3.1502 \(\int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=177 \[ -\frac{3 (a+b) \left (a^2+7 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac{3 (a-b) \left (a^2-7 a b+8 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac{29 a b^2 \sin (c+d x)}{8 d}-\frac{\sec ^2(c+d x) (5 a \sin (c+d x)+8 b) (a+b \sin (c+d x))^2}{8 d}+\frac{\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac{b^3 \sin ^2(c+d x)}{2 d} \]

[Out]

(-3*(a + b)*(a^2 + 7*a*b + 8*b^2)*Log[1 - Sin[c + d*x]])/(16*d) + (3*(a - b)*(a^2 - 7*a*b + 8*b^2)*Log[1 + Sin
[c + d*x]])/(16*d) - (29*a*b^2*Sin[c + d*x])/(8*d) - (b^3*Sin[c + d*x]^2)/(2*d) - (Sec[c + d*x]^2*(8*b + 5*a*S
in[c + d*x])*(a + b*Sin[c + d*x])^2)/(8*d) + (Sec[c + d*x]^3*(a + b*Sin[c + d*x])^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.367833, antiderivative size = 177, normalized size of antiderivative = 1., 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, 1629, 633, 31} \[ -\frac{3 (a+b) \left (a^2+7 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac{3 (a-b) \left (a^2-7 a b+8 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac{29 a b^2 \sin (c+d x)}{8 d}-\frac{\sec ^2(c+d x) (5 a \sin (c+d x)+8 b) (a+b \sin (c+d x))^2}{8 d}+\frac{\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac{b^3 \sin ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(a + b)*(a^2 + 7*a*b + 8*b^2)*Log[1 - Sin[c + d*x]])/(16*d) + (3*(a - b)*(a^2 - 7*a*b + 8*b^2)*Log[1 + Sin
[c + d*x]])/(16*d) - (29*a*b^2*Sin[c + d*x])/(8*d) - (b^3*Sin[c + d*x]^2)/(2*d) - (Sec[c + d*x]^2*(8*b + 5*a*S
in[c + d*x])*(a + b*Sin[c + d*x])^2)/(8*d) + (Sec[c + d*x]^3*(a + b*Sin[c + d*x])^3*Tan[c + d*x])/(4*d)

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]

Rule 12

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

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 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, 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 \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{x^4 (a+x)^3}{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)^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{\operatorname{Subst}\left (\int \frac{(a+x)^2 \left (-a b^4-4 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) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{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+x) \left (b^4 \left (3 a^2+16 b^2\right )+21 a b^4 x+8 b^4 x^2\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^3 d}\\ &=-\frac{\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{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 \left (-29 a b^4-8 b^4 x+\frac{3 \left (a b^4 \left (a^2+15 b^2\right )+8 b^4 \left (a^2+b^2\right ) x\right )}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^3 d}\\ &=-\frac{29 a b^2 \sin (c+d x)}{8 d}-\frac{b^3 \sin ^2(c+d x)}{2 d}-\frac{\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}+\frac{3 \operatorname{Subst}\left (\int \frac{a b^4 \left (a^2+15 b^2\right )+8 b^4 \left (a^2+b^2\right ) x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^3 d}\\ &=-\frac{29 a b^2 \sin (c+d x)}{8 d}-\frac{b^3 \sin ^2(c+d x)}{2 d}-\frac{\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}-\frac{\left (3 (a-b) \left (a^2-7 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac{\left (3 (a+b) \left (a^2+7 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=-\frac{3 (a+b) \left (a^2+7 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac{3 (a-b) \left (a^2-7 a b+8 b^2\right ) \log (1+\sin (c+d x))}{16 d}-\frac{29 a b^2 \sin (c+d x)}{8 d}-\frac{b^3 \sin ^2(c+d x)}{2 d}-\frac{\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{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 = 0.575711, size = 174, normalized size = 0.98 \[ \frac{3 \left (a^2-7 a b+8 b^2\right ) (a-b) \log (\sin (c+d x)+1)-3 (a+b) \left (a^2+7 a b+8 b^2\right ) \log (1-\sin (c+d x))-48 a b^2 \sin (c+d x)-\frac{(a-b)^3}{(\sin (c+d x)+1)^2}+\frac{(5 a-11 b) (a-b)^2}{\sin (c+d x)+1}+\frac{(a+b)^2 (5 a+11 b)}{\sin (c+d x)-1}+\frac{(a+b)^3}{(\sin (c+d x)-1)^2}-8 b^3 \sin ^2(c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(a + b)*(a^2 + 7*a*b + 8*b^2)*Log[1 - Sin[c + d*x]] + 3*(a - b)*(a^2 - 7*a*b + 8*b^2)*Log[1 + Sin[c + d*x]
] + (a + b)^3/(-1 + Sin[c + d*x])^2 + ((a + b)^2*(5*a + 11*b))/(-1 + Sin[c + d*x]) - 48*a*b^2*Sin[c + d*x] - 8
*b^3*Sin[c + d*x]^2 - (a - b)^3/(1 + Sin[c + d*x])^2 + ((5*a - 11*b)*(a - b)^2)/(1 + Sin[c + d*x]))/(16*d)

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Maple [B]  time = 0.086, size = 385, normalized size = 2.2 \begin{align*}{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{3\,{a}^{3}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{3\,{a}^{2}b \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{3\,{a}^{2}b \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-3\,{\frac{{a}^{2}b\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{9\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{9\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d}}-{\frac{15\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{45\,a{b}^{2}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{45\,a{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{2\,d}}-{\frac{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}{b}^{3}}{4\,d}}-{\frac{3\,{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-3\,{\frac{{b}^{3}\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)^4*(a+b*sin(d*x+c))^3,x)

[Out]

1/4/d*a^3*sin(d*x+c)^5/cos(d*x+c)^4-1/8/d*a^3*sin(d*x+c)^5/cos(d*x+c)^2-1/8*a^3*sin(d*x+c)^3/d-3/8*a^3*sin(d*x
+c)/d+3/8/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/4/d*a^2*b*tan(d*x+c)^4-3/2/d*a^2*b*tan(d*x+c)^2-3/d*a^2*b*ln(cos(d
*x+c))+3/4/d*a*b^2*sin(d*x+c)^7/cos(d*x+c)^4-9/8/d*a*b^2*sin(d*x+c)^7/cos(d*x+c)^2-9/8/d*a*b^2*sin(d*x+c)^5-15
/8/d*a*b^2*sin(d*x+c)^3-45/8*a*b^2*sin(d*x+c)/d+45/8/d*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*b^3*sin(d*x+c)^8/
cos(d*x+c)^4-1/2/d*b^3*sin(d*x+c)^8/cos(d*x+c)^2-1/2/d*b^3*sin(d*x+c)^6-3/4/d*sin(d*x+c)^4*b^3-3/2*b^3*sin(d*x
+c)^2/d-3/d*b^3*ln(cos(d*x+c))

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Maxima [A]  time = 0.998339, size = 257, normalized size = 1.45 \begin{align*} -\frac{8 \, b^{3} \sin \left (d x + c\right )^{2} + 48 \, a b^{2} \sin \left (d x + c\right ) - 3 \,{\left (a^{3} - 8 \, a^{2} b + 15 \, a b^{2} - 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (a^{3} + 8 \, a^{2} b + 15 \, a b^{2} + 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left ({\left (5 \, a^{3} + 27 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} - 18 \, a^{2} b - 10 \, b^{3} + 12 \,{\left (2 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )^{2} - 3 \,{\left (a^{3} + 7 \, 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} \end{align*}

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))^3,x, algorithm="maxima")

[Out]

-1/16*(8*b^3*sin(d*x + c)^2 + 48*a*b^2*sin(d*x + c) - 3*(a^3 - 8*a^2*b + 15*a*b^2 - 8*b^3)*log(sin(d*x + c) +
1) + 3*(a^3 + 8*a^2*b + 15*a*b^2 + 8*b^3)*log(sin(d*x + c) - 1) - 2*((5*a^3 + 27*a*b^2)*sin(d*x + c)^3 - 18*a^
2*b - 10*b^3 + 12*(2*a^2*b + b^3)*sin(d*x + c)^2 - 3*(a^3 + 7*a*b^2)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x
+ c)^2 + 1))/d

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Fricas [A]  time = 2.23466, size = 508, normalized size = 2.87 \begin{align*} \frac{8 \, b^{3} \cos \left (d x + c\right )^{6} - 4 \, b^{3} \cos \left (d x + c\right )^{4} + 3 \,{\left (a^{3} - 8 \, a^{2} b + 15 \, a b^{2} - 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (a^{3} + 8 \, a^{2} b + 15 \, 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} - 24 \,{\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (24 \, a b^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{3} - 6 \, a b^{2} +{\left (5 \, a^{3} + 27 \, 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}} \end{align*}

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))^3,x, algorithm="fricas")

[Out]

1/16*(8*b^3*cos(d*x + c)^6 - 4*b^3*cos(d*x + c)^4 + 3*(a^3 - 8*a^2*b + 15*a*b^2 - 8*b^3)*cos(d*x + c)^4*log(si
n(d*x + c) + 1) - 3*(a^3 + 8*a^2*b + 15*a*b^2 + 8*b^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 12*a^2*b + 4*b^
3 - 24*(2*a^2*b + b^3)*cos(d*x + c)^2 - 2*(24*a*b^2*cos(d*x + c)^4 - 2*a^3 - 6*a*b^2 + (5*a^3 + 27*a*b^2)*cos(
d*x + c)^2)*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)**4*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.28398, size = 298, normalized size = 1.68 \begin{align*} -\frac{8 \, b^{3} \sin \left (d x + c\right )^{2} + 48 \, a b^{2} \sin \left (d x + c\right ) - 3 \,{\left (a^{3} - 8 \, a^{2} b + 15 \, a b^{2} - 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \,{\left (a^{3} + 8 \, a^{2} b + 15 \, a b^{2} + 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (18 \, a^{2} b \sin \left (d x + c\right )^{4} + 18 \, b^{3} \sin \left (d x + c\right )^{4} + 5 \, a^{3} \sin \left (d x + c\right )^{3} + 27 \, a b^{2} \sin \left (d x + c\right )^{3} - 12 \, a^{2} b \sin \left (d x + c\right )^{2} - 24 \, b^{3} \sin \left (d x + c\right )^{2} - 3 \, a^{3} \sin \left (d x + c\right ) - 21 \, a b^{2} \sin \left (d x + c\right ) + 8 \, b^{3}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}

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))^3,x, algorithm="giac")

[Out]

-1/16*(8*b^3*sin(d*x + c)^2 + 48*a*b^2*sin(d*x + c) - 3*(a^3 - 8*a^2*b + 15*a*b^2 - 8*b^3)*log(abs(sin(d*x + c
) + 1)) + 3*(a^3 + 8*a^2*b + 15*a*b^2 + 8*b^3)*log(abs(sin(d*x + c) - 1)) - 2*(18*a^2*b*sin(d*x + c)^4 + 18*b^
3*sin(d*x + c)^4 + 5*a^3*sin(d*x + c)^3 + 27*a*b^2*sin(d*x + c)^3 - 12*a^2*b*sin(d*x + c)^2 - 24*b^3*sin(d*x +
c)^2 - 3*a^3*sin(d*x + c) - 21*a*b^2*sin(d*x + c) + 8*b^3)/(sin(d*x + c)^2 - 1)^2)/d