3.1504 \(\int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=144 \[ \frac {\left (a^3-9 a b^2-8 b^3\right ) \log (1-\sin (c+d x))}{16 d}-\frac {\left (a^3-9 a b^2+8 b^3\right ) \log (\sin (c+d x)+1)}{16 d}-\frac {\sec ^2(c+d x) \left (\left (a^2+4 b^2\right ) \sin (c+d x)+5 a b\right ) (a+b \sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^3}{4 d} \]

[Out]

1/16*(a^3-9*a*b^2-8*b^3)*ln(1-sin(d*x+c))/d-1/16*(a^3-9*a*b^2+8*b^3)*ln(1+sin(d*x+c))/d-1/8*sec(d*x+c)^2*(a+b*
sin(d*x+c))*(5*a*b+(a^2+4*b^2)*sin(d*x+c))/d+1/4*sec(d*x+c)^3*(a+b*sin(d*x+c))^3*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2837, 12, 1645, 819, 633, 31} \[ \frac {\left (a^3-9 a b^2-8 b^3\right ) \log (1-\sin (c+d x))}{16 d}-\frac {\left (a^3-9 a b^2+8 b^3\right ) \log (\sin (c+d x)+1)}{16 d}-\frac {\sec ^2(c+d x) \left (\left (a^2+4 b^2\right ) \sin (c+d x)+5 a b\right ) (a+b \sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^3 - 9*a*b^2 - 8*b^3)*Log[1 - Sin[c + d*x]])/(16*d) - ((a^3 - 9*a*b^2 + 8*b^3)*Log[1 + Sin[c + d*x]])/(16*d
) - (Sec[c + d*x]^2*(a + b*Sin[c + d*x])*(5*a*b + (a^2 + 4*b^2)*Sin[c + d*x]))/(8*d) + (Sec[c + d*x]^3*(a + b*
Sin[c + d*x])^3*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 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

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 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 ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {x^2 (a+x)^3}{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)^3}{\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))^3 \tan (c+d x)}{4 d}+\frac {b \operatorname {Subst}\left (\int \frac {(a+x)^2 \left (-a b^2-4 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) (a+b \sin (c+d x)) \left (5 a b+\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}-\frac {\operatorname {Subst}\left (\int \frac {a b^2 \left (a^2-9 b^2\right )-8 b^4 x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b d}\\ &=-\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) \left (5 a b+\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}-\frac {\left (a^3-9 a b^2-8 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {\left (a^3-9 a b^2+8 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=\frac {\left (a^3-9 a b^2-8 b^3\right ) \log (1-\sin (c+d x))}{16 d}-\frac {\left (a^3-9 a b^2+8 b^3\right ) \log (1+\sin (c+d x))}{16 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) \left (5 a b+\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.39, size = 140, normalized size = 0.97 \[ \frac {\left (a^3-9 a b^2-8 b^3\right ) \log (1-\sin (c+d x))-\left (a^3-9 a b^2+8 b^3\right ) \log (\sin (c+d x)+1)-\frac {(a-b)^3}{(\sin (c+d x)+1)^2}+\frac {(a-7 b) (a-b)^2}{\sin (c+d x)+1}+\frac {(a+b)^2 (a+7 b)}{\sin (c+d x)-1}+\frac {(a+b)^3}{(\sin (c+d x)-1)^2}}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^3 - 9*a*b^2 - 8*b^3)*Log[1 - Sin[c + d*x]] - (a^3 - 9*a*b^2 + 8*b^3)*Log[1 + Sin[c + d*x]] + (a + b)^3/(-1
+ Sin[c + d*x])^2 + ((a + b)^2*(a + 7*b))/(-1 + Sin[c + d*x]) - (a - b)^3/(1 + Sin[c + d*x])^2 + ((a - 7*b)*(
a - b)^2)/(1 + Sin[c + d*x]))/(16*d)

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 156, normalized size = 1.08 \[ -\frac {{\left (a^{3} - 9 \, a b^{2} + 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 9 \, a b^{2} - 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 12 \, a^{2} b - 4 \, b^{3} + 8 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{3} + 6 \, a b^{2} - {\left (a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((a^3 - 9*a*b^2 + 8*b^3)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (a^3 - 9*a*b^2 - 8*b^3)*cos(d*x + c)^4*l
og(-sin(d*x + c) + 1) - 12*a^2*b - 4*b^3 + 8*(3*a^2*b + 2*b^3)*cos(d*x + c)^2 - 2*(2*a^3 + 6*a*b^2 - (a^3 + 15
*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

________________________________________________________________________________________

giac [A]  time = 0.27, size = 168, normalized size = 1.17 \[ -\frac {{\left (a^{3} - 9 \, a b^{2} + 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (a^{3} - 9 \, a b^{2} - 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, b^{3} \sin \left (d x + c\right )^{4} + a^{3} \sin \left (d x + c\right )^{3} + 15 \, a b^{2} \sin \left (d x + c\right )^{3} + 12 \, a^{2} b \sin \left (d x + c\right )^{2} - 4 \, b^{3} \sin \left (d x + c\right )^{2} + a^{3} \sin \left (d x + c\right ) - 9 \, a b^{2} \sin \left (d x + c\right ) - 6 \, a^{2} b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((a^3 - 9*a*b^2 + 8*b^3)*log(abs(sin(d*x + c) + 1)) - (a^3 - 9*a*b^2 - 8*b^3)*log(abs(sin(d*x + c) - 1))
- 2*(6*b^3*sin(d*x + c)^4 + a^3*sin(d*x + c)^3 + 15*a*b^2*sin(d*x + c)^3 + 12*a^2*b*sin(d*x + c)^2 - 4*b^3*si
n(d*x + c)^2 + a^3*sin(d*x + c) - 9*a*b^2*sin(d*x + c) - 6*a^2*b)/(sin(d*x + c)^2 - 1)^2)/d

________________________________________________________________________________________

maple [A]  time = 0.32, size = 263, normalized size = 1.83 \[ \frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {a^{3} \sin \left (d x +c \right )}{8 d}-\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 a^{2} b \left (\sin ^{4}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {3 a \,b^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {3 a \,b^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {3 a \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {9 a \,b^{2} \sin \left (d x +c \right )}{8 d}+\frac {9 a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {b^{3} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {b^{3} \ln \left (\cos \left (d x +c \right )\right )}{d} \]

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))^3,x)

[Out]

1/4/d*a^3*sin(d*x+c)^3/cos(d*x+c)^4+1/8/d*a^3*sin(d*x+c)^3/cos(d*x+c)^2+1/8*a^3*sin(d*x+c)/d-1/8/d*a^3*ln(sec(
d*x+c)+tan(d*x+c))+3/4/d*a^2*b*sin(d*x+c)^4/cos(d*x+c)^4+3/4/d*a*b^2*sin(d*x+c)^5/cos(d*x+c)^4-3/8/d*a*b^2*sin
(d*x+c)^5/cos(d*x+c)^2-3/8/d*a*b^2*sin(d*x+c)^3-9/8*a*b^2*sin(d*x+c)/d+9/8/d*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+1
/4/d*b^3*tan(d*x+c)^4-1/2/d*b^3*tan(d*x+c)^2-1/d*b^3*ln(cos(d*x+c))

________________________________________________________________________________________

maxima [A]  time = 0.33, size = 151, normalized size = 1.05 \[ -\frac {{\left (a^{3} - 9 \, a b^{2} + 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 9 \, a b^{2} - 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (a^{3} + 15 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} - 6 \, a^{2} b - 6 \, b^{3} + 4 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{3} - 9 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((a^3 - 9*a*b^2 + 8*b^3)*log(sin(d*x + c) + 1) - (a^3 - 9*a*b^2 - 8*b^3)*log(sin(d*x + c) - 1) - 2*((a^3
+ 15*a*b^2)*sin(d*x + c)^3 - 6*a^2*b - 6*b^3 + 4*(3*a^2*b + 2*b^3)*sin(d*x + c)^2 + (a^3 - 9*a*b^2)*sin(d*x +
c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

________________________________________________________________________________________

mupad [B]  time = 12.14, size = 299, normalized size = 2.08 \[ \frac {b^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (-\frac {a^3}{8}+\frac {9\,a\,b^2}{8}+b^3\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {a^3}{8}-\frac {9\,a\,b^2}{8}+b^3\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {9\,a\,b^2}{4}-\frac {a^3}{4}\right )+2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {9\,a\,b^2}{4}-\frac {a^3}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {7\,a^3}{4}+\frac {33\,a\,b^2}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {7\,a^3}{4}+\frac {33\,a\,b^2}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (12\,a^2\,b+8\,b^3\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^2*(a + b*sin(c + d*x))^3)/cos(c + d*x)^5,x)

[Out]

(b^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (log(tan(c/2 + (d*x)/2) - 1)*((9*a*b^2)/8 - a^3/8 + b^3))/d - (log(tan
(c/2 + (d*x)/2) + 1)*(a^3/8 - (9*a*b^2)/8 + b^3))/d - (tan(c/2 + (d*x)/2)*((9*a*b^2)/4 - a^3/4) + 2*b^3*tan(c/
2 + (d*x)/2)^2 + 2*b^3*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^7*((9*a*b^2)/4 - a^3/4) - tan(c/2 + (d*x)/2)^
3*((33*a*b^2)/4 + (7*a^3)/4) - tan(c/2 + (d*x)/2)^5*((33*a*b^2)/4 + (7*a^3)/4) - tan(c/2 + (d*x)/2)^4*(12*a^2*
b + 8*b^3))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^
8 + 1))

________________________________________________________________________________________

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)**2*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________