3.1495 \(\int \sec ^2(c+d x) (a+b \sin (c+d x))^2 \tan ^3(c+d x) \, dx\)

Optimal. Leaf size=116 \[ -\frac {b (3 a+4 b) \log (1-\sin (c+d x))}{8 d}+\frac {b (3 a-4 b) \log (\sin (c+d x)+1)}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (2 a+3 b \sin (c+d x))}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2837, 12, 1645, 633, 31} \[ -\frac {b (3 a+4 b) \log (1-\sin (c+d x))}{8 d}+\frac {b (3 a-4 b) \log (\sin (c+d x)+1)}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (2 a+3 b \sin (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.38, size = 129, normalized size = 1.11 \[ \frac {a^2 \tan ^4(c+d x)}{4 d}+\frac {2 a b \tan ^3(c+d x) \sec (c+d x)}{d}-\frac {a b \left (6 \tan (c+d x) \sec ^3(c+d x)-3 \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )\right )}{4 d}-\frac {b^2 \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*Sec[c + d*x]*Tan[c + d*x]^3)/d + (a^2*Tan[c + d*x]^4)/(4*d) - (b^2*(4*Log[Cos[c + d*x]] + 2*Tan[c + d*x
]^2 - Tan[c + d*x]^4))/(4*d) - (a*b*(6*Sec[c + d*x]^3*Tan[c + d*x] - 3*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*T
an[c + d*x])))/(4*d)

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fricas [A]  time = 0.46, size = 127, normalized size = 1.09 \[ \frac {{\left (3 \, a b - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 4 \, {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, b^{2} - 2 \, {\left (5 \, a b \cos \left (d x + c\right )^{2} - 2 \, a b\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.28, size = 130, normalized size = 1.12 \[ \frac {{\left (3 \, a b - 4 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (3 \, a b + 4 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (3 \, b^{2} \sin \left (d x + c\right )^{4} + 5 \, a b \sin \left (d x + c\right )^{3} + 2 \, a^{2} \sin \left (d x + c\right )^{2} - 2 \, b^{2} \sin \left (d x + c\right )^{2} - 3 \, a b \sin \left (d x + c\right ) - a^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((3*a*b - 4*b^2)*log(abs(sin(d*x + c) + 1)) - (3*a*b + 4*b^2)*log(abs(sin(d*x + c) - 1)) + 2*(3*b^2*sin(d*
x + c)^4 + 5*a*b*sin(d*x + c)^3 + 2*a^2*sin(d*x + c)^2 - 2*b^2*sin(d*x + c)^2 - 3*a*b*sin(d*x + c) - a^2)/(sin
(d*x + c)^2 - 1)^2)/d

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maple [A]  time = 0.30, size = 168, normalized size = 1.45 \[ \frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {a b \left (\sin ^{5}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{4}}-\frac {a b \left (\sin ^{5}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{2}}-\frac {a b \left (\sin ^{3}\left (d x +c \right )\right )}{4 d}-\frac {3 a b \sin \left (d x +c \right )}{4 d}+\frac {3 a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {b^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {b^{2} \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 123, normalized size = 1.06 \[ \frac {{\left (3 \, a b - 4 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a b + 4 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (5 \, a b \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right ) + 2 \, {\left (a^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} - 3 \, b^{2}\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 12.09, size = 247, normalized size = 2.13 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {3\,a\,b}{4}-b^2\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (b^2+\frac {3\,a\,b}{4}\right )}{d}+\frac {b^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {2\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,a^2+8\,b^2\right )+2\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {11\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {11\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}+\frac {3\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+\frac {3\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(tan(c/2 + (d*x)/2) + 1)*((3*a*b)/4 - b^2))/d - (log(tan(c/2 + (d*x)/2) - 1)*((3*a*b)/4 + b^2))/d + (b^2*l
og(tan(c/2 + (d*x)/2)^2 + 1))/d - (2*b^2*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^4*(4*a^2 + 8*b^2) + 2*b^2*t
an(c/2 + (d*x)/2)^6 - (11*a*b*tan(c/2 + (d*x)/2)^3)/2 - (11*a*b*tan(c/2 + (d*x)/2)^5)/2 + (3*a*b*tan(c/2 + (d*
x)/2)^7)/2 + (3*a*b*tan(c/2 + (d*x)/2))/2)/(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))

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

[Out]

Timed out

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