Optimal. Leaf size=295 \[ -\frac {a b n \left (a^2 (2-n)-b^2 (n+2)\right ) \sin ^{n+2}(c+d x) \, _2F_1\left (1,\frac {n+2}{2};\frac {n+4}{2};\sin ^2(c+d x)\right )}{2 d (n+2)}-\frac {\left (-\left (a^4 \left (n^2-4 n+3\right )\right )+6 a^2 b^2 \left (1-n^2\right )-b^4 \left (n^2+4 n+3\right )\right ) \sin ^{n+1}(c+d x) \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};\sin ^2(c+d x)\right )}{8 d (n+1)}+\frac {\sec ^4(c+d x) \sin ^{n+1}(c+d x) \left (a^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)+6 a^2 b^2+b^4\right )}{4 d}+\frac {\sec ^2(c+d x) \sin ^{n+1}(c+d x) \left (a^4 (3-n)+4 a b \left (a^2 (2-n)-b^2 (n+2)\right ) \sin (c+d x)-6 a^2 b^2 (n+1)-b^4 (n+5)\right )}{8 d} \]
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Rubi [A] time = 0.53, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2837, 1806, 808, 364} \[ -\frac {\left (6 a^2 b^2 \left (1-n^2\right )+a^4 \left (-\left (n^2-4 n+3\right )\right )-b^4 \left (n^2+4 n+3\right )\right ) \sin ^{n+1}(c+d x) \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};\sin ^2(c+d x)\right )}{8 d (n+1)}-\frac {a b n \left (a^2 (2-n)-b^2 (n+2)\right ) \sin ^{n+2}(c+d x) \, _2F_1\left (1,\frac {n+2}{2};\frac {n+4}{2};\sin ^2(c+d x)\right )}{2 d (n+2)}+\frac {\sec ^4(c+d x) \sin ^{n+1}(c+d x) \left (4 a b \left (a^2+b^2\right ) \sin (c+d x)+6 a^2 b^2+a^4+b^4\right )}{4 d}+\frac {\sec ^2(c+d x) \sin ^{n+1}(c+d x) \left (4 a b \left (a^2 (2-n)-b^2 (n+2)\right ) \sin (c+d x)-6 a^2 b^2 (n+1)+a^4 (3-n)-b^4 (n+5)\right )}{8 d} \]
Antiderivative was successfully verified.
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Rule 364
Rule 808
Rule 1806
Rule 2837
Rubi steps
\begin {align*} \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n (a+x)^4}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\sec ^4(c+d x) \sin ^{1+n}(c+d x) \left (a^4+6 a^2 b^2+b^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)\right )}{4 d}-\frac {b^3 \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n \left (-a^4 (3-n)+6 a^2 b^2 (1+n)+b^4 (1+n)-4 a \left (a^2 (2-n)-b^2 (2+n)\right ) 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) \sin ^{1+n}(c+d x) \left (a^4+6 a^2 b^2+b^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)\right )}{4 d}+\frac {\sec ^2(c+d x) \sin ^{1+n}(c+d x) \left (a^4 (3-n)-6 a^2 b^2 (1+n)-b^4 (5+n)+4 a b \left (a^2 (2-n)-b^2 (2+n)\right ) \sin (c+d x)\right )}{8 d}+\frac {b \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n \left (8 b^4+(1-n) \left (a^4 (3-n)-6 a^2 b^2 (1+n)-b^4 (5+n)\right )-4 a n \left (a^2 (2-n)-b^2 (2+n)\right ) x\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=\frac {\sec ^4(c+d x) \sin ^{1+n}(c+d x) \left (a^4+6 a^2 b^2+b^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)\right )}{4 d}+\frac {\sec ^2(c+d x) \sin ^{1+n}(c+d x) \left (a^4 (3-n)-6 a^2 b^2 (1+n)-b^4 (5+n)+4 a b \left (a^2 (2-n)-b^2 (2+n)\right ) \sin (c+d x)\right )}{8 d}-\frac {\left (a b^2 n \left (a^2 (2-n)-b^2 (2+n)\right )\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^{1+n}}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 d}-\frac {\left (b \left (6 a^2 b^2 \left (1-n^2\right )-a^4 \left (3-4 n+n^2\right )-b^4 \left (3+4 n+n^2\right )\right )\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=-\frac {\left (6 a^2 b^2 \left (1-n^2\right )-a^4 \left (3-4 n+n^2\right )-b^4 \left (3+4 n+n^2\right )\right ) \, _2F_1\left (1,\frac {1+n}{2};\frac {3+n}{2};\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{8 d (1+n)}-\frac {a b n \left (a^2 (2-n)-b^2 (2+n)\right ) \, _2F_1\left (1,\frac {2+n}{2};\frac {4+n}{2};\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{2 d (2+n)}+\frac {\sec ^4(c+d x) \sin ^{1+n}(c+d x) \left (a^4+6 a^2 b^2+b^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)\right )}{4 d}+\frac {\sec ^2(c+d x) \sin ^{1+n}(c+d x) \left (a^4 (3-n)-6 a^2 b^2 (1+n)-b^4 (5+n)+4 a b \left (a^2 (2-n)-b^2 (2+n)\right ) \sin (c+d x)\right )}{8 d}\\ \end {align*}
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Mathematica [A] time = 0.20, size = 164, normalized size = 0.56 \[ \frac {\sin ^{n+1}(c+d x) \left (6 \left (a^2-b^2\right )^2 \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};\sin ^2(c+d x)\right )+2 (a-b)^4 \, _2F_1(3,n+1;n+2;-\sin (c+d x))+(3 a+5 b) (a-b)^3 \, _2F_1(2,n+1;n+2;-\sin (c+d x))+(3 a-5 b) (a+b)^3 \, _2F_1(2,n+1;n+2;\sin (c+d x))+2 (a+b)^4 \, _2F_1(3,n+1;n+2;\sin (c+d x))\right )}{16 d (n+1)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (4 \, {\left (a b^{3} \cos \left (d x + c\right )^{2} - a^{3} b - a b^{3}\right )} \sec \left (d x + c\right )^{5} \sin \left (d x + c\right ) - {\left (b^{4} \cos \left (d x + c\right )^{4} + a^{4} + 6 \, a^{2} b^{2} + b^{4} - 2 \, {\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sec \left (d x + c\right )^{5}\right )} \sin \left (d x + c\right )^{n}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{4} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 5.40, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{4}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{4} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\sin \left (c+d\,x\right )}^n\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^4}{{\cos \left (c+d\,x\right )}^5} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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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.
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