac {16 \, b^{2} \sin \left (d x + c\right ) - {\left (3 \, a^{2} - 16 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + {\left (3 \, a^{2} + 16 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (12 \, a b \sin \left (d x + c\right )^{4} + 5 \, a^{2} \sin \left (d x + c\right )^{3} + 9 \, b^{2} \sin \left (d x + c\right )^{3} - 8 \, a b \sin \left (d x + c\right )^{2} - 3 \, a^{2} \sin \left (d x + c\right ) - 7 \, b^{2} \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

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

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maple [A]  time = 0.32, size = 262, normalized size = 1.75 \[ \frac {a^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {a^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {3 a^{2} \sin \left (d x +c \right )}{8 d}+\frac {3 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {a b \left (\tan ^{4}\left (d x +c \right )\right )}{2 d}-\frac {a b \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {2 a b \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {b^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {3 b^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {3 b^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d}-\frac {5 b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {15 b^{2} \sin \left (d x +c \right )}{8 d}+\frac {15 b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

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 = 0.33, size = 148, normalized size = 0.99 \[ -\frac {16 \, b^{2} \sin \left (d x + c\right ) - {\left (3 \, a^{2} - 16 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, a^{2} + 16 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (16 \, a b \sin \left (d x + c\right )^{2} + {\left (5 \, a^{2} + 9 \, b^{2}\right )} \sin \left (d x + c\right )^{3} - 12 \, a b - {\left (3 \, a^{2} + 7 \, b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

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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mupad [B]  time = 12.21, size = 332, normalized size = 2.21 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {3\,a^2}{8}-2\,a\,b+\frac {15\,b^2}{8}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {3\,a^2}{8}+2\,a\,b+\frac {15\,b^2}{8}\right )}{d}+\frac {\left (-\frac {3\,a^2}{4}-\frac {15\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (2\,a^2+10\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {11\,a^2}{2}-\frac {9\,b^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (2\,a^2+10\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (-\frac {3\,a^2}{4}-\frac {15\,b^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2\,a\,b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^4*(a + b*sin(c + d*x))^2)/cos(c + d*x)^5,x)

[Out]

(log(tan(c/2 + (d*x)/2) + 1)*((3*a^2)/8 - 2*a*b + (15*b^2)/8))/d - (log(tan(c/2 + (d*x)/2) - 1)*(2*a*b + (3*a^
2)/8 + (15*b^2)/8))/d + (tan(c/2 + (d*x)/2)^3*(2*a^2 + 10*b^2) + tan(c/2 + (d*x)/2)^7*(2*a^2 + 10*b^2) + tan(c
/2 + (d*x)/2)^5*((11*a^2)/2 - (9*b^2)/2) - tan(c/2 + (d*x)/2)^9*((3*a^2)/4 + (15*b^2)/4) - tan(c/2 + (d*x)/2)*
((3*a^2)/4 + (15*b^2)/4) - 4*a*b*tan(c/2 + (d*x)/2)^2 + 12*a*b*tan(c/2 + (d*x)/2)^4 + 12*a*b*tan(c/2 + (d*x)/2
)^6 - 4*a*b*tan(c/2 + (d*x)/2)^8)/(d*(2*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^6
- 3*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) + (2*a*b*log(tan(c/2 + (d*x)/2)^2 + 1))/d

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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