3.192 \(\int \frac{\tan ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=321 \[ -\frac{a^5}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}-\frac{a^4 \left (a^2+5 b^2\right )}{d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}+\frac{a^3 \left (13 a^2 b^2+a^4+10 b^4\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^5}-\frac{\left (8 a^2-5 a b-b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^5}-\frac{\left (8 a^2+5 a b-b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^5}+\frac{\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d \left (a^2-b^2\right )^3}-\frac{\sec ^2(c+d x) \left (8 a^3 \left (a^2+5 b^2\right )-b \left (22 a^2 b^2+27 a^4-b^4\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^4} \]

[Out]

-((8*a^2 - 5*a*b - b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^5*d) - ((8*a^2 + 5*a*b - b^2)*Log[1 + Sin[c + d*x]]
)/(16*(a - b)^5*d) + (a^3*(a^4 + 13*a^2*b^2 + 10*b^4)*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^5*d) - a^5/(2*(a^2
- b^2)^3*d*(a + b*Sin[c + d*x])^2) - (a^4*(a^2 + 5*b^2))/((a^2 - b^2)^4*d*(a + b*Sin[c + d*x])) + (Sec[c + d*
x]^4*(a*(a^2 + 3*b^2) - b*(3*a^2 + b^2)*Sin[c + d*x]))/(4*(a^2 - b^2)^3*d) - (Sec[c + d*x]^2*(8*a^3*(a^2 + 5*b
^2) - b*(27*a^4 + 22*a^2*b^2 - b^4)*Sin[c + d*x]))/(8*(a^2 - b^2)^4*d)

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Rubi [A]  time = 0.877422, antiderivative size = 321, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2721, 1647, 1629} \[ -\frac{a^5}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}-\frac{a^4 \left (a^2+5 b^2\right )}{d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}+\frac{a^3 \left (13 a^2 b^2+a^4+10 b^4\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^5}-\frac{\left (8 a^2-5 a b-b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^5}-\frac{\left (8 a^2+5 a b-b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^5}+\frac{\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d \left (a^2-b^2\right )^3}-\frac{\sec ^2(c+d x) \left (8 a^3 \left (a^2+5 b^2\right )-b \left (22 a^2 b^2+27 a^4-b^4\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^4} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(a + b*Sin[c + d*x])^3,x]

[Out]

-((8*a^2 - 5*a*b - b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^5*d) - ((8*a^2 + 5*a*b - b^2)*Log[1 + Sin[c + d*x]]
)/(16*(a - b)^5*d) + (a^3*(a^4 + 13*a^2*b^2 + 10*b^4)*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^5*d) - a^5/(2*(a^2
- b^2)^3*d*(a + b*Sin[c + d*x])^2) - (a^4*(a^2 + 5*b^2))/((a^2 - b^2)^4*d*(a + b*Sin[c + d*x])) + (Sec[c + d*
x]^4*(a*(a^2 + 3*b^2) - b*(3*a^2 + b^2)*Sin[c + d*x]))/(4*(a^2 - b^2)^3*d) - (Sec[c + d*x]^2*(8*a^3*(a^2 + 5*b
^2) - b*(27*a^4 + 22*a^2*b^2 - b^4)*Sin[c + d*x]))/(8*(a^2 - b^2)^4*d)

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 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]

Rubi steps

\begin{align*} \int \frac{\tan ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5}{(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) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^3 d}+\frac{\operatorname{Subst}\left (\int \frac{\frac{a^3 b^6 \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^3}-\frac{a^2 b^4 \left (4 a^4+3 a^2 b^2-3 b^4\right ) x}{\left (a^2-b^2\right )^3}-\frac{a b^6 \left (23 a^2-3 b^2\right ) x^2}{\left (a^2-b^2\right )^3}-\frac{b^2 \left (4 a^6-12 a^4 b^2+21 a^2 b^4-b^6\right ) x^3}{\left (a^2-b^2\right )^3}}{(a+x)^3 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b^2 d}\\ &=\frac{\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^3 d}-\frac{\sec ^2(c+d x) \left (8 a^3 \left (a^2+5 b^2\right )-b \left (27 a^4+22 a^2 b^2-b^4\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^4 d}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{a^3 b^6 \left (21 a^4+26 a^2 b^2+b^4\right )}{\left (a^2-b^2\right )^4}+\frac{a^2 b^4 \left (8 a^6+a^4 b^2-54 a^2 b^4-3 b^6\right ) x}{\left (a^2-b^2\right )^4}+\frac{a b^6 \left (65 a^4-14 a^2 b^2-3 b^4\right ) x^2}{\left (a^2-b^2\right )^4}+\frac{b^6 \left (27 a^4+22 a^2 b^2-b^4\right ) x^3}{\left (a^2-b^2\right )^4}}{(a+x)^3 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 b^4 d}\\ &=\frac{\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^3 d}-\frac{\sec ^2(c+d x) \left (8 a^3 \left (a^2+5 b^2\right )-b \left (27 a^4+22 a^2 b^2-b^4\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^4 d}+\frac{\operatorname{Subst}\left (\int \left (-\frac{b^4 \left (-8 a^2+5 a b+b^2\right )}{2 (a+b)^5 (b-x)}+\frac{8 a^5 b^4}{\left (a^2-b^2\right )^3 (a+x)^3}+\frac{8 a^4 b^4 \left (a^2+5 b^2\right )}{\left (a^2-b^2\right )^4 (a+x)^2}+\frac{8 a^3 b^4 \left (a^4+13 a^2 b^2+10 b^4\right )}{\left (a^2-b^2\right )^5 (a+x)}-\frac{b^4 \left (8 a^2+5 a b-b^2\right )}{2 (a-b)^5 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^4 d}\\ &=-\frac{\left (8 a^2-5 a b-b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^5 d}-\frac{\left (8 a^2+5 a b-b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^5 d}+\frac{a^3 \left (a^4+13 a^2 b^2+10 b^4\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^5 d}-\frac{a^5}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}-\frac{a^4 \left (a^2+5 b^2\right )}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac{\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^3 d}-\frac{\sec ^2(c+d x) \left (8 a^3 \left (a^2+5 b^2\right )-b \left (27 a^4+22 a^2 b^2-b^4\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^4 d}\\ \end{align*}

Mathematica [A]  time = 6.34144, size = 304, normalized size = 0.95 \[ -\frac{a^5}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}-\frac{a^4 \left (a^2+5 b^2\right )}{d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}+\frac{a^3 \left (13 a^2 b^2+a^4+10 b^4\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^5}-\frac{\left (8 a^2-5 a b-b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^5}-\frac{\left (8 a^2+5 a b-b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^5}-\frac{7 a+b}{16 d (a+b)^4 (1-\sin (c+d x))}-\frac{7 a-b}{16 d (a-b)^4 (\sin (c+d x)+1)}+\frac{1}{16 d (a+b)^3 (1-\sin (c+d x))^2}+\frac{1}{16 d (a-b)^3 (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/(a + b*Sin[c + d*x])^3,x]

[Out]

-((8*a^2 - 5*a*b - b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^5*d) - ((8*a^2 + 5*a*b - b^2)*Log[1 + Sin[c + d*x]]
)/(16*(a - b)^5*d) + (a^3*(a^4 + 13*a^2*b^2 + 10*b^4)*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^5*d) + 1/(16*(a +
b)^3*d*(1 - Sin[c + d*x])^2) - (7*a + b)/(16*(a + b)^4*d*(1 - Sin[c + d*x])) + 1/(16*(a - b)^3*d*(1 + Sin[c +
d*x])^2) - (7*a - b)/(16*(a - b)^4*d*(1 + Sin[c + d*x])) - a^5/(2*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])^2) - (a
^4*(a^2 + 5*b^2))/((a^2 - b^2)^4*d*(a + b*Sin[c + d*x]))

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Maple [A]  time = 0.116, size = 465, normalized size = 1.5 \begin{align*} -{\frac{{a}^{5}}{2\,d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{7}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{5} \left ( a-b \right ) ^{5}}}+13\,{\frac{{a}^{5}\ln \left ( a+b\sin \left ( dx+c \right ) \right ){b}^{2}}{d \left ( a+b \right ) ^{5} \left ( a-b \right ) ^{5}}}+10\,{\frac{{a}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ){b}^{4}}{d \left ( a+b \right ) ^{5} \left ( a-b \right ) ^{5}}}-{\frac{{a}^{6}}{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4} \left ( a+b\sin \left ( dx+c \right ) \right ) }}-5\,{\frac{{a}^{4}{b}^{2}}{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4} \left ( a+b\sin \left ( dx+c \right ) \right ) }}+{\frac{1}{16\,d \left ( a+b \right ) ^{3} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{b}{16\,d \left ( a+b \right ) ^{4} \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{7\,a}{16\,d \left ( a+b \right ) ^{4} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ){a}^{2}}{2\,d \left ( a+b \right ) ^{5}}}+{\frac{5\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) ab}{16\,d \left ( a+b \right ) ^{5}}}+{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ){b}^{2}}{16\,d \left ( a+b \right ) ^{5}}}+{\frac{1}{16\,d \left ( a-b \right ) ^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{b}{16\,d \left ( a-b \right ) ^{4} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{7\,a}{16\,d \left ( a-b \right ) ^{4} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ){a}^{2}}{2\,d \left ( a-b \right ) ^{5}}}-{\frac{5\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) ab}{16\,d \left ( a-b \right ) ^{5}}}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ){b}^{2}}{16\,d \left ( a-b \right ) ^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*a^5/(a+b)^3/(a-b)^3/(a+b*sin(d*x+c))^2+1/d*a^7/(a+b)^5/(a-b)^5*ln(a+b*sin(d*x+c))+13/d*a^5/(a+b)^5/(a-b
)^5*ln(a+b*sin(d*x+c))*b^2+10/d*a^3/(a+b)^5/(a-b)^5*ln(a+b*sin(d*x+c))*b^4-1/d*a^6/(a+b)^4/(a-b)^4/(a+b*sin(d*
x+c))-5/d*a^4/(a+b)^4/(a-b)^4/(a+b*sin(d*x+c))*b^2+1/16/d/(a+b)^3/(sin(d*x+c)-1)^2+1/16/d/(a+b)^4/(sin(d*x+c)-
1)*b+7/16/d/(a+b)^4/(sin(d*x+c)-1)*a-1/2/d/(a+b)^5*ln(sin(d*x+c)-1)*a^2+5/16/d/(a+b)^5*ln(sin(d*x+c)-1)*a*b+1/
16/d/(a+b)^5*ln(sin(d*x+c)-1)*b^2+1/16/d/(a-b)^3/(1+sin(d*x+c))^2+1/16/d/(a-b)^4/(1+sin(d*x+c))*b-7/16/d/(a-b)
^4/(1+sin(d*x+c))*a-1/2/d/(a-b)^5*ln(1+sin(d*x+c))*a^2-5/16/d/(a-b)^5*ln(1+sin(d*x+c))*a*b+1/16/d/(a-b)^5*ln(1
+sin(d*x+c))*b^2

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Maxima [B]  time = 1.30265, size = 986, normalized size = 3.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*(a^7 + 13*a^5*b^2 + 10*a^3*b^4)*log(b*sin(d*x + c) + a)/(a^10 - 5*a^8*b^2 + 10*a^6*b^4 - 10*a^4*b^6 +
5*a^2*b^8 - b^10) - (8*a^2 + 5*a*b - b^2)*log(sin(d*x + c) + 1)/(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*
a*b^4 - b^5) - (8*a^2 - 5*a*b - b^2)*log(sin(d*x + c) - 1)/(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4
+ b^5) - 2*(18*a^7 + 72*a^5*b^2 + 6*a^3*b^4 + (8*a^6*b + 67*a^4*b^3 + 22*a^2*b^5 - b^7)*sin(d*x + c)^5 + 2*(6*
a^7 + 41*a^5*b^2 + 2*a^3*b^4 - a*b^6)*sin(d*x + c)^4 - (5*a^6*b + 159*a^4*b^3 + 27*a^2*b^5 + b^7)*sin(d*x + c)
^3 - 4*(8*a^7 + 37*a^5*b^2 + 4*a^3*b^4 - a*b^6)*sin(d*x + c)^2 - (a^6*b - 86*a^4*b^3 - 11*a^2*b^5)*sin(d*x + c
))/(a^10 - 4*a^8*b^2 + 6*a^6*b^4 - 4*a^4*b^6 + a^2*b^8 + (a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*
sin(d*x + c)^6 + 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*sin(d*x + c)^5 + (a^10 - 6*a^8*b^2 + 14
*a^6*b^4 - 16*a^4*b^6 + 9*a^2*b^8 - 2*b^10)*sin(d*x + c)^4 - 4*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*
b^9)*sin(d*x + c)^3 - (2*a^10 - 9*a^8*b^2 + 16*a^6*b^4 - 14*a^4*b^6 + 6*a^2*b^8 - b^10)*sin(d*x + c)^2 + 2*(a^
9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*sin(d*x + c)))/d

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Fricas [B]  time = 6.42709, size = 2206, normalized size = 6.87 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4*a^9 - 16*a^7*b^2 + 24*a^5*b^4 - 16*a^3*b^6 + 4*a*b^8 - 4*(6*a^9 + 35*a^7*b^2 - 39*a^5*b^4 - 3*a^3*b^6
+ a*b^8)*cos(d*x + c)^4 - 16*(a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*cos(d*x + c)^2 - 16*((a^7*b^2 + 13*a^5*b
^4 + 10*a^3*b^6)*cos(d*x + c)^6 - 2*(a^8*b + 13*a^6*b^3 + 10*a^4*b^5)*cos(d*x + c)^4*sin(d*x + c) - (a^9 + 14*
a^7*b^2 + 23*a^5*b^4 + 10*a^3*b^6)*cos(d*x + c)^4)*log(b*sin(d*x + c) + a) + ((8*a^7*b^2 + 45*a^6*b^3 + 104*a^
5*b^4 + 125*a^4*b^5 + 80*a^3*b^6 + 23*a^2*b^7 - b^9)*cos(d*x + c)^6 - 2*(8*a^8*b + 45*a^7*b^2 + 104*a^6*b^3 +
125*a^5*b^4 + 80*a^4*b^5 + 23*a^3*b^6 - a*b^8)*cos(d*x + c)^4*sin(d*x + c) - (8*a^9 + 45*a^8*b + 112*a^7*b^2 +
170*a^6*b^3 + 184*a^5*b^4 + 148*a^4*b^5 + 80*a^3*b^6 + 22*a^2*b^7 - b^9)*cos(d*x + c)^4)*log(sin(d*x + c) + 1
) + ((8*a^7*b^2 - 45*a^6*b^3 + 104*a^5*b^4 - 125*a^4*b^5 + 80*a^3*b^6 - 23*a^2*b^7 + b^9)*cos(d*x + c)^6 - 2*(
8*a^8*b - 45*a^7*b^2 + 104*a^6*b^3 - 125*a^5*b^4 + 80*a^4*b^5 - 23*a^3*b^6 + a*b^8)*cos(d*x + c)^4*sin(d*x + c
) - (8*a^9 - 45*a^8*b + 112*a^7*b^2 - 170*a^6*b^3 + 184*a^5*b^4 - 148*a^4*b^5 + 80*a^3*b^6 - 22*a^2*b^7 + b^9)
*cos(d*x + c)^4)*log(-sin(d*x + c) + 1) - 2*(2*a^8*b - 8*a^6*b^3 + 12*a^4*b^5 - 8*a^2*b^7 + 2*b^9 + (8*a^8*b +
59*a^6*b^3 - 45*a^4*b^5 - 23*a^2*b^7 + b^9)*cos(d*x + c)^4 - (11*a^8*b - 36*a^6*b^3 + 42*a^4*b^5 - 20*a^2*b^7
+ 3*b^9)*cos(d*x + c)^2)*sin(d*x + c))/((a^10*b^2 - 5*a^8*b^4 + 10*a^6*b^6 - 10*a^4*b^8 + 5*a^2*b^10 - b^12)*
d*cos(d*x + c)^6 - 2*(a^11*b - 5*a^9*b^3 + 10*a^7*b^5 - 10*a^5*b^7 + 5*a^3*b^9 - a*b^11)*d*cos(d*x + c)^4*sin(
d*x + c) - (a^12 - 4*a^10*b^2 + 5*a^8*b^4 - 5*a^4*b^8 + 4*a^2*b^10 - b^12)*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(tan(d*x+c)**5/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 3.70755, size = 790, normalized size = 2.46 \begin{align*} \frac{\frac{16 \,{\left (a^{7} b + 13 \, a^{5} b^{3} + 10 \, a^{3} b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{10} b - 5 \, a^{8} b^{3} + 10 \, a^{6} b^{5} - 10 \, a^{4} b^{7} + 5 \, a^{2} b^{9} - b^{11}} - \frac{{\left (8 \, a^{2} + 5 \, a b - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}} - \frac{{\left (8 \, a^{2} - 5 \, a b - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}} - \frac{2 \,{\left (8 \, a^{6} b \sin \left (d x + c\right )^{5} + 67 \, a^{4} b^{3} \sin \left (d x + c\right )^{5} + 22 \, a^{2} b^{5} \sin \left (d x + c\right )^{5} - b^{7} \sin \left (d x + c\right )^{5} + 12 \, a^{7} \sin \left (d x + c\right )^{4} + 82 \, a^{5} b^{2} \sin \left (d x + c\right )^{4} + 4 \, a^{3} b^{4} \sin \left (d x + c\right )^{4} - 2 \, a b^{6} \sin \left (d x + c\right )^{4} - 5 \, a^{6} b \sin \left (d x + c\right )^{3} - 159 \, a^{4} b^{3} \sin \left (d x + c\right )^{3} - 27 \, a^{2} b^{5} \sin \left (d x + c\right )^{3} - b^{7} \sin \left (d x + c\right )^{3} - 32 \, a^{7} \sin \left (d x + c\right )^{2} - 148 \, a^{5} b^{2} \sin \left (d x + c\right )^{2} - 16 \, a^{3} b^{4} \sin \left (d x + c\right )^{2} + 4 \, a b^{6} \sin \left (d x + c\right )^{2} - a^{6} b \sin \left (d x + c\right ) + 86 \, a^{4} b^{3} \sin \left (d x + c\right ) + 11 \, a^{2} b^{5} \sin \left (d x + c\right ) + 18 \, a^{7} + 72 \, a^{5} b^{2} + 6 \, a^{3} b^{4}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )}{\left (b \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right ) - a\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*(a^7*b + 13*a^5*b^3 + 10*a^3*b^5)*log(abs(b*sin(d*x + c) + a))/(a^10*b - 5*a^8*b^3 + 10*a^6*b^5 - 10*
a^4*b^7 + 5*a^2*b^9 - b^11) - (8*a^2 + 5*a*b - b^2)*log(abs(sin(d*x + c) + 1))/(a^5 - 5*a^4*b + 10*a^3*b^2 - 1
0*a^2*b^3 + 5*a*b^4 - b^5) - (8*a^2 - 5*a*b - b^2)*log(abs(sin(d*x + c) - 1))/(a^5 + 5*a^4*b + 10*a^3*b^2 + 10
*a^2*b^3 + 5*a*b^4 + b^5) - 2*(8*a^6*b*sin(d*x + c)^5 + 67*a^4*b^3*sin(d*x + c)^5 + 22*a^2*b^5*sin(d*x + c)^5
- b^7*sin(d*x + c)^5 + 12*a^7*sin(d*x + c)^4 + 82*a^5*b^2*sin(d*x + c)^4 + 4*a^3*b^4*sin(d*x + c)^4 - 2*a*b^6*
sin(d*x + c)^4 - 5*a^6*b*sin(d*x + c)^3 - 159*a^4*b^3*sin(d*x + c)^3 - 27*a^2*b^5*sin(d*x + c)^3 - b^7*sin(d*x
+ c)^3 - 32*a^7*sin(d*x + c)^2 - 148*a^5*b^2*sin(d*x + c)^2 - 16*a^3*b^4*sin(d*x + c)^2 + 4*a*b^6*sin(d*x + c
)^2 - a^6*b*sin(d*x + c) + 86*a^4*b^3*sin(d*x + c) + 11*a^2*b^5*sin(d*x + c) + 18*a^7 + 72*a^5*b^2 + 6*a^3*b^4
)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(b*sin(d*x + c)^3 + a*sin(d*x + c)^2 - b*sin(d*x + c) - a)^
2))/d