3.1505 \(\int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x) \, dx\)

Optimal. Leaf size=90 \[ -\frac {3 b \left (a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {3 \sec ^2(c+d x) (a+b \sin (c+d x)) \left (a b \sin (c+d x)+b^2\right )}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2837, 12, 805, 723, 206} \[ -\frac {3 b \left (a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {3 \sec ^2(c+d x) (a+b \sin (c+d x)) \left (a b \sin (c+d x)+b^2\right )}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + b*Sin[c + d*x])^3*Tan[c + d*x],x]

[Out]

(-3*b*(a^2 - b^2)*ArcTanh[Sin[c + d*x]])/(8*d) + (Sec[c + d*x]^4*(a + b*Sin[c + d*x])^3)/(4*d) - (3*Sec[c + d*
x]^2*(a + b*Sin[c + d*x])*(b^2 + a*b*Sin[c + d*x]))/(8*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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 723

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[((2*p + 3)*(c*d^2 + a*e^2))/(2*a*c*(p + 1)), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

Rule 805

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

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

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Mathematica [B]  time = 1.47, size = 370, normalized size = 4.11 \[ \frac {2 b \left (a^2-b^2\right ) \sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^5+2 a \left (a^2-b^2\right )^2 \sec ^4(c+d x) (a+b \sin (c+d x))^4+b \sec ^2(c+d x) \left (-3 a \left (a^2+b^2\right ) \sin (c+d x)+5 a^2 b+b^3\right ) (a+b \sin (c+d x))^5-a b \left (a^2+b^2\right ) \left (60 a b^2 \left (2 a^2+b^2\right ) \sin (c+d x)+6 b^3 \left (10 a^2+b^2\right ) \sin ^2(c+d x)+20 a b^4 \sin ^3(c+d x)-6 (a-b)^5 \log (\sin (c+d x)+1)+6 (a+b)^5 \log (1-\sin (c+d x))+3 b^5 \sin ^4(c+d x)\right )+\frac {1}{2} \left (5 a^4 b+10 a^2 b^3+b^5\right ) \left (6 b^2 \left (6 a^2+b^2\right ) \sin (c+d x)+12 a b^3 \sin ^2(c+d x)+3 \left ((a+b)^4 \log (1-\sin (c+d x))-(a-b)^4 \log (\sin (c+d x)+1)\right )+2 b^4 \sin ^3(c+d x)\right )}{8 d \left (a^2-b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + b*Sin[c + d*x])^3*Tan[c + d*x],x]

[Out]

(2*a*(a^2 - b^2)^2*Sec[c + d*x]^4*(a + b*Sin[c + d*x])^4 + 2*b*(a^2 - b^2)*Sec[c + d*x]^4*(b - a*Sin[c + d*x])
*(a + b*Sin[c + d*x])^5 + b*Sec[c + d*x]^2*(a + b*Sin[c + d*x])^5*(5*a^2*b + b^3 - 3*a*(a^2 + b^2)*Sin[c + d*x
]) + ((5*a^4*b + 10*a^2*b^3 + b^5)*(3*((a + b)^4*Log[1 - Sin[c + d*x]] - (a - b)^4*Log[1 + Sin[c + d*x]]) + 6*
b^2*(6*a^2 + b^2)*Sin[c + d*x] + 12*a*b^3*Sin[c + d*x]^2 + 2*b^4*Sin[c + d*x]^3))/2 - a*b*(a^2 + b^2)*(6*(a +
b)^5*Log[1 - Sin[c + d*x]] - 6*(a - b)^5*Log[1 + Sin[c + d*x]] + 60*a*b^2*(2*a^2 + b^2)*Sin[c + d*x] + 6*b^3*(
10*a^2 + b^2)*Sin[c + d*x]^2 + 20*a*b^4*Sin[c + d*x]^3 + 3*b^5*Sin[c + d*x]^4))/(8*(a^2 - b^2)^3*d)

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fricas [A]  time = 0.45, size = 143, normalized size = 1.59 \[ -\frac {3 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 24 \, a b^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{3} - 12 \, a b^{2} - 2 \, {\left (6 \, a^{2} b + 2 \, b^{3} - {\left (3 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(3*(a^2*b - b^3)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(a^2*b - b^3)*cos(d*x + c)^4*log(-sin(d*x + c)
+ 1) + 24*a*b^2*cos(d*x + c)^2 - 4*a^3 - 12*a*b^2 - 2*(6*a^2*b + 2*b^3 - (3*a^2*b + 5*b^3)*cos(d*x + c)^2)*si
n(d*x + c))/(d*cos(d*x + c)^4)

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giac [A]  time = 0.26, size = 142, normalized size = 1.58 \[ -\frac {3 \, {\left (a^{2} b - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (a^{2} b - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a^{2} b \sin \left (d x + c\right )^{3} + 5 \, b^{3} \sin \left (d x + c\right )^{3} + 12 \, a b^{2} \sin \left (d x + c\right )^{2} + 3 \, a^{2} b \sin \left (d x + c\right ) - 3 \, b^{3} \sin \left (d x + c\right ) + 2 \, a^{3} - 6 \, a b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(3*(a^2*b - b^3)*log(abs(sin(d*x + c) + 1)) - 3*(a^2*b - b^3)*log(abs(sin(d*x + c) - 1)) - 2*(3*a^2*b*si
n(d*x + c)^3 + 5*b^3*sin(d*x + c)^3 + 12*a*b^2*sin(d*x + c)^2 + 3*a^2*b*sin(d*x + c) - 3*b^3*sin(d*x + c) + 2*
a^3 - 6*a*b^2)/(sin(d*x + c)^2 - 1)^2)/d

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maple [B]  time = 0.31, size = 231, normalized size = 2.57 \[ \frac {a^{3}}{4 d \cos \left (d x +c \right )^{4}}+\frac {3 a^{2} b \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {3 a^{2} b \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {3 a^{2} b \sin \left (d x +c \right )}{8 d}-\frac {3 a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 a \,b^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {b^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {b^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {3 b^{3} \sin \left (d x +c \right )}{8 d}+\frac {3 b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)*(a+b*sin(d*x+c))^3,x)

[Out]

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

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maxima [A]  time = 0.38, size = 140, normalized size = 1.56 \[ -\frac {3 \, {\left (a^{2} b - b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{2} b - b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (12 \, a b^{2} \sin \left (d x + c\right )^{2} + {\left (3 \, a^{2} b + 5 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + 2 \, a^{3} - 6 \, a b^{2} + 3 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(3*(a^2*b - b^3)*log(sin(d*x + c) + 1) - 3*(a^2*b - b^3)*log(sin(d*x + c) - 1) - 2*(12*a*b^2*sin(d*x + c
)^2 + (3*a^2*b + 5*b^3)*sin(d*x + c)^3 + 2*a^3 - 6*a*b^2 + 3*(a^2*b - b^3)*sin(d*x + c))/(sin(d*x + c)^4 - 2*s
in(d*x + c)^2 + 1))/d

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mupad [B]  time = 16.99, size = 228, normalized size = 2.53 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a^2\,b}{4}-\frac {3\,b^3}{4}\right )+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {3\,a^2\,b}{4}-\frac {3\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {21\,a^2\,b}{4}+\frac {11\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {21\,a^2\,b}{4}+\frac {11\,b^3}{4}\right )+12\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {3\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-b^2\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((3*a^2*b)/4 - (3*b^3)/4) + 2*a^3*tan(c/2 + (d*x)/2)^2 + 2*a^3*tan(c/2 + (d*x)/2)^6 + tan(
c/2 + (d*x)/2)^7*((3*a^2*b)/4 - (3*b^3)/4) + tan(c/2 + (d*x)/2)^3*((21*a^2*b)/4 + (11*b^3)/4) + tan(c/2 + (d*x
)/2)^5*((21*a^2*b)/4 + (11*b^3)/4) + 12*a*b^2*tan(c/2 + (d*x)/2)^4)/(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)) - (3*b*atanh(tan(c/2 + (d*x)/2))*(a^2 - b^2))/
(4*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)*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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