3.2.93 \(\int \frac {\tan ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [193]

Optimal. Leaf size=232 \[ \frac {(2 a-b) \log (1-\sin (c+d x))}{4 (a+b)^4 d}+\frac {(2 a+b) \log (1+\sin (c+d x))}{4 (a-b)^4 d}-\frac {a \left (a^4+8 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4 d}+\frac {a^3}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac {a^2 \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {\sec ^2(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d} \]

[Out]

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

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Rubi [A]
time = 0.40, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2800, 1661, 1643} \begin {gather*} \frac {a^2 \left (a^2+3 b^2\right )}{d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {\sec ^2(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 )}{2 d \left (a^2-b^2\right )^3}-\frac {a \left (a^4+8 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^4}+\frac {a^3}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {(2 a-b) \log (1-\sin (c+d x))}{4 d (a+b)^4}+\frac {(2 a+b) \log (\sin (c+d x)+1)}{4 d (a-b)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1643

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 1661

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 2800

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]

Rubi steps

\begin {align*} \int \frac {\tan ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\frac {\text {Subst}\left (\int \frac {x^3}{(a+x)^3 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\sec ^2(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 )}{2 \left (a^2-b^2\right )^3 d}+\frac {\text {Subst}\left (\int \frac {\frac {a^3 b^4 \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^3}-\frac {a^2 b^2 \left (2 a^4-3 a^2 b^2-3 b^4\right ) x}{\left (a^2-b^2\right )^3}-\frac {a b^4 \left (7 a^2-3 b^2\right ) x^2}{\left (a^2-b^2\right )^3}-\frac {b^4 \left (3 a^2+b^2\right ) x^3}{\left (a^2-b^2\right )^3}}{(a+x)^3 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac {\sec ^2(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 )}{2 \left (a^2-b^2\right )^3 d}+\frac {\text {Subst}\left (\int \left (\frac {b^2 (-2 a+b)}{2 (a+b)^4 (b-x)}-\frac {2 a^3 b^2}{\left (a^2-b^2\right )^2 (a+x)^3}-\frac {2 a^2 b^2 \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3 (a+x)^2}-\frac {2 a b^2 \left (a^4+8 a^2 b^2+3 b^4\right )}{\left (a^2-b^2\right )^4 (a+x)}+\frac {b^2 (2 a+b)}{2 (a-b)^4 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac {(2 a-b) \log (1-\sin (c+d x))}{4 (a+b)^4 d}+\frac {(2 a+b) \log (1+\sin (c+d x))}{4 (a-b)^4 d}-\frac {a \left (a^4+8 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4 d}+\frac {a^3}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac {a^2 \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {\sec ^2(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 )}{2 \left (a^2-b^2\right )^3 d}\\ \end {align*}

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Mathematica [A]
time = 1.47, size = 196, normalized size = 0.84 \begin {gather*} \frac {\frac {(2 a-b) \log (1-\sin (c+d x))}{(a+b)^4}+\frac {(2 a+b) \log (1+\sin (c+d x))}{(a-b)^4}-\frac {4 a \left (a^4+8 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4}-\frac {1}{(a+b)^3 (-1+\sin (c+d x))}+\frac {1}{(a-b)^3 (1+\sin (c+d x))}+\frac {2 a^3}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {4 a^2 \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*a - b)*Log[1 - Sin[c + d*x]])/(a + b)^4 + ((2*a + b)*Log[1 + Sin[c + d*x]])/(a - b)^4 - (4*a*(a^4 + 8*a^2
*b^2 + 3*b^4)*Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^4 - 1/((a + b)^3*(-1 + Sin[c + d*x])) + 1/((a - b)^3*(1 + S
in[c + d*x])) + (2*a^3)/((a^2 - b^2)^2*(a + b*Sin[c + d*x])^2) + (4*a^2*(a^2 + 3*b^2))/((a^2 - b^2)^3*(a + b*S
in[c + d*x])))/(4*d)

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Maple [A]
time = 0.62, size = 197, normalized size = 0.85

method	result	size
derivativedivides	\(\frac {-\frac {1}{4 \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (2 a -b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{4}}+\frac {1}{4 \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (2 a +b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{4}}+\frac {a^{3}}{2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {a \left (a^{4}+8 a^{2} b^{2}+3 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {a^{2} \left (a^{2}+3 b^{2}\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} \left (a +b \sin \left (d x +c \right )\right )}}{d}\)	\(197\)
default	\(\frac {-\frac {1}{4 \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (2 a -b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{4}}+\frac {1}{4 \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (2 a +b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{4}}+\frac {a^{3}}{2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {a \left (a^{4}+8 a^{2} b^{2}+3 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {a^{2} \left (a^{2}+3 b^{2}\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} \left (a +b \sin \left (d x +c \right )\right )}}{d}\)	\(197\)
risch	\(\text {Expression too large to display}\)	\(1339\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 441, normalized size = 1.90 \begin {gather*} -\frac {\frac {4 \, {\left (a^{5} + 8 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} - \frac {{\left (2 \, a + b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac {{\left (2 \, a - b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {2 \, {\left (4 \, a^{5} + 8 \, a^{3} b^{2} - {\left (2 \, a^{4} b + 9 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{3} - {\left (3 \, a^{5} + 10 \, a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{4} b + 11 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )\right )}}{a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6} - {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \sin \left (d x + c\right )^{4} - 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (d x + c\right )^{3} - {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sin \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (d x + c\right )}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*(a^5 + 8*a^3*b^2 + 3*a*b^4)*log(b*sin(d*x + c) + a)/(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8) -
(2*a + b)*log(sin(d*x + c) + 1)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - (2*a - b)*log(sin(d*x + c) - 1)/
(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - 2*(4*a^5 + 8*a^3*b^2 - (2*a^4*b + 9*a^2*b^3 + b^5)*sin(d*x + c)^
3 - (3*a^5 + 10*a^3*b^2 - a*b^4)*sin(d*x + c)^2 + (a^4*b + 11*a^2*b^3)*sin(d*x + c))/(a^8 - 3*a^6*b^2 + 3*a^4*
b^4 - a^2*b^6 - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*sin(d*x + c)^4 - 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*
b^7)*sin(d*x + c)^3 - (a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*sin(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 +
3*a^3*b^5 - a*b^7)*sin(d*x + c)))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 788 vs. \(2 (224) = 448\).
time = 0.53, size = 788, normalized size = 3.40 \begin {gather*} -\frac {2 \, a^{7} - 6 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - 2 \, a b^{6} + 2 \, {\left (3 \, a^{7} + 7 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (a^{5} b^{2} + 8 \, a^{3} b^{4} + 3 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{6} b + 8 \, a^{4} b^{3} + 3 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - {\left (a^{7} + 9 \, a^{5} b^{2} + 11 \, a^{3} b^{4} + 3 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left ({\left (2 \, a^{5} b^{2} + 9 \, a^{4} b^{3} + 16 \, a^{3} b^{4} + 14 \, a^{2} b^{5} + 6 \, a b^{6} + b^{7}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (2 \, a^{6} b + 9 \, a^{5} b^{2} + 16 \, a^{4} b^{3} + 14 \, a^{3} b^{4} + 6 \, a^{2} b^{5} + a b^{6}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - {\left (2 \, a^{7} + 9 \, a^{6} b + 18 \, a^{5} b^{2} + 23 \, a^{4} b^{3} + 22 \, a^{3} b^{4} + 15 \, a^{2} b^{5} + 6 \, a b^{6} + b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (2 \, a^{5} b^{2} - 9 \, a^{4} b^{3} + 16 \, a^{3} b^{4} - 14 \, a^{2} b^{5} + 6 \, a b^{6} - b^{7}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (2 \, a^{6} b - 9 \, a^{5} b^{2} + 16 \, a^{4} b^{3} - 14 \, a^{3} b^{4} + 6 \, a^{2} b^{5} - a b^{6}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - {\left (2 \, a^{7} - 9 \, a^{6} b + 18 \, a^{5} b^{2} - 23 \, a^{4} b^{3} + 22 \, a^{3} b^{4} - 15 \, a^{2} b^{5} + 6 \, a b^{6} - b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7} - {\left (2 \, a^{6} b + 7 \, a^{4} b^{3} - 8 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} b^{2} - 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - 4 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - {\left (a^{10} - 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} + 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 + 2*(3*a^7 + 7*a^5*b^2 - 11*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + 4*
((a^5*b^2 + 8*a^3*b^4 + 3*a*b^6)*cos(d*x + c)^4 - 2*(a^6*b + 8*a^4*b^3 + 3*a^2*b^5)*cos(d*x + c)^2*sin(d*x + c
) - (a^7 + 9*a^5*b^2 + 11*a^3*b^4 + 3*a*b^6)*cos(d*x + c)^2)*log(b*sin(d*x + c) + a) - ((2*a^5*b^2 + 9*a^4*b^3
+ 16*a^3*b^4 + 14*a^2*b^5 + 6*a*b^6 + b^7)*cos(d*x + c)^4 - 2*(2*a^6*b + 9*a^5*b^2 + 16*a^4*b^3 + 14*a^3*b^4
+ 6*a^2*b^5 + a*b^6)*cos(d*x + c)^2*sin(d*x + c) - (2*a^7 + 9*a^6*b + 18*a^5*b^2 + 23*a^4*b^3 + 22*a^3*b^4 + 1
5*a^2*b^5 + 6*a*b^6 + b^7)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - ((2*a^5*b^2 - 9*a^4*b^3 + 16*a^3*b^4 - 14*a
^2*b^5 + 6*a*b^6 - b^7)*cos(d*x + c)^4 - 2*(2*a^6*b - 9*a^5*b^2 + 16*a^4*b^3 - 14*a^3*b^4 + 6*a^2*b^5 - a*b^6)
*cos(d*x + c)^2*sin(d*x + c) - (2*a^7 - 9*a^6*b + 18*a^5*b^2 - 23*a^4*b^3 + 22*a^3*b^4 - 15*a^2*b^5 + 6*a*b^6
- b^7)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 - (2*a^6*b + 7*a^4*b^3
- 8*a^2*b^5 - b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d
*x + c)^4 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)^2*sin(d*x + c) - (a^10 - 3*a^
8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10)*d*cos(d*x + c)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3/(a+b*sin(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)**3/(a + b*sin(c + d*x))**3, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 464 vs. \(2 (224) = 448\).
time = 12.51, size = 464, normalized size = 2.00 \begin {gather*} -\frac {\frac {4 \, {\left (a^{5} b + 8 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}} - \frac {{\left (2 \, a + b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac {{\left (2 \, a - b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {2 \, {\left (a^{5} \sin \left (d x + c\right )^{2} + 8 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} + 3 \, a b^{4} \sin \left (d x + c\right )^{2} - 3 \, a^{4} b \sin \left (d x + c\right ) + 2 \, a^{2} b^{3} \sin \left (d x + c\right ) + b^{5} \sin \left (d x + c\right ) - 6 \, a^{3} b^{2} - 6 \, a 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 (\sin \left (d x + c\right )^{2} - 1\right )}} - \frac {2 \, {\left (3 \, a^{5} b^{2} \sin \left (d x + c\right )^{2} + 24 \, a^{3} b^{4} \sin \left (d x + c\right )^{2} + 9 \, a b^{6} \sin \left (d x + c\right )^{2} + 8 \, a^{6} b \sin \left (d x + c\right ) + 52 \, a^{4} b^{3} \sin \left (d x + c\right ) + 12 \, a^{2} b^{5} \sin \left (d x + c\right ) + 6 \, a^{7} + 26 \, a^{5} b^{2} + 4 \, 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 ) + a\right )}^{2}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*(a^5*b + 8*a^3*b^3 + 3*a*b^5)*log(abs(b*sin(d*x + c) + a))/(a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7
+ b^9) - (2*a + b)*log(abs(sin(d*x + c) + 1))/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - (2*a - b)*log(abs(
sin(d*x + c) - 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) + 2*(a^5*sin(d*x + c)^2 + 8*a^3*b^2*sin(d*x + c
)^2 + 3*a*b^4*sin(d*x + c)^2 - 3*a^4*b*sin(d*x + c) + 2*a^2*b^3*sin(d*x + c) + b^5*sin(d*x + c) - 6*a^3*b^2 -
6*a*b^4)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(sin(d*x + c)^2 - 1)) - 2*(3*a^5*b^2*sin(d*x + c)^2
+ 24*a^3*b^4*sin(d*x + c)^2 + 9*a*b^6*sin(d*x + c)^2 + 8*a^6*b*sin(d*x + c) + 52*a^4*b^3*sin(d*x + c) + 12*a^2
*b^5*sin(d*x + c) + 6*a^7 + 26*a^5*b^2 + 4*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) + a)^2))/d

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Mupad [B]
time = 7.46, size = 690, normalized size = 2.97 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (2\,a-b\right )}{2\,d\,{\left (a+b\right )}^4}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (-a^5+6\,a^3\,b^2+7\,a\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^5+2\,a^3\,b^2+9\,a\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (3\,a^4\,b+13\,a^2\,b^3-4\,b^5\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-a^5+6\,a^3\,b^2+7\,a\,b^4\right )}{\left (a^2-b^2\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (7\,a^4+5\,a^2\,b^2\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^4+13\,a^2\,b^2-4\,b^4\right )}{\left (a^2-b^2\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (7\,a^3\,b+5\,a\,b^3\right )}{\left (a^2-b^2\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^2+8\,b^2\right )+4\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+a^2-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (2\,a+b\right )}{2\,d\,{\left (a-b\right )}^4}-\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^5+8\,a^3\,b^2+3\,a\,b^4\right )}{d\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(tan(c/2 + (d*x)/2) - 1)*(2*a - b))/(2*d*(a + b)^4) - ((2*tan(c/2 + (d*x)/2)^6*(7*a*b^4 - a^5 + 6*a^3*b^2)
)/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (4*tan(c/2 + (d*x)/2)^4*(9*a*b^4 + a^5 + 2*a^3*b^2))/(a^6 - b^6 + 3*a^
2*b^4 - 3*a^4*b^2) - (tan(c/2 + (d*x)/2)^5*(3*a^4*b - 4*b^5 + 13*a^2*b^3))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)
+ (2*tan(c/2 + (d*x)/2)^2*(7*a*b^4 - a^5 + 6*a^3*b^2))/((a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + (b*tan(c/2 + (
d*x)/2)^7*(7*a^4 + 5*a^2*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (b*tan(c/2 + (d*x)/2)^3*(3*a^4 - 4*b^4 +
13*a^2*b^2))/((a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + (a*tan(c/2 + (d*x)/2)*(5*a*b^3 + 7*a^3*b))/((a^2 - b^2)*(
a^4 + b^4 - 2*a^2*b^2)))/(d*(a^2*tan(c/2 + (d*x)/2)^8 - tan(c/2 + (d*x)/2)^4*(2*a^2 + 8*b^2) + 4*b^2*tan(c/2 +
(d*x)/2)^2 + 4*b^2*tan(c/2 + (d*x)/2)^6 + a^2 - 4*a*b*tan(c/2 + (d*x)/2)^3 - 4*a*b*tan(c/2 + (d*x)/2)^5 + 4*a
*b*tan(c/2 + (d*x)/2)^7 + 4*a*b*tan(c/2 + (d*x)/2))) + (log(tan(c/2 + (d*x)/2) + 1)*(2*a + b))/(2*d*(a - b)^4)
- (log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2)*(3*a*b^4 + a^5 + 8*a^3*b^2))/(d*(a^8 + b^8 - 4*a^
2*b^6 + 6*a^4*b^4 - 4*a^6*b^2))

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