Optimal. Leaf size=298 \[ -\frac {d^2 (a+b x)^2 (b c-a d) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^4}-\frac {2 d^2 (b c-a d) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi ^2 b^4}-\frac {(b c-a d)^4 C(a+b x)}{4 b^4 d}+\frac {3 d (b c-a d)^2 S(a+b x)}{2 \pi b^4}-\frac {(b c-a d)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^4}-\frac {3 d (a+b x) (b c-a d)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 \pi b^4}+\frac {3 d^3 C(a+b x)}{4 \pi ^2 b^4}-\frac {d^3 (a+b x)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 \pi b^4}-\frac {3 d^3 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 \pi ^2 b^4}+\frac {(c+d x)^4 C(a+b x)}{4 d} \]
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Rubi [A] time = 0.37, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {6429, 3434, 3352, 3380, 2637, 3386, 3351, 3296, 2638, 3385} \[ -\frac {d^2 (a+b x)^2 (b c-a d) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^4}-\frac {2 d^2 (b c-a d) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi ^2 b^4}-\frac {(b c-a d)^4 \text {FresnelC}(a+b x)}{4 b^4 d}+\frac {3 d (b c-a d)^2 S(a+b x)}{2 \pi b^4}-\frac {(b c-a d)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^4}-\frac {3 d (a+b x) (b c-a d)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 \pi b^4}+\frac {3 d^3 \text {FresnelC}(a+b x)}{4 \pi ^2 b^4}-\frac {d^3 (a+b x)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 \pi b^4}-\frac {3 d^3 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 \pi ^2 b^4}+\frac {(c+d x)^4 \text {FresnelC}(a+b x)}{4 d} \]
Antiderivative was successfully verified.
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Rule 2637
Rule 2638
Rule 3296
Rule 3351
Rule 3352
Rule 3380
Rule 3385
Rule 3386
Rule 3434
Rule 6429
Rubi steps
\begin {align*} \int (c+d x)^3 C(a+b x) \, dx &=\frac {(c+d x)^4 C(a+b x)}{4 d}-\frac {b \int (c+d x)^4 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) \, dx}{4 d}\\ &=\frac {(c+d x)^4 C(a+b x)}{4 d}-\frac {\operatorname {Subst}\left (\int \left (b^4 c^4 \left (1+\frac {a d \left (-4 b^3 c^3+6 a b^2 c^2 d-4 a^2 b c d^2+a^3 d^3\right )}{b^4 c^4}\right ) \cos \left (\frac {\pi x^2}{2}\right )+4 b^3 c^3 d \left (1-\frac {a d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right )}{b^3 c^3}\right ) x \cos \left (\frac {\pi x^2}{2}\right )+6 b^2 c^2 d^2 \left (1+\frac {a d (-2 b c+a d)}{b^2 c^2}\right ) x^2 \cos \left (\frac {\pi x^2}{2}\right )+4 b c d^3 \left (1-\frac {a d}{b c}\right ) x^3 \cos \left (\frac {\pi x^2}{2}\right )+d^4 x^4 \cos \left (\frac {\pi x^2}{2}\right )\right ) \, dx,x,a+b x\right )}{4 b^4 d}\\ &=\frac {(c+d x)^4 C(a+b x)}{4 d}-\frac {d^3 \operatorname {Subst}\left (\int x^4 \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4}-\frac {\left (d^2 (b c-a d)\right ) \operatorname {Subst}\left (\int x^3 \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^4}-\frac {\left (3 d (b c-a d)^2\right ) \operatorname {Subst}\left (\int x^2 \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^4}-\frac {(b c-a d)^3 \operatorname {Subst}\left (\int x \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^4}-\frac {(b c-a d)^4 \operatorname {Subst}\left (\int \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4 d}\\ &=-\frac {(b c-a d)^4 C(a+b x)}{4 b^4 d}+\frac {(c+d x)^4 C(a+b x)}{4 d}-\frac {3 d (b c-a d)^2 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac {d^3 (a+b x)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }-\frac {\left (d^2 (b c-a d)\right ) \operatorname {Subst}\left (\int x \cos \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^4}-\frac {(b c-a d)^3 \operatorname {Subst}\left (\int \cos \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^4}+\frac {\left (3 d^3\right ) \operatorname {Subst}\left (\int x^2 \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4 \pi }+\frac {\left (3 d (b c-a d)^2\right ) \operatorname {Subst}\left (\int \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^4 \pi }\\ &=-\frac {3 d^3 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2}-\frac {(b c-a d)^4 C(a+b x)}{4 b^4 d}+\frac {(c+d x)^4 C(a+b x)}{4 d}+\frac {3 d (b c-a d)^2 S(a+b x)}{2 b^4 \pi }-\frac {(b c-a d)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac {3 d (b c-a d)^2 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac {d^2 (b c-a d) (a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac {d^3 (a+b x)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }+\frac {\left (3 d^3\right ) \operatorname {Subst}\left (\int \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4 \pi ^2}+\frac {\left (d^2 (b c-a d)\right ) \operatorname {Subst}\left (\int \sin \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{b^4 \pi }\\ &=-\frac {2 d^2 (b c-a d) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi ^2}-\frac {3 d^3 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2}-\frac {(b c-a d)^4 C(a+b x)}{4 b^4 d}+\frac {3 d^3 C(a+b x)}{4 b^4 \pi ^2}+\frac {(c+d x)^4 C(a+b x)}{4 d}+\frac {3 d (b c-a d)^2 S(a+b x)}{2 b^4 \pi }-\frac {(b c-a d)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac {3 d (b c-a d)^2 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac {d^2 (b c-a d) (a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac {d^3 (a+b x)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }\\ \end {align*}
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Mathematica [A] time = 0.95, size = 424, normalized size = 1.42 \[ \frac {\pi a^3 d^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-4 \pi a^2 b c d^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-\pi a^2 b d^3 x \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+C(a+b x) \left (d^3 \left (-\pi ^2 a^4+\pi ^2 b^4 x^4+3\right )+4 \pi ^2 b c d^2 \left (a^3+b^3 x^3\right )+6 \pi ^2 b^2 c^2 d \left (b^2 x^2-a^2\right )+4 \pi ^2 b^3 c^3 (a+b x)\right )-4 \pi b^3 c^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-6 \pi b^3 c^2 d x \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-4 \pi b^3 c d^2 x^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+\pi b^3 \left (-d^3\right ) x^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+6 \pi a b^2 c^2 d \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+4 \pi a b^2 c d^2 x \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+\pi a b^2 d^3 x^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-8 b c d^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+6 \pi d (b c-a d)^2 S(a+b x)-3 b d^3 x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+5 a d^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 \pi ^2 b^4} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} {\rm fresnelc}\left (b x + a\right ), x\right ) \]
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 \[ \int {\left (d x + c\right )}^{3} {\rm fresnelc}\left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 397, normalized size = 1.33 \[ \frac {\frac {\FresnelC \left (b x +a \right ) \left (\left (b x +a \right ) d -a d +b c \right )^{4}}{4 b^{3} d}-\frac {\frac {d^{4} \left (b x +a \right )^{3} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {3 d^{4} \left (-\frac {\left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {\FresnelC \left (b x +a \right )}{\pi }\right )}{\pi }+\frac {\left (-4 a \,d^{4}+4 b c \,d^{3}\right ) \left (b x +a \right )^{2} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {2 \left (-4 a \,d^{4}+4 b c \,d^{3}\right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}+\frac {\left (6 a^{2} d^{4}-12 a b c \,d^{3}+6 b^{2} c^{2} d^{2}\right ) \left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\left (6 a^{2} d^{4}-12 a b c \,d^{3}+6 b^{2} c^{2} d^{2}\right ) \mathrm {S}\left (b x +a \right )}{\pi }+\frac {\left (-4 a^{3} d^{4}+12 a^{2} b c \,d^{3}-12 a \,b^{2} c^{2} d^{2}+4 b^{3} c^{3} d \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+a^{4} d^{4} \FresnelC \left (b x +a \right )-4 a^{3} b c \,d^{3} \FresnelC \left (b x +a \right )+6 a^{2} b^{2} c^{2} d^{2} \FresnelC \left (b x +a \right )-4 a \,b^{3} c^{3} d \FresnelC \left (b x +a \right )+b^{4} c^{4} \FresnelC \left (b x +a \right )}{4 b^{3} d}}{b} \]
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 \[ \int {\left (d x + c\right )}^{3} {\rm fresnelc}\left (b x + a\right )\,{d x} \]
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 \[ \int \mathrm {FresnelC}\left (a+b\,x\right )\,{\left (c+d\,x\right )}^3 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{3} C\left (a + b x\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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